Ellipse
Area & Perimeter Calculator
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Unit |
Number |
Required Data Entry |
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Major Axis Length |
Units |
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Minor Axis Length |
Units |
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AnyCoordinate(x)along Major axis |
Units |
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AnyCoordinate(y)along Minor axis |
Units |
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Angle (A) of any Point P in the perimeter in degree
so that x=acosA and
y=bsinA |
Degree |
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n (any positive real no. , preferably integer) |
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Angle (B) of any Point P in the perimeter between major
axis & radius
vector (r) from Focus in degree
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Find Results |
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sinA |
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cosA |
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cosA =(e-cosf)/(1-ecosf) |
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tanA |
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cotA |
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secA |
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cosecA |
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sin2A |
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(b2cos2A
+a2sin2A) |
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(1+e)/(1-e)=(radf0/radf1) max =|(KE/TE)|max |
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(1-e)/(1+e)=(radf0/radf1) min =|(KE/TE)|min |
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2/(1-e) =( 2a /radf1)max=|(PE/TE)|max |
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2/(1+e) =( 2a /radf1)min=|(PE/TE)|min |
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at tip of semi latus rectum,( radf0/radf1) =(1+e2)
/ (1-e2) = -(KE/TE) |
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at tip of other semi latus rectum,( radf0/radf1) =(1-e2)
/ (1+e2)=-(KE/TE) |
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At the tip of semi minor axis, KE=-TE and the ratio
will be -1.
If x,y are 2 variable numbers/functions such that x+y=2z
and 2z/x - y/x=1 is equn. of ellipse. |
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(1-e2) |
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√(1-e2) |
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1-e2cos2A |
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1-e2sin2A |
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√(1-e2sin2A) |
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sinr2a=[1-√(1-e2sin2A)]
/2 |
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Angle r2a in degree |
in degree |
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Angle r2a max sin value=0.5(1-b/a) |
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Angle r2a max |
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Angle r2a min cos value(cr2amin)= |
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integral sum=2a(1-cr2amin ) |
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PP1=2a*sinr2a |
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2a*cosr2a |
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(-b/ae)=disc1=-√(1/e2 -
1) |
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x-coordinate of P with center as origin (x=acosA) |
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x-coordinate of P with center
as origin (a2/c)(1-radf/a) |
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y-coordinate
of P with center as origin (y=bsinA) |
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Slope of tangent with Center as origin=-(b/a)cotA |
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Slope of tangent with Center
as origin=-(e+cosf)/sinf |
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Slope of tangent with Center
as origin=-(b2/a2)cotA1 |
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Angle A at which slope of tangent is 1 : tanA=-b/a=-√(1-e2) |
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Angle B1(f) at which slope of tangent is 1 : sin2f1=-(1-e2),f=-(90+f1) |
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Slope
Radius vector of P with major axis from center=(b/a)tanA |
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Angle A1(Latitude)
of Radius Vector of P
from center with major axis=tanA*(b/a) |
degree |
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angle between radius vec. of latitude & radf1=Angle
η1=f
-A1 (or f=A1+η1) |
degree |
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A
(eccentric anomaly E)
: tan(A/2)=√[(1-e)/(1+e)]*tan(B1/2)
A converted to radian shall actually give E |
degree |
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Angle B1(true
anomaly) of Radius Vector of P from focus with
major axis: sin B1/sinA1=(rf0/rf1) |
degree |
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Angle B1(true
anomaly) of Radius Vector of P
from focus with major axis: cosB1=(cosA-e)/(1-ecosA)
deduced from tan B1/2 =√[(1+e)/(1-e)]
tan A/2, finding 2cos2 B1/2 -1 and
cosB1 |
degree |
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Mean Anomaly M=E-esinE=[t-T]n where n is av.
angular speed, T is time period of revolution and (t-T)is the time interval after the planet has crossed
perehelion position anti clockwise.The equation in bold
is called Kepler Equation. This equation cannot be
solved for E in terms of elementary functions but the
Lagrange Reversion Theorem gives the solution as a power
series in e. This series converges for small values of e
but diverges for any value of M other than multiple of
π
if e exceeds a certain value that does not depend on
M. The Laplace limit is this value. It is the radius of
convergence of the power series. The value is
0.66274 34193 49181 58097 47420 97109 25290. |
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Angle
φ1 between radius vector of P from center & tangent |
degree |
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Radius vector of P from center |
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Radius vector of P from center=√
[a2b2/(b2
+c2sin2A1)] |
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x-coordinate of P with focus as origin (x1=acosB1) |
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y-coordinate
of P with focus as origin (y1=bsinB1) |
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Slope of tangent with Focus as origin |
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Slope of Radius vector of P from focus=(b/a)sinA/(cosA-e) |
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Angle
φ
between radius vector of P from focus & tangent |
degree |
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cosφ
=esinf /
√[(1+e2+2ecosf)] |
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sinφ
(radf * sinφ =hf) |
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sinφ =(1+ecosf) /√[(1+e2+2ecosf)] |
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tanφ
=(1+ecosf) /esinf |
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tanφ=-[b/ae
]*cosecA =disc1*cosecA |
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(1+ecosf) |
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(1-ecosf) |
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√(1+e2+2ecosf) |
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Radius vector of P from focus (radf)=√[yP2
+(xP-c)2] |
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Radius vector of P from focus
(radf11)=a(1-ecosA)=r0(1-ecosA) |
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Radius vector of P from focus (radf1)=a(1-e2)
/(1+ecosf) |
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Radius vector of P from
2nd focus radf0= S=SP=(2a-radf)=a(1+e2+2ecosf)/(1+ecosf) |
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Radius vector of P from
2nd focus radf0=
S=SP=(2a-radf)=a(1+ecosA)=r0(1+ecosA) |
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radf0-radf1=2ae(e+cosf)/(1+ecosf) |
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radf0-radf1=2aecosA |
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d=√[(radf1*radf0/b2)
-1] |
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area of triangle formed by radf1,radf0 and
focus2focus line=b2d |
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radf10b=radf1-radf0=-2a(e+cosf)/(1/e + cosf) |
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radf10=radf1-radf0 |
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radf1/radf0=(1-e2)
/(1+e2+2ecosf)=-TE/KE=-h2/b2 |
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radf0/radf1=(1+e2+2ecosf)
/(1-e2)=-KE/TE=-b2/h2 |
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Angle f for which radius vector of
P from 1st focus has magnitude a(1-eccenn):
rf1=a(1-eccenn)
=a(1-e2)
/(1+ecosf) |
degree |
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Angle between 2 radius vectors from 2 foci
when rf1=a(1-en)
and rf2=a(1+en)
angle=cos-1 [(1+e2n
-2e2)/(1-e2n)] |
degree |
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Angle between 2 radius vectors from 2 foci when rf1=a
and rf2=a,
angle=cos-1 [1
-2e2] |
degree |
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x-coordinate of P1 in the circle
(see figure below)=a/√[1+(1-e2)
*tan2A] |
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y-coordinate
of P1 in the circle (see figure below)=a√(1-e2)/√[(1-e2)
+cot2A] |
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cos of angle of triangle by 2 st.lines from foci
up to length a of minor
axis(1-e2)
/(1+e2) |
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cos F1P1F2(see related figure)=(1-e2)
/√[(1+e2)2
- 4e2/(1+(1-e2)tan2A)] |
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sum of cos of angles of the triangle by 2
radius vectors from foci and major axis at minor
axis vertex
= 1+ 2e(1-e) (value only
for blue part which is a variable) |
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sum of cos of angles of the triangle by 2 st. lines from foci
up to length a of minor axis and
line between 2 foci =[(1-e2)
+2e √(1+e2)]/(1+e2)
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sum of cos of angles of the triangle by 2 st. lines from
apogee & perigee of major axis to vertex of
minor axis and 2a =[2√(2-e2)
-e2 ]/(2-e2)
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cos of the angle by 2 st.lines from
perigee and apogee at P of ellipse
cosAPB=(-e2) / √[e4 +4(1-e2)/sin2A
] |
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Length of perpendicular from 2nd focus S
to radius vector from 1st focus SK=2ae*mr1/√(mr12
+1) |
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PK=√(SP2 -SK2
) |
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2(SP+KP) -Latus Rectum = When difference
is zero, curve is a parabola with the other focus
receding to infinity |
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Angle f=B1 between radius vector from focus
and major axis |
degree |
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sinf |
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cosf |
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tanf |
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e+cosf |
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e-cosf |
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x=√(2-e2)
-1 |
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y=e(e-1) |
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tanf=-[xy/(x+y)]/e.......when slope of
tangent is 45 degree. |
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Angle f in degree.......when slope of tangent is 45
degree. |
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Another formula of tanf=(1-e2)
/[e√(2-e2)
-1]when slope of tangent is 45 degree. |
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Vertical Distance of tangent from focus (hf) |
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hf=b *
√[(a2-cx)/(a2+cx)].....Another
formula |
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hf=semi latusrectum/√(1+e2+2ecosf) |
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b2/h2
- 2a/radf (orbit equation, should be equal
to -1) |
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[1+e(e+2cosf)] /(1-e2)
=b2/h2=-
(Kinetic energy/total energy)=v2(P)/(GM/a) |
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2*(1+ecosf) /
(1-e2) =2a/radf =(Potential
energy/total energy) |
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Point outside/inside Ellipse |
Square
Units |
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Ellipse Area=
πab |
Square
Units |
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Area swept from 0 to A =∫ r2dA
=abM/2 |
Square
Units |
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Area / perimeter(Mclurin) |
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Ellipse Eccentricity (e) |
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1/√2 |
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x'=b/ae=√(1/e2
- 1) |
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d(p) =a(1-e)=perihelion distance |
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d(a) =a(1+e)=apehelion distance |
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Q(min)=√[(1-e) /(1+e)]=√(perihelion
dist./aphelion dist) |
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Q(max)=√(1+e) /(1-e)=√(aphelion
dist./perihelion dist)=1/Q(min) |
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√[(1+e2)/(1-e2)] |
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Ellipticity or flattening γ : 1 -b/a |
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ellipticity/eccentricity =√
[(a-b)/(a+b)]=√λ |
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sin-1 e =α : |
degree |
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cos-1 e =f1 : |
degree |
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Semi-Latus Rectum Length=l = a(1-e2)=b2/a=b√(1-e2)
; |
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Radius of curvature at P=ρ=l[1+{esinf/(1+ecosf)2}]3/2
;=l/sin3φ
; |
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Curvature=cu1=1/ρ |
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Curvature=cu2=ab/(bcosA+asinA)3/2 |
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Curvature(max)=cu1a= a/b2 at (a,0) |
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Curvature(min)=cu1b=b/a2 at
(0,b) |
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Distance from Center to Focus =c =ae=√(a2-b2) |
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Focus to Focus Length =2ae |
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Length of DB= line joining perehelion and vertex
of minor axis=a√(2-e2) |
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Area of triangle DOB (See figure below) |
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Area of arc DOB =1/4 th of area of ellipse |
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Area of arc DOB /Area of triangle DOB= π/2 |
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Δ1=Area of arc DOB - Area of triangle DOB=ab(π/4
- 1/2)=a2√(1-e2)* |
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height=H1=2Δ1/DB
=(π/2
- 1)a√(1-e2)/√(2-e2) |
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Equ. of DB =bx+ay-ab=0 |
x+y-=0 |
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cosA+sinA-1 |
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length of perpendicular from P to DB ={a√(1-e2)/√(2-e2)
}*|(cosA+sinA-1)|=H1*|(cosA+sinA-1)|/(π/2
- 1) |
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(length of perpendicular from P to DB) / H1= plratio |
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(plratio)max |
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intersection of ellipse and st line y=-x+a that
joins perihelion to vertex of
extended minor axis with length a |
(x') |
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x' , y'
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(y') |
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angle AA=tan-1 (2ab/{a2-b2}) |
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angle of inflection of curve f(A)=a*√(1-cos2A)
: angle=cos-1 [√(1-b/a)
/ e] |
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Radius of Auxiliary Circle |
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Radius of Minor Auxiliary Circle |
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Area of Auxiliary circle |
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Perimeter of auxiliary circle |
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Area of minor Auxiliary circle |
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Perimeter of minor auxiliary circle |
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Radius of Equivalent Circle |
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Radius of Director Circle |
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Hill Radius constant=a(1-e) |
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Length of each EquiConjugate Diameter |
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√[1+(a2/b2 - 1)sin2A] |
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v(x) / v(y) =b/a |
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v(x)/v(z) =sinφ1
/√[1+(a2/b2 - 1)sin2A] |
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Obliquity of the Ecliptic
ε=23 degree 26
minute=23.433333 degree |
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speed of revolving mass m in
elliptical orbit with fixed point at focus at perihelion
be under a conservative central force=v(x) |
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speed at apehelion: v(y)=v(x)(1-e)/(1+e) |
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√[1+e2+2ecosB1] |
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speed at P=√[1+e2+2ecosB1]/(1+e)]
*v(x) |
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speed at P=(1+ecosB1)*v(x)/((1+e)sinφ) |
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At P, ratio of potential and kinetic
energy(Epp / Ekp =-2 (1+ecosB1)/[1+e2+2ecosB1]
) |
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Let GM=1 where G=universal const. of
Gravitation, M=mass of central body SI unit |
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Gravitational Field Strength at
P=E=central acceleration=GM/(radf1)2
; |
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GM/a=(-2)TE |
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√(GM/a)=velocity
of point mass around a circle of radius a |
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v(max)=v(x)=√(GM)
* √((1+e)/(1-e))
*√(1/a) |
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v(min)=v(y)=√(GM)
* √((1-e)/(1+e))
*√(1/a) |
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v(P)=√(GM/a)*√(radf0/radf1) |
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v(P) =√(GM/a)
*[Qmin,Qmax] velocity changes continuously from
a constant times i.e.√(GM/a)
and Qmin to Qmax=1/Qmin |
(Qmin)(Qmax) |
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v(P)=√(GM)
*(1+ecosB1)*√(1/(1-e2)
)*√(1/a)*(1/ sinφ) |
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v(P)=√(GM/a)√(2a/rf1
- 1)=√[GM/a(1-e2)]*√[1+e2+2ecosB1]=√[GM/l]*√[1+e2+2ecosB1]=√(GM/a)*√[1+e2+2ecosB1]/√(1-e2) |
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ω(P) =(L/m)(1+ecosB1)2
/[a(1-e2)]2 |
rad/sec |
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r2(P)ω(P) |
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linear acceleration towards radius
of curvature =a1= v2(P)/ρ
where is radius of curvature |
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Resultant
linear acceleration towards focus =a= GM/radf12 |
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Radial acceleration is not towards
the focus. It is towards the center of instantaneous
radius of curvature It is towards the focus only at
apogee and perigee and its value a1=
v2(P)/ρ =a at apogee/perigee.
Actually, acceleration towards the focus is the
resultant acceleration. |
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TE=Total Energy per unit mass =-GM/2a.........(1) |
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Kinetic Energy Ek per unit mass at
P=v2(P)
/2 .....(2) |
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Potential
Energy Ep per unit mass at P =-GM/radf1 ........(3) |
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Ep /TE=(3) / (1)=energy ratio1=2a/radf1=2a/r
=(1+ecosf)/(1-e2) |
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Ek /TE=(2) / (1)=energy ratio2 =(-b2/h2)=1-
(2a/r)=-v2 /(GM/a)=-[1+e2+2ecosf]
/(1-e2)=-radf0/radf1 |
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√(Ep /TE)=secA3 |
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√(Ek /-TE)=tanA3 |
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A3 in degree |
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secA3=√[2(1+ecosf)/(1-e2)]=√(2a/r) |
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secA3=√[2/(1-ecosA)] |
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tanA3=√[(1+e2+2ecosf)
/(1-e2)] |
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sinA3=tanA3/secA3 |
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x' = a*secA3 |
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y' = b* tanA3 |
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x'2
/ a2
-y'2
/b2
=(This is the equation of hyperbola where x' represents
ratio of root of PE /TE if a =1 and y' represents b
times root of -KE/TE) |
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x4=√[2(1+ecosf)/(1-e2)]=√(PE/TE) |
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y4= √[(1+e2+2ecosf)/(1-e2)]
=√(KE/-TE) |
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(x4)2-(y4)2= |
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K1=2/(1-e2)=tanψ0 |
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K1e=M1=slope=tanψ |
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ψ0 |
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ψ |
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Y=K+M1x
(equn. of st.line with y-intercept K and slope M1).
Here Y is Ep/TE and x is
cosf.The minimum value of K is 2 and maximum is
infinity. x varies from -1 to +1 . If the
y-intercept K is say 2, value y is 2.
K is say 3,maximum
value y is (3+√3)
and minimum value is 3.
K is say 4,maximum value y is 4+2√2
and minimum value is 4. Spread is
2/(1/e - e).
When e=0, spread=0
when e=1, spread =infinity
Hence numerically potential energy in a closed path
is minimum 2 times the total energy
i,e for a circle. As the circle flattens to ellipse,
the minimum ratio increases with e.
The ratio varies linearly with cos of real anomaly
f. |
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J=Specific Orbital Angular Momentum =√GM
*
√a *√(1-e2)
=√(GM*semi
latus rectum) |
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| J2
/2*Total
Energy per unit mass | =b2 =a2(1-e2)
L2
=-2mEb2 =-2mEa2(1-e2).
Compare this with the conventional relation of a free
particle or J2
=-2E'b2
between linear momentum p & total energy E where
p2=2mE
and E'=E/m |
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linear momentum2=-2mE(1+ecosB1)2(1-e2cos2A)/(1-e2)2
;where m=1 |
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4*(1-e2)/GM =K1a=constant |
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J2
/ TE2
=t2
=4a2l/GM=4[(1-e2)/GM]*a3
=K1a*a3
; |
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J/TE=-2*b/√(GM/a)= |
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t |
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angular momentum will be numerically
equal to TE when e=√(1
- GM/4a3) |
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When a=(GM/4)1/3,angular
momentum will be numerically equal to TE and trajectory
is a circle. a= |
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Ratio of angular momentum & TE is
independent of mass. Its dimension is time but with
a 1/ 2π
factor where as ratio of linear momentum to TE
has dimension time/length or inverse of linear velocity. |
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4π2 /GM=K1b |
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(Time period=T)2
=K1b*a3
; T=2π√(a3/GM) |
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T/t =π
/
√(1-e2) |
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time period of revolution T depends
on semi major axis and is independent of eccentricity
which is a measure of curvature (For all bounded
trajectories, total energy is < 0 ) |
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GM * semi latus rectum =J2
; |
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GM /
semi latus rectum=(GM/a)*1/(1-e2)=[1/(1-e2)](vcircle with radius a)2 |
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(GM/l)*(6π/c2)
=[ v2/(1-e2)
]*6π/c2) = (v2/ c2)*6π/(1-e2) |
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(6π/c2) |
*10-10 |
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Perihelion shift angle ψ
per revolution =(6π/c2)*GM/semi latus
rectum) |
*10-10 |
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Earth Time period of revolutionT(e)=31.5576*106
sec |
*106
sec |
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n=Mean Motion=Average angular speed
=2π
/T(e) |
*10-6
rad /sec |
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tan OPB |
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angle OPB in degree |
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B1=TE/(1-e2) |
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C1=(1+e2)
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B1C1 |
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D1=2e*B1 |
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D1cosf |
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PE= 2B1 +
D1cosf |
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KE=-B1C1 - D1cosf |
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TE=PE+KE |
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(Fixed component of PE)/(Fixed
component of KE)=-2/C1=-2/(1+e2)
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Both potential and kinetic energy
have 2 components, a fixed component i,e trajectory
specific and is
independent of the position of point
mass in the orbit. The second component is position
specific and is
a variable component. The most important point is
that at any position, the variable parts of potential
and kinetic energy cancel out and net variable
component is zero. Moreover , each variable component
is a cosine (periodic function) wave with amplitude
D1. If we draw graphically the functions with that of
PE in red and KE in black, it will be
as above.|KE/PE|fixed =C1 / 2
minimum value of which will be 1/2
and maximum value will be 1.
|KE/PE|variable =1 |
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Lagrangian Function Calculation |
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lag1a=(3+e2)/(1-e2) |
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lag1b=-TE*(3+e2)/(1-e2)
joules |
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lag1c=4e/(3+e2) |
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lag1d=4e cosf/(3+e2) |
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Lagrangian1 = lag1=(-TE) *[(3+e2)/(1-e2)]*(1
+ 4ecosf/(3+e2)) = lag1b(1+lag1d)
joules |
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(Lagrangian1)max =(-TE) *[(3+e2)/(1-e2)]*(1
+ 4e/(3+e2))
=lag1b(1+lag1c)
joules |
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(Lagrangian1)min =(-TE) *[(3+e2)/(1-e2)]*(1
- 4e/(3+e2) )=lag1b(1 - lag1c)
joules |
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(Lagrangian1)av =(-TE) *[(3+e2)/(1-e2)]=lag1b |
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(Lagrangian1)e=0 =(-TE) *3 |
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new e' = |
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z2=(any number showing ratio --new TE / old TE |
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(1-e'2) |
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√(1-e'2) |
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(1-e2) / (1-e'2)= |
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b'=a√(1-e'2) |
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new TE'= x *TE |
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lag2a=(3+e'2)/(1-e'2) |
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lag2b=-TE*(3+e'2)/(1-e'2)
joules |
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lag2c=4e'/(3+e'2) |
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lag2d=4e'cosf/(3+e'2) |
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Lagrangian2 = lag2=lag2b(1+lag2d)
joules |
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What should be the value of z2 so
that Lag2=lag1; z20=[lag1a*(1+lag1d)] /[lag2a*(1+lag2d)] |
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Given |
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coordinate of x on an ellipse curve
x: (x=[0,±a] ) Center is(0,0) and a is in x-axis |
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Find y=b√(1-
x2/a2) |
± |
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ga=(geometric mean)2
/ (arithmetic mean)2=b2/a2
: |
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slope of tangent at(x,y)=dy/dx=-ga
*(x/y)=(e2-1)*(x/y) |
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A st.line y=mx+c1 to touch the
ellipse--- given slope m= |
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c1=±√(a2m2
+b2)
c1 has to be real |
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In elliptical orbit, total energy TE
is a function of a and so also time period T and both do
not exclusively depend on b or e. But the fact remains
when we change e, b changes and the focal distance ae
also
changes and the location of sun which is in focus
shifts so as to keep T and TE constant and a is already
kept as a constant. In summary,
by keeping M,a
constant => TE, time period also remains const even
when the ellipse becomes
different and the focus coordinates gets adjusted
accordingly .
Then , change in focal distance is δ
da = a(e'-e) where e' is new
eccentricity and change in eccentricity
δ de =(e'-e), then change in focal
distance per unit change in eccentricity= δ
da / δ de =
a= constant. Now, if we map a point in b
identical to focus and call it bfocus , then
δ db = b'e' -be =
ae'√(1-e'2) - ae√(1-e2)
=a[e'√(1-e'2) - e√(1-e2)]
δ db / δ de =
a[e'√(1-e'2) - e√(1-e2)]
/(e'-e)
δ da / δ de
=a
δ db /
δ da =[e'√(1-e'2)
- e√(1-e2)] /(e'-e) As
e' =0,
δ db / δ da
=b/a |
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δ de =e'-e =delde |
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[e'√(1-e'2)
- e√(1-e2)] =eccendiff |
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δ db=a*eccendiff=a*[e'√(1-e'2)
- e√(1-e2)] |
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δ
da = a(e'-e)=a*δ de |
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δ db /
δ de=a*[e'√(1-e'2)
- e√(1-e2)] /(e'-e) |
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δ db /
δda=[e'√(1-e'2)
- e√(1-e2)] /(e'-e)=eccendiff /
δ de |
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J=√GM
*
√a *√(1-e2)
=√(GM/a)*
b =(velocity in circle of radius a) *b
J2=b2(velocity in circle of radius
a)2 ;(1/2)J2=(1/2)b2
*(velocity in circle of radius a)2
; The structure
is analogus to kinetic energy structure with b2
analogous to m.
In elliptical orbit, Total specific
angular momentum J is a function of M,a, e. Hence to
keep J constant, if
we do not change a, then M and e
to be changed such that M(1-e2)
=M'(1-e'2) or Ma(1-e2) =M'a(1-e'2)
Ml=M'l' or Mb2/a
=M'b'2/a
or Mb2
/2 = M'b'2
/2. This has the structure of Kinetic Energy conserva
tion if we assume b
analogous to velocity. |
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M'=
M(1-e2) / (1-e'2) or GM'=GM(1-e2)
/ (1-e'2)
GM'= |
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GMb2 /2 |
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GM'b'2
/2 |
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Lemniscate |
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Equation : (x2
+ y2
)2
=a2(x2
- y2)
where x=asint/(1+cos2t)
y=asint*cost /(1+cos2t)
Lemniscate
constant =
⊼= 2 0∫1 dt/√(1-t4)
.....The integral is elliptic integral of first kind.
In comparison,
2 0∫1 dt/√(1-t2)
=π Also
⊼= π / M (1,√2) =π / AGM (1,√2)
where AGM is the
arithmetic-geometric mean. |
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In geometry,
the lemniscate
of Bernoulli is
a plane
curve defined
from two given points F1 and F2,
known as foci,
at distance 2c from
each other as the locus of points P so
that PF1·PF2 = c2.
The curve has a shape
similar to the numeral
8 and
to the ∞ symbol.
Its name is fromlemniscatus,
which is Latin for
"decorated with hanging ribbons". It is a special case
of the Cassini
oval and
is a rational algebraic
curve of
degree 4.
This curve can be obtained as the inverse
transform of
a hyperbola,
with the inversion circle centered
at the center of the hyperbola (bisector of its two
foci). |
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Perimeter: |
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BC=√2a |
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DB=a√(2-e2) |
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Expt perimeter harmonic mean of BC and DB |
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Ramanujan Formula |
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Ramanujan Revised Formula |
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Hudson Formula |
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Euler Formula |
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Kepler Formula |
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Sipos Formula |
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Naive Formula |
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Peano Formula |
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Another formula:π*(6a/5
+ 3b/4) |
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quadratic mean of da and dp:
a√
(1+e2) |
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Bijan Formula |
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Maclaurin Formula |
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a=semi major axis;
b=semi minor
axis;
c=ae=√(a2-b2)=length
from center to focus
e=√[1-b2/a2] ; e is eccentricity.
e=0 for circle, e=1 for parabola; e > 1 for hyperbola
b2
=a2
(1-e2)
then
a2=
b2/(1-e2)
Taking
e2 = -x
1/(1-e2) =1/(1+x)=(1+x)-1 ;
(1+x)-1=1-x+x2-x3+x4-......
infinite series. Putting the value of x,
(1-e2)-1
=1+e2+e4+e6.......
infinite series where e < 1
So
a2=b2(1+e2+e4+e6.......
infinite series)
and
a2=b2(1+e2) is only approximations of
first 2 degrees
Area =
π*a*b ;
Perimeter Formulae
:
Ramanujan Formula- P =~
π*[3(a+b)-√{(3a+b)(a+3b)}] ;
Ramanujan Revised
Formula* :- P =π(a+b)(1 + 3h/(10 +√(4-3h)))
Hudson Formula : - P =
π(a+b)(64-3h2)/(64-16h) ;
where h=(a-b)2/(a+b)2 ;
Euler Formula : - P
=2π*√[(a2+b2)/2] ;
Kepler's Formula :- P
=2π*√(ab) ;
Sipos Formula:- P =2π*(a+b)2
/(√a +
√b)2 ;
Naive Formula:- P =π*(a + b)
;
Peano Formula :- P =π*[3(a+b)/2 -
√(ab) ]
;Another formula:P
=π*(6a/5 + 3b/4)
;
Colin Maclaurin Formula ;- P/2π*a
= 1-(1/2)2
e2
-(1.3/2.4)2
*(e^4)/3 -(1.3.5/2.4.6)2 * (e6) /5 - ......
infinite convergent series.
The ellipse perimeter is
about solving the integral [(a sin t)2 + (b cos t)2]^1/2 dt
over t=0 to t=2π which due to symmetry becomes 4 times the
value over t=0 to t=PI/2;
or a√(1 - (e*cos t)2)
integrated over t=0 to 2π yields the value.e=eccentricity Let
Integral of a√(1 - (e*cos
t)2) = I
This is equal to infinite series
1-(1/2)2 e2
-(1.3/2.4)2 *(e4)/3
-(1.3.5/2.4.6)2
*(e6) /5 - .. ....-[(2n-1)!!/2n!!]2
* e2n
/(2n-1)
whose refined value as per Ramanujan is
Circumference of the ellipse can be written as 4aE(e) where E is the
complete elliptic integral of the second kind from 0 to π/2
for eccentricity e.
Finding PP1:
PP1=a -√(a2cos2A
+b2sin2A)
=a [1 -
√(1 - e2sin2A)]
Let
e2sin2A=sin2r1,
PP1=a(1-cosr1)=2asin2(r1/2) =2a*sinr2
e2sin2A=sin2r1=(2sinr1/2 cosr1/2)2
=4sinr2(1-sinr2) or
sin2r2
-sinr2+ e2sin2A/4
=0, This is a qiadratic equn. and we solve for sinr2
sinr2=[1+√(1 - e2sin2A)]
/2 and
sinr2=[1-√(1 - e2sin2A)] /2
we take the second value.
x2/a2
+ y2/b2
=1 is equation of ellipse and
x2/a2 - y2/b2 =1 is equation
of ellipse if y is imaginary axis.
Latus Rectum = 2b.b/a
Focus to Focus Length = 2ae
Radius of Auxiliary Circle = a
Radius of Minor Auxiliary Circle = b
Radius of Equivalent Circle = √(ab)
Radius of Director Circle =
√( a.a + b.b )
Length of each Equi Conjugate circle =√( (a.a +b.b )/2 )
Homo focal ellipses are those sharing common foci.
Circle is a degenerate ellipse. A st.line y=mx+c touches the ellipse when x2
/a2
+(mx+c)2
/b2
=1 is satisfied. This is a quadratic equn of x of the type
ax2
+bx+c=0 whose both roots are to be real and same i.e. b2
-4ac=0. The condition is c =
±√(a2m2
+b2) If semi minor axis is
rotated 90 degree clockwise so that it lies along major axis, where the
tip of b shall be ? since distance from center to focus is ae, let
b=aex' where x' > 0 and < 1 OR x'=1 OR x' >1 but less than/equal to
1/e. e2
=1 - b2/a2
=1 -e2x'2
or x' =√(1/e2
- 1) =b/ae if x'=0, b=aex' =0 if x=1, b=ae and tip of b coincides with focus.
if 0<x<1, b <ae and
tip of b shall be between center and focus if x >1,
b >ae and tip of b shall be between focus and perihelion. Since (1+x'2) e2
=1 If x'=1, then b=ae, e =1/√2
and tip of b coincides with focus.
If x' >1 ,b > ae, e <
1/√2 and tip of b shall be between focus and perihelion.
If x' < 1, then b < ae,
e > 1/√2 and
tip of b shall be between center and focus
Since center to focus distance =ae, as e --> 0 , distance -->0 and focus
shifts towards the center.
as e --> 1, distance --->a, and focus shifts towards the circumference. Point of inflection in
ellipse: f(A)=a√[1-e2cos2A]
f '(A)=(ae2/2)
* sin2A /
√[1-e2cos2A] f "(A) =( ae2
/√[1-e2cos2A]
) * [cos2A - e2sin 22A/4(1-e2cos2A)
] At inflection point ,
f "(A) = 0 or [cos2A - e2sin 22A/4(1-e2cos2A)
] =0 simplifying, we get cosA=√(1
±√(1 - e2)) /e Beyond this
point, the sign of curvature changes . f " (A) > 0, concave
f " (A) < 0, convex Intersection of y=-x+a and x2/a2
+ y2/b2
=1 Put value of y in equation of ellipse.we get the following
quadratic equation- x2(a2
+ b2)
-2a3x + (a4 -a2b2)
=0 solving, x=a(a2
± b2) /
(a2 + b2)
hence x=a or a(a2
- b2) /(a2
+ b2) =a cos A'
y=0 or
2ab2 / (a2 + b2)
= b sinA' cosA'= 1 or
(a2 - b2) /(a2
+ b2) sinA' = 0 or 2ab/
(a2 + b2) Area swept by
Ellipse : The total area swept in an elliptical orbit of semi major
axis a is √(1-e2)
times the total area swept in a circle of radius a. In ellipse, you get arithmetic mean, geometric mean and
also harmonic mean. If perihelion distance(dp) is one number and
aphelion distance(da) is another number,
the arithmetic mean=AM= semi major axis=(dp
+da) /2 the geometric
mean=GM=semi minor axis=√(dp*da) the harmonic mean=HM=semi latus rectum=([1/da
+1/dp]*(1/2))-1 the
quadratic mean=QM= root mean square =√( (a.a +b.b )/2 )
used in averaging radius AM*
HM=(GM)2
; da/dp= vp/va or da*va=dp*vp=(1-e)/(1+e) (da - dp) /
(da + dp) =e da-dp=2ae
Similarly, radf1=a(1-e2)
/(1+ecosf) radf0=a(1+e2+2ecosf)/(1+ecosf)
radfo+radf1=2a radf0-radf1=2ae*(e+cosf)
/(1+ecosf) (radf0-radf1)
/(radfo+radf1)
=e*(e+cosf)
/(1+ecosf) Equivalently
radf1=a(1-ecosA) radf0=a(1+ecosA)
radfo+radf1=2a
radfo-radf1=2aecosA
(radf0-radf1)
/(radfo+radf1)
=e*cosA
So cosA=(e+cosf)
/(1+ecosf)
cosf=(cosA-e)/(1-ecosA) which we already know Example of HM/harmonic summations: 1) when n no. of resistances are connected in
parallel , total resistance is 1/n times the harmonic mean. When n no. of
capacitances are joined in series, total capacitance is 1/n times the HM.
(harmonic sum)
2) For 2 body system, the reduced mass is 1/2 times the HM.(harmonic
sum) 3)When one
traverses a particular distance at a fixed speed and returns at another
fixed speed, the average speed is the harmonic mean of 2 speeds. 4)In
a triangle, the radius of the in circle is 1/3 of the harmonic mean of
the height of 3 altitudes. 5) In thin lens equation of 1/f = 1/u
+ 1/v, the focal length is 1/2 of the harmonic mean.(harmonic
sum) 6) Massless Springs in series , with spring constant
k1 and k2 1/k = 1/k1 + 1/k2
7) Conducting rectangular slabs of equal dimension i.e. area of cross
section and length being same, in parallel having thermal conductivity
k1,k2 respectively,
1/k = 1/k1 + 1/k2 *If you drag your point (x,y) so that the angle to
the origin changes at a constant rate, then the curve is an
ellipse. If on the other hand, you move along the ellipse at
a constant speed, then (x,y) as a function of time are
fairly complicated. However in a circle, the
constant speed and
constant angle change are the
same. * In a circle of radius r, x=acosθ , y=asinθ
. x2 + y2 =r2 ; In an ellipse,
x=acosθ , y=bsinθ . Suppose we
scale the x-axis by m . x(new) =xn =mx and y(new)=yn=y
(xn/m)2
+(yn/1)2=r2;
(xn)2/m2r2 +(yn)2/r2 =
1 which is the equation of an ellipse with center at (0,0).
*Let P be any point on the perimeter of the ellipse. Let the
radius vector of P from center of ellipse be r(A') where A' is
the angle between radius vector and major axis from center
Then r=√
(a2b2 / (b2 +c2sin2 A'))
where c2
= a2
-b2
;
*Let P be any point on the perimeter of the ellipse. Let the
radius vector of P from any focus of ellipse be r(B1) where
B1 is
the angle between radius vector and major axis from focus
Then r=a(1-e2)
/(1+ecosB1) ; *
Equation of tangent at P to the ellipse with center as
origin is xx1/a2 +yy1/b2
=1 and slope is mt =(- b2x1/a2y1).
slope of radius vector of P from center is mr=tanA. Angle
between the two is tan-1 [(mt-mr) /(1+mt*mr)]
* Length of Perpendicular d
from focus to tangent at P( h) : Let the
equation of tangent be Ax + By + C=0 where A=b2x1
, B=a2y1,
C=-a2b2
; d =(A*0 +B*0+C) /
√
(A2
+B2)
where co-ordinate of focus is (0,0) Slope of tangent:
tan A=-√(1-e2)=-b/a
when slope angle is 45 degree tanA=2tanA/2 / (1-tan2
A/2)=-b/a. Solving for tanA/2, tan A/2 =-1/√(1-e2)
± √(2-e2)/√(1-e2)
Since tan A/2=√[(1-e)/(1+e)]tan
f/2 we get tan f/2=[√(2-e2)
- 1] /(1-e) tan f=[√(2-e2)
-1][e(e-1)]/{[√(2-e2)
-1]+[e(e-1)]}(-1/e) *General relation
between angle f (real anomaly) and angle of slope of tangent referred to
here as k tanf1=tan(180-f)=[(-b2/a2)*cotk]
/ ( 1-e√(1+{b2/a2
}cot2k)
) when slope is 45 degree, cotk=1 and tanf1=((1-e2))
/[e√(2-e2)
- 1] * Orbit Equation for
Ellipse : h2/b2
-2a/r =-1 where r is the radius vector with focus as origin
* The truncated edge of an ellipse in a cylinder with
slanted cut is a sine wave. However, to view it , you have
to unroll the truncated cylinder.
Ratio of radf1/radf0 =
(1-e2)
/(1+e2+2ecosf)
The ratio is 1 when cosf=-e or
cosf1=cos(180-f) =e and
radf1=radf0=a
The 2 radius vecors from 2 foci make an angle at the apex of
semi minor axis 180-2f1 which is 60 degree when f1=60 or
e=1/2 and distance between 2 foci become a and together they
form an equilateral triangle.
At latus rectum apex,
rf1=a(1-e2)
rf0=a(1+e2)
angle between 2nd radius vector and major axis=sin-1
[(1-e2)/(1+e2)].
When e=1/2, the sides of the triangle formed by 2 rfs and
the st.line joining 2 foci form pythagorean triplets
(5,3,4)
Length of first radius vector from focus to
perihelion : a(1-e)
vertex of latus rectum : a(1-e2):
another point:a(1-e3)
vertex of minor axis:a(1-einfinity)=a(1-0)
=a since e is less than 1.
Angle between 2 radius vectors from respective foci:
We find that
at perihilion,
rf1=a(1-e) , rf2=a(1+e)
at latus rectum rf1=a(1-e2),
rf2= a(1+e2)
at another point
rf1=a(1-e3), rf2= a(1+e3)
at another point rf1=a(1-e4),
rf2= a(1+e4)
.....
.....
at nth point
rf1=a(1-en), rf2= a(1+en) where n is a
positive integer
Now the 2 radius vectors at side between 2 foci form a
triangle.
We know that if a,b,c are 3 sides of a triangle, a2
+ b2
- c2
=2abcosA where A is the angle between a,b.
Putting this formula, the angle K between 2 radius vectors
with n th integer is given by
cosK = [a2(1-en)2
+a2(1+en)2
-4a2e2]
/ [2*a(1-en)*a(1+en)]
=[2(1+e2n) -4e2] /[ 2*(1+en)(1-en)]
=(1+e2n -2e2)/(1-e2n)
when n--> ∞ , e2n
---> 0 as e < 1 and
cosK =1 -2e2
;
Sum of cos of angles(Σi=1 to 3cosf1i) of the triangle formed by 2 radius
vectors from foci and line joining 2 foci =1+2e(1-e)
One extreme is e=0 when
2e(1-e) =0, two angles become zero and the other 180
other extreme e=1
when 2e(1-e) =0
two angles become 90 and the other zero
2e(1-e) is 0 when e=0 and e=1 and its maximum value is
1/2. If we take e=sinα, then the expression is
f(α)=2sinα
-2sin2α
=2sinα+cos2α -1. when f(x) is extremum, f'(x) =0;
Differentiating, 2cosα
-2sin2α
=0 or cosα
=cos(90-2α) or 3α=90 and α=30 hence
e=sin30=1/2.
Maximum value of f(x)= 1/2
We address the issue from another perspective.
Let the angles formed by 2 radius vectors joined at
vertex of semi major axis and the line joining 2 foci be
θi with i=1,2,3.
Σ cosθi =180=1+2e(1-e)=1+2e(1-sinα
)=1+2cos2γ(1-cos2γ )=1
+ sin22γ/2=5/4 -
(cos4γ)/4
;Maximum value of cos4γ/4 =1/4 and minimum value is
-1/4
If we take
Σ cosθi =y and
5/4=(n+1)/n and cos4γ/4 =cosnγ / n
and assume y=
(n+1)/n - cosnγ / n and keep a fixed value of
n, the equation becomes
y=mx+c which is the equation of a straight line with
gamma=x. As n --> infinity, y --> 1. m=-cosn / n
Since we have assumed sinα
= cos2γ
and value of sinα can be both positive and negative,
for +Ve values of
sinα, cosγ can be either positive or
negative but real
for -Ve values of sinα, cosγ can be either
positive or negative but imaginary which means one of the
axes will be imaginary.
Sum
of cos of angles of the triangle by 2 st.
lines from foci up to length
a of minor axis and the line
connecting 2 foci.
=cos of angle formed at minor axis extended to length a +2* cos of
angles formed at focus =((1-e2)/(1+e2))
+2e/√(1+e2) =[(1-e2)
+2e √(1+e2)]/(1+e2)
When e=0, sum=1; when e=1, sum =√2.What is
the maximum value of the sum ? We have to maximize f(e)
=[(1-e2)
+2e √(1+e2)]/(1+e2)
or[ cos2α +2sinα√(1+sin2α)]/(1+sin2α)
This is pretty complicated. So we take help of geometry. We know that
the triangle is an isoceles triangle and maximum value of sum of cos of
the angles is 3/2 when each angle is 60 degree and it condenses into
an equilateral triangle. To satisfy that condition, a√(1+e2)
=2ae or e=1/√3 = sinα
Suppose we extend OP so that it touches the
circumference of circle with radius a at P1. Let F2P1=y, F1P1=x
y2
=a2e2
+a2 -2a2ecosf
=a2e2
+a2 -2a2ecosA1=a2e2
+a2 -2a2ecosA1. But cosA1= 1/√[1+(1-e2)tan2A]
y2
=a2 [(1+e2) - 2e/√(1+(1-e2)tan2A)]
Similarly x2=a2 [(1+e2) + 2e/√(1+(1-e2)tan2A)]
x2+y2
=2a2(1+e2)
x2-y2
=4e/√(1+(1-e2)tan2A)]
x=a√[(1+e2) + 2e/√(1+(1-e2)tan2A)]
y=a√[(1+e2) - 2e/√(1+(1-e2)tan2A)]
cos F1P1F2 =(F1P12
+F2P22 -F1F22)
/2*F1P1*F2P2
Putting the values, we get
cos F1P1F2 =(1-e2)
/√[(1+e2)2
- 4e2/(1+(1-e2)tan2A)]
Sum of cos of angles of the triangle by 2
st.lines from 2 ends of major axis a to the vertex of minor axis and the
line 2a. -1/(2/e2
- 1) +2* √(2-e2)/(2-e2) =[2√(2-e2)
-e2 ] /(2-e2) =(b2-a2) /(b2+a2)
+ 2a /√(b2+a2) when e=0, sum =√2
when e=1,sum = 1 But to maximize the sum , we have to
differentiate f(e)=[2√(2-e2) -e2 ] /(2-e2)
and set it to zero which is pretty messy. We know that for an isoceles
triangle, sum (maximum)=3/2 or the triangle becomes equilateral.
or
√(b2+a2) =2a or, b=√3a and e2
=-2 which means the ellipse rotates 90 degree with major axis lying in y
axis and minor axis lying along x-axis. b expands to become a s and a
contracts to become the semi minor axis. if Length of semi minor axis
is b, length of major axis is √3b and eccentricity is 1- a2
/b2=√(-2) with major
axis lying in y axis and minor axis lying along x-axis. If we rotate
this ellipse by 90 degree clockwise, the triangle is formed by 2 extreme
points of minor axis on the perihelion and the minor axis . (see
figure). Then e=√[1-( b2 / 3b2)] =√(2/3)
cos of the angle formed at any point on the ellipse perimeter by the
lines emanating from 2 ends of major axis : BP2=Square of
Line joining (acosA,bsinA) and (a,0) =a2(1-cosA)2
+b2sin2A AP2=Square of Line
joining (acosA,bsinA) and (-a,0) =a2(1+cosA)2
+b2sin2A AP2-BP2
=4a2cosA AP2+BP2 =2a2(2-e2sin2A)
AP2 *BP2=sin4A[(a2-b2)2
+ 4a2b2/sin2A] 2AP*BP =2sin2A
√ [(a2-b2)2 + 4a2b2/sin2A]
cosAPB =(2a2(2-e2sin2A) -4a2)
/2sin2A √ [(a2-b2)2
+ 4a2b2/sin2A] = (-a2e2)
/ √ [(a2-b2)2 + 4a2b2/sin2A]
=(-a2e2) /√[a4e4 +4a2b2/sin2A
] =(-e2) / √[e4 +4b2/a2sin2A
]
cosAPB=(-e2) / √[e4
+4(1-e2)/sin2A ]
Angle between radius vector1 and tangent to radius
vector :
tanφ
=(m1-m2)/(1+m1m2)
m1=alope of radius vector1=tanf =bsinA / (acosA=ae)
m2=(-x1/y1)*(b2/a2)=(-b/a)cotA
tanφ =(-b/ae)*cosecA=-√(1/e2 - 1) *
cosecA
Constructing an ellipse from a triangle ABC:
conditions-- AB+BC=constant=k, AC=fixed value. Length of AB,
BC can be changed subject to the constraint.
Let AB=x, BC=k-x, AC=ke where e > 0 but less than 1 ,
otherwise triangle cannot be formed.
What value of x will fit in ?
BC+AC > AB or k-x +ke > x or
x >k(1-e)/2 =k'(1-e)
Similarly, it can be proved that x <
k'(1+e)
Area =√
[s(s-a)(s-b)(s-c)]
=k'2 (1-e2)√ [x(k-x)/k'2
(1-e2) - 1] = b2
*√ [AB*BC/b2 - 1] where b2=k'2
(1-e2)
We have obtained all parameters of ellipse from this
triangle.
k' = semi major axis
b=semi minor axis
e=eccentricity
AB, BC are the 2 radius vectors from respective foci.
AC= 2ae= focus 2 focus distance
One parameter equation of ellipse:
x2
+k2y
=1 where k >1
Relation of focal distance c with dp :
c=ae=dp=a(1-e) or e=1/2
when e=1/2,c =dp i.e focus lies midway between center and
perigee
when e>1/2,c >dp i.e focus lies away from midway towards
perigee
when e<1/2,c >dp i.e focus lies away from midway towards
center
When radf1 =1 ?
radf1=a(1-e2)/(1+ecosf)
when radf1=1,
cosf=[a(1-e2)
-1] /e or ae2
+cosf*e+(1-a) =0 or
e =(-cosf ± √[4a2
-4a+cos2f])
/ 2a
explore whether we can interchange e, cosf and find the
outcome.
case 1: cosf =0
e=√(1-
1/a) for a > 1 or a=1
case 2:cosf=1
e=(-1±(2a-1))/2a
=(1 - 1/a)
case 3: cosf =-1
e=1/a - 1 Here a =(1/2 , 1)
.
* When the cylinder is truncated perpendicular to its axis
(m=0), the truncated region is a circle. When ellipse
unrolls to a sine wave, the circle unrolls to a straight
line. The arc length of ellipse and sine wave have no closed
form solution. This relationship between ellipse and sine
wave also means that all planets are riding a sine wave on
different cylindrical surfaces around the sun (sun may
not be at the center of these cylinders)
* The arc length of a curve with f(x) =
b∫a√(1 +[ f'(x)]2)dx.
For sine wave, it is 2π∫0√(1 +cos2
x )dx =4√2 π/2∫0√(1 +(1/2)sin2
x )dx
we do it for a cos curve with
f(x) =4√2 π/2∫0√(1
-(1/2)sin2
x )dx
This is elliptical integral of second kind of the
form E(m) =π/2∫0√(1
- m*sin2
x )dx =4√2E(1/2)
The result to be obtained by computational
methods. The value approximately given by l(x) =121x/100
+sin(2x)/10 with error of 0.5% to 1.5%. For full sin
wave L=4√2E(1/2)=7.640395578055424
* Area of ellipse =A =πab. If a circle of radius b is
drawn inside, area of circle=C =πb2
.
Residual area of ellipse=B =πb(a-b)=πa'b
which is another ellipse.
B/A =1 -b/a . If we keep a constant, b varies between
0-a i.e. b =[0,1] B/A =[0,1] =sinα
Oscillation of b gives rise to a sinusoidal wave.
Length of the perpendicular from
2nd focus to the first radius vector from first focus
=-2aey' /√[y'2
+(ae-x')2]Where
(x' , y') are coordinates of point P in the ellipse with
origin at the center.
Elliptical Rotation Matrix R(E):
cosη (a/b)(-sinη)
where
η is angle of rotation in anti clockwise direction.,a and b
are semi major and semi minor axis respectively.
(b/a)sinη cosη
=
cosη
secα* (-sinη)
cosα*sinη
cosη
det =1
Derivation--
x' =a[cos(θ+η)]
y'=b[sin(θ+η)]
rearranging , we get the result.
cos F2CB= (1-e2)/(1+e2)]
; This angle is same as the angle between the latus rectum
and 2nd focus.
we use tan X formula, and tan2X =2tan X / (1-tan2
X) and then calculate cos2X.
perimeter of F2DF1 / perimeter of F2CF1=(1+e)/[1+√
(1+e2)]
angle ADB=2angle ODB
cos ADB=(b2-a2
)/(b2+a2
) =-1/(2/e2
- 1)
when e=0, cos ADB=0 and ADB =90 degree
when e=1 cos ADB=-1 and ADB=180 degree
So when a circle is flattened along major axis 2a to become
stretched ellipse, angle ADB moves from 90 to 180 degree.
when a circle is flattened along minor axis 2b to become
stretched ellipse, angle ADB moves from 90 to 0 degree.
CD= a[1-√
(1-e2)]
When e=0, CD=0
e=1 CD=a
DB2
=a2
+a2(1-e2)
=a2(2-e2)
CF22= =a2+c2)
=a2(1+e2)
DB2=CF22
when e2
=1/2 and DB=CF2 when e=1/√2
DB2
> CF22 when e2
>1/2 and DB>CF2 when e>1/√2
DB2<CF22 when e2
<1/2 and DB<CF2 when e<1/√2
A point mass m revolving in elliptical orbit with a much
bigger mass M being at the center :
At aphelion x, angular momentum is mv(x)a sin 90 =mv(x)a
At perihelion z, angular momentum is mv(y)b sin 90 =mv(y)b
Since angular momentum is conserved , mv(x)a=mv(y)b or
v(x) / v(y) =b/a
At any other point z,
v(x) / v(z) =r(z)sinφ1
/ a =
bsinφ1/√[b2+(a2 -
b2)sin2A]
But
v(x) =√
[2GM /(a+b)] *
√
(b/a)
v(y) =√
[2GM /(a+b)] *
√
(a/b)
So v(z) =√
[2GM /((a+b)ab)] * √[b2+(a2 -
b2)sin2A] / sinφ1
= { √[b2+(a2 -
b2)sin2A] /
√[ab*(a+b)/2]}
*(1/sinφ1)
*√(GM) ={√(a2sin2A
+b2cos2A)
/√[ab*(a+b)/2]}
*(1/sinφ1)
*√(GM)
First 3 factors are related to geometry of
ellipse and the last one represents
physics i.e. the mass at the center and Gravitational
constant.
The second factor represents the product of
Geometric mean & square root of arithmetic mean.
A point mass m revolving in elliptical orbit with a much
bigger mass M being at the focus :
v(x) =√
[GM /a] *
√
((1+e)/(1-e)) ......at perihelion
v(y) =√
[GM /a] *
√
((1-e)/(1+e)) ..... at apehelion
v(x) / v(y) = (1+e)/(1-e)
v(z) =√
[2GM /r - GM/a] =√[(1+e2+2ecosB1)/(1-e2)]
*√(GM/a) =√[1+e2+2ecosB1]/(1+e)]
*v(x)= √[1+e2+2ecosB1]/(1+e)]
*v(max) where B1 refers to
true anomaly.
= √[1+e2+2ecosB1]/(1-e)]
*v(min)
when √[(1+e2+2ecosB1)/(1-e2)]
=1, or e=cosB1, v(z) =√(GM/a)
v(min)*v(max) =GM / a and v2(at
tip of minor axis) =GM/a
If we assume v(max) as an arithmatic mean
AM, v(min) as a harmonic mean HM, then v(at
tip of minor axis) is a Geometric mean GM
Relation between Angle A and angle B/B1
sinB/sinA=radc/radf ;
dA/dt =L/2m or A=LT / 2m or T=2mA/L or T2
=4m2A2/L2
= 4m2π2a2b2/L2
= 4m2π2a4(1-e2)
/L2
L =mv(x)a(1-e)
L2 =m2v2(x)a2(1-e)2
=GMm2a(1-e2)
Hence T2
=(4π2/GM)a3
; (Kepler's 3rd law)
Time period of revolution around an
ellipse with semi major axis a in which the heavier mass M
is at the focus is same as the Time period of revolution
around a circle of radius a in which the heavier mass M is
at the center. This is irrespective of eccentricity.
Kepler's 3rd Law when heavier
mass is at the center of the ellipse instead of at the focus
:
Kepler's 2nd law is obeyed in the above case
and dA/dt(aerial velocity) =L/2m So A=(L/2m )* T +const. of
integration where L is angular momentum and T is time period
of revolution
When T=0, A=0 and hence constant=0
T=2mA/L= 2mπ*ab/L where
A =πab=area of ellipse
T2 =
4m2π2a2b2/L2
;
L is constant as per law of conservation of angular momentum
and L(x)=mr(x)v(x)sin 90=ma√
[2GM /(a+b)] *
√
(b/a)
L2
=2GMm2
*ab/(a+b)
Putting this in
T2, we get
T2=
(2π2)[(1-e2)
+
√(1-e2)]
*(1/GM) * a3 ;
Hence T2
is proportional to
a3 .
When sun is at the focus, the constant of proportionality
becomes 4π2
/GM
When
[(1-e2) +
√(1-e2)]
=2 , both constants of proportionality become same . Solving
the equations,
we get either e=0 when the ellipse
becomes a circle or
or e2
=-3 which means b=2a and e2
= 1-b2/a2
=1-4 =-3
Sometimes, question is asked as to why square of time period
is proportional to cube of a and not that of b.
Kepler's 3rd law states that square of the time period is
proportional to cube of the average distance covered during
one revolution as reckoned from focus to the point on
perimeter. Since the perigee is a(1+e) and apogee is
a(1-e) , the average is a and not b.
Energy in elliptical orbit :
Let Epp --- potential energy at perihelion = -GMm/a(1-e)
Ekp --- kinetic energy at
perehelion = (1/2)(GMm/a) (1+e)/(1-e)
Epp /Ekp=-2/(1+e)
Let Epr ---potential
energy at P = -GMm/rf1 =Epp *(1+ecosf)/(1+e)=(-GMm/a)*(1+ecosf)/(1-e2)
Ekr ---kinetic energy at P =(1/2)mv2r=
Ekp (1+e2+2ecosf)/(1+e)2
=(GMm/2a)*(1+e2+2ecosf)/(1-e2)
Epr+Ekr=-GMm/2a =constant
x= Epr/Ekr=(Epp/Ekp) *(1+ecosf)/(1+e)
/(1+e2+2ecosf) =-2 (1+ecosf)/(1+e2+2ecosf)
if dx/df=0 for ratio to be extremum, [ 2esinf /(1+e2+2ecosf)]
*[-1-ecosf]=0
which implies sinf =0 ....(1) or
cosf=-1/2e .......(2) the lower and upper bound of x =[-2,
-1] for e=0 and e=1 respectively
At f=0, x=-2/(1+e)
At cosf=-1/2e, x=-1/e2
;
Epr/Ekr - 1 =(1-e2) /(1+e2+2ecosf)
or,
Epr/Ekr = 1+(1-e2) /(1+e2+2ecosf)
neglecting the minus sign
When e=0, Epr=2Ekr magnitude wise and because
potential energy is negative, total energy is negative
e=1 Epr=Ekr,
total energy is zero.
How to prove that the trajectory of
planets around the sun are ellipses ?
Total energy =E=mv2/2 - GMm/r
-----(1). Since the
force is attractive, E < 0
Angular momentum =L=mvrsinφ=mvh----(2)
where h=rsinφ and h <= r L/m is positive.
Putting the value of v=L/mh in 1st equation
E=mL2
/2m2h2
-GMm/r
Rearranging, we get
b2/h2
-2a/r=-1 (this is the equation of ellipse known
as orbit equation in general)
where a=GM/(-2E/m) =GM/(-2E'), since E=-GMm/2a ;E'=E/m
b=(L/m) /
√(-2E/m)
=J/√(-2E')
semi- latus rectum=l=J2
/GM
a*l=b2=- J2/2E'
or L =-2mEb2=-2mE
*a(1-e2)
Thus E is linked to a and L is linked to l i.e. semi
latus rectum.
Dimensionally, angular momentum* frequency=energy. So
let us assume that there exists an arbitrary frequency f in
an arbitrary elliptical orbit such that Lf =-E or L =-E/f
Thus 2mf * a(1-e2)
=1 or mf*a(1-e2)=1/2
which is the dimension of momentum. This probably
defines the half integer value of momentum of fermions.
Since merging of particles and antiparticles result in
energy,
momentum of energy particles is integral and they are termed
as bosons. Since energy does not occupy space , there is no
restriction in number of bosons accommodated in a given
space where as there is restriction on fermions.We can work
how to explain Pauli exclusion principle based on this
model.
In the above figure, l*a=l(OF2+F2B)=2*l/2(OF2+F2B)=2*Area of
triangle OPB =- J2/2E'
Area of triangle OPB =- J2/4E'
Then we try to find out angle OPB :
tanOPF2=e/(1-e2)
tanBPF2=1/(1+e)
tan OPB =(tanOPF2 +tanBPF2)/(1-tanOPF2*tanBPF2) =(1+e) /(1-e2-e3)
When tan OPB =90 degree ,e3
+e2 -1=0. Solving this
cubic equation,
we get e=0.754877666
and e=-0.87745883 + 0.7448617666i
e=-0.87745883 - 0.7448617666i
For OPB =90 degree , l2
=OF2 *BF2
e=√[1
+ 2E'J2/(GM)2]
where E'=E/m and E' < 0 for bounded orbits
if 2E'J2/(GM)2
=-1, e=0
if -1< 2E'J2/(GM)2<0
, then 0<e <1
if 2E'J2/(GM)2=
1, e=1
if 2E'J2/(GM)2>
1, e > 1
We have thus defined the physical equivalents of a,b,l,e in
terms of G,M,E' and J or their combination. Like a,b,l,e
which are invariant for a particular ellipse, G, M, E' and J
are also invariant.
Mapping a,b,l,e into G,M,J, E' and also to
statistical averages :
By this mapping, a curved figure has been mapped into a
non-curved figure.
1) TE' = TE/m = E' =-GM/2a or
a =-GM /2E' = am (arithmetic mean of
da and dp )
2)J2
= -2E'b2
or
b =J*√(-
1/2E') =gm (geometric mean of da and
dp )
3)GM*l=J2
or
l = J2/GM
=hm (harmonic mean of da and
dp )
4) e = √(1-
b2/a2)
or
e =√[1
- 2*(E'/GM)*(J2/GM)]
=√[(am)2
-(gm)2]
/ am
5) c =ae =√(a2-b2)
=(1/2E')√[(GM)2
+2E'J2]
=√[(am)2
-(gm)2]
focus2focus=2ae=(1/E')√[(GM)2
+2E'J2]
=2√[(am)2
-(gm)2]
6)dp =a(1-e) = (-GM/2E')[1 - √(1
-2E'J2
/(GM)2
)] =am[1 -√[((am)2
-(gm)2
) /am ]
7)da =a(1+e) = (-GM/2E')[1
+ √(1 -2E'J2
/(GM)2
)] =am[1 +√[((am)2
-(gm)2
) /am ]
For computational purpose ,
we assign L= a constant arbitray value
At aphelion, E=L2/2ma2
-GMm/a
with condition that GM >L2/2ma
;
GM * semi latus rectum =(L/m)2
=J2
;
{GM(g) / semi latus rectum(cm)}
*(6 π / c2)=
Perehelion angle shift per revolution
r=a(1-e2)
/(1+ecosB1) ; since B1=f
U=1/r=(1+ecosf)
/a(1-e2)
d2u/df2 +u
=1/a(1-e2)=1/semi
latus rectum=GM/J2
;
So d2u/df2 +u
- GM/J2
=0 OR d2u/df2 +u
-1/ l
=0 .......(a)where l is the semi latus rectum (this is a
linear differential equation)
The above is the Newtonian Orbit Equation with solution
u=(1+ecosf) / l . Einstein modified it to
d2u/df2 +u
-1/ l
=3GMu2/c2
=3GM/l2c2*(1+e2cos2f
+2ecosf)....(b) where c is the velocity of light
in empty space with solution
u=(1+ecosf) / l +
3GM/l2c2*(1+e2/2
+e2cos2f/6+efsinf)
* The precession of planetary orbit (Perehelion Shift) as a consequence of
curvature of space - time as per General Theory of
Relativity (GTR)
Perehelion Shift is the angular rotation of the major
axis of the ellipse around the focus which is δf where
f is the real anomaly.
assumptions: 1) All planetary orbits lie in
same plane which is a first order approximation in GTR.
2) Gravitational Potential φ(r) =-GM/rf1
or
The precession of planetary orbits as a consequence of solar
spin (proposed by G G Nyambuya in 2009)
The sun spins about its spin axis once about every ∼ 25.38
(roughly) days and the spin axis makes an angle of about 83
degree with the ecliptic plane.
* From a general non-relativistic standpoint, We solve the
empty space Poisson Equation ∇2ψ
=0 for a symmetrical azimuthal setting (for a spinning
gravitational body),
In spherical co-ordinate system, the gravitational
potential
ψ is a
function of (r,θ,φ).
When something is spherically symmetrical, it is f(r)
only , (θ,φ))
symmetry automatically follow. However,
azimuthal symmetry which is f(r,θ)
does not imply spherical symmetry. Here
θ is the azimuthal angle. Assuming azimuthal
symmetry, we solve empty space Poisson equation.
we seek general solution for ψ(r,θ)
where general solution is constrained such that at the
zeroth order approximation, it reduces to inverse square law
of gravitation. We use
the separation of variables technique so that
ψ(r,θ) =ψ1(r) *ψ2(θ)
Rewriting the Poisson Equation after separation of
variables,
[1/ψ1]
∂/∂r [r2
∂ψ1/∂r] +[1/ψ2]*(1/sinθ)∂/∂θ[sin2θ*∂ψ2/∂θ] =0
Red part is the radial part and blue part is the angular
part and they must equal same constant so that the sum is
zero.
[1/ψ1]
∂/∂r [r2
∂ψ1/∂r] =const=l(l+1) where l=0,1,2,3.......
[1/ψ2]*(1/sinθ)∂/∂θ[sin2θ*∂ψ2/∂θ]
=const=- l(l+1) where l=0,1,2,3.......
The general solution of azimuthally symmetric Poisson
Equation of empty space in its zeroeth order yields Newton's
Laws of Gravitation of inverse square law and in its 2nd
order approximation
leads to -- perehelion shift, variation(increase ) in mean
sun-planet distance, and change (decrease ) in Spin time
period.
Planetary Orbital Equation of Motion as per
Newton d2u/df2
+ u - GM/J2
=0 where U=1/rf=(1+ecosf) / a(1-e2)
and
d2u/df2
+ u =1/a(1-e2) = 1/semi latus rectum . But
J =GM*semi latus rectum
and rf is radius vector and f=real
anomaly and in spherical coordinate is akin to
φ which is polar coordinate.
Einstein
d2u/df2
+ u - GM/J2 =3GMu2/c2
*We write Einstein Field equation as
rμν- (1/2) Rgμν
=kTμν
+Λgμν
;where
rμν is Ricci tensor,R is ricci
scalar,
gμν is Minkwoski flat space-time
metric, Tμνis
stress energy tensor and
Λ
is cosmological constant. At low energy and low space time
curvature, the equation reduces to Poisson Equation. (k=8πG/c4)
While dealing with the 2 body system in elliptical
orbit, It is customary to work out with reduced mass. But in
those cases where M >> m, the reduced mass is approximated
to m.
Mean Anomaly : M=E - esinE where E is eccentric anomaly .
Jk is the kth Bessel function of the
First Kind. Bessel functions are also known as cylindrical
functions. They are the canonical solution y(x) of Bessel
differential equation
x2 d2y/dx2 + x dy/dx+(x2-α2)y=0
for an arbitrary complex number
α . The most important cases are when α is integer or half
integer.
The Bessel function is a generalization of the sine function. It can be
interpreted as the vibration of a string with variable thickness,
variable tension (or both conditions simultaneously); vibrations in a
medium with variable properties; vibrations of the disc membrane, etc.
Bessel's equation arises when finding separable solutions to Laplace's
equation and the Helmholtz
equation in cylindrical
or spherical
coordinates. Bessel functions are therefore especially important for
many
problems of wave
propagation and static
potentials.
radf1c+radf1=2a(1-e2)
/(1-e2cos2f)
radf1c-radf1=a(1-e2)2ecosf
/ (1-e2cos2f)
(radf1c-radf1)
/(radf1c+radf1)
= ecosf
Let u=1/(radf1c+radf1)=(1-e2cos2f)/2a(1-e2)
∫ udf =1/2a(1-e2)
[f - sin2f *e2/4]
d2u/df2
+4u =2/a
Conservative Central Force:
A feature of conservative central force is
∇x F =0 or
∇xE=0 where E is intensity.
Since Gravitational Force is conservative central force, curl of
gravitational intensity is zero. When we talk of Electrical force and
electrical intensity,
From Maxwell's law--
∇ x E = -∂B/∂t which indicates that electro dynamic force is
not a conservative force. Only for static case,∂B/∂t =0, and hence
electrostatic force is a conservative force.
Elliptic Equations : 2nd order partial differential
equations are of the form
A
∂2u/∂x2
+2B∂2u/∂x
∂y +C
∂2u/∂y2
+D
∂u/∂x
+E∂u/∂y
+Fu +G=0 where A,B,C,D,E,F,G are functions of x and y. This
equation is called Elliptic Equation
if B2
-AC < 0.
Example-- Laplace Equation which is
f(x,y) =
∆u=∇2u
=∂2u/∂x2
+ ∂2u/∂y2=0
Since elliptic equations have no real characteristic curves,
there is no meaningful sense of information propagation for
elliptic equations. This makes elliptic equations better
suited to describe static, rather than dynamic,
processes.The twice continuously differentiable solutions of
Laplace's equation are the harmonic
functions,[1] which
are important in multiple branches of physics, notably
electrostatics, gravitation, and fluid
dynamics.
In the study of heat
conduction,
the Laplace equation is the steady-state heat
equation.[2] In
general, Laplace's equation describes situations of
equilibrium, or those that do not depend explicitly on time.
Suppose u=potential (electrostatic or gravitational), E=-∇u
for
electrostatics and g=-∇u for gravity
∇.E=ρ/ε
Laplace's equation possesses two properties that are particularly
important, The first is that its solutions are unique once
a suitable number of boundary conditions are specified. The second is
that its solutions satisfy the superposition
principle.
The formulation of Laplace's equation in a typical
application involves a number of boundaries, on which the potential u
is specified. Examples of such formulations, known as boundary-value
problems, are abundant in electrostatics. Here, a typical
boundary-value problem asks for u between conductors, on which uis
necessarily constant. In such applications, the surface of each
conductor is a boundary, and by specifying the constant value of each
boundary, we can find a unique solution to Laplace's equation in the
space between the conductors. In other situations the boundary may not
be a conducting surface, and u may not be
constant on the boundary. But the property remains, that once is
specified on each boundary, the solution to Laplace's equation between
boundaries is unique. We shall see this uniqueness property confirmed
again and again in the boundary-value problems examined in this chapter.
A general proof of the uniqueness
theorem is not difficult to construct, but we shall not pursue this
here.
The superposition principle follows directly from the
fact that Laplace's equation is linear in the potential
U
Acceleration
in Elliptical Orbit :
Acceleration is
not normal to instantaneous velocity except in circular cases. It has a
component tangential to the velocity and another normal to the velocity.
Neither of them is centripetal. In fact, the normal component is
directed towards the center of curvature and not towards the focus.
Centripetal acceleration is the resultant acceleration.
Comparison
between
centripetal
acceleration
and
Tangential acceleration
1.arises due to
change in direction of tangential velocity.
1.arises
due to change in magnitude of tangential velocity.
2. Direction
towards the center.
2. Direction towards the tangent.
To calculate,
tangential
acceleration =aT =(v.a) /
||v||
normal
acceleration =
aN
=(vxa) / ||v||
resultant
acceleration a is the vector sum of
aT
and
aN .
Example--
suppose radius vector r(t) = t2
i +(2t-3)j + (3t2-3t)k
then v=2ti +2j
+(6t-3)k and
||v|| =√(4t2
+4 +36t2
+9 -36t )=√(40t2
-36t+13) ;
a=2i +6k
and
||a|| =√(4+36)
=
√40
aT
=(v.a) / ||v|| = (40t-18) /
√(40t2
-36t+13)
aN
=(vxa) / ||v|| =√[
(||a|)2
-(|| aT||)2]
=
√[√40
- (40t-18)2
/
(40t2
-36t+13)]
Lagrangian
ℒ of the system :
ℒ
= T - V where T=KE and
V=PE
T=(GMm /2a) *
(1+e2+2ecosf)/(1-e2)
V=(-GMm
/a) * (1+ecosf)/(1-e2)
ℒ
=(-TE)* [(3+e2)
/(1-e2)
]*[1 + 4ecosf /(3+e2)
]
(ℒ)max
=(-TE)*
[(3+e2)
/(1-e2)
]*[1
+ 4e/(3+e2)
]
(ℒ)min
=(-TE)* [(3+e2)
/(1-e2)
]*[1
- 4e/(3+e2)
]
(ℒ)av
=(-TE)* [(3+e2)
/(1-e2)
]
(ℒ)e=0
=(-TE)*3
*
with total energy of the system remaining the same
(a)
ℒ = constant if e=0
(b)
ℒ = not constant if e≠0 and
lagrangian varies with angle f.
*
Suppose e remains the same but TE changes and becomes xTE
where x≠0
Then new Lagrangian =ℒ'
=(-xTE)*
[(3+e2)
/(1-e2)
]*[1 + 4ecosf /(3+e2)
]
ℒ'/ℒ
=x *[(3+e2)
/(1-e2)
]*[1 + 4ecosf /(3+e2)
] / [(3+e'2)
/(1-e'2)
]*[1 + 4e'cosf /(3+e'2)
]
if
ℒ'/ℒ
=1, then
x=
[(3+e'2)
/(1-e'2)
]*[1 + 4e'cosf /(3+e'2)
] / [(3+e2)
/(1-e2)
]*[1 + 4ecosf /(3+e2)
]
Lagrangian has 2 components , one constant
and the other variable.
Lagrange
Equn in polar coordinates is d/dt (∂ℒ
/∂θ')
=0 where
θ' =dθ/dt
and ℒ =(1/2)[r'2
+rθ'2]
-V(r) and r'
=dr/dt
In General, Lagrangian= T-V where T is the kinetic term and
V is the interaction term. V has 3 or more fields whereas T
has only 2 fields.
Ratio of Kinetic Energy to Potential Energy in elliptical
orbit :
KE /
PE= radf0 / 2a . Since the denominator is major axis which
is fixed for a specific ellopse, we can rescale major axis
to make it 1.
(KE/PE)min = (1-e) / 2
and
it has range (1/2 , 0)
and
(KE/PE)max = (1+e)/2 and it has range (1/2 , 1)
(KE/PE)max +(KE/PE)min
= 1 irrespective of value of e , a .
(KE/PE)max
- (KE/PE)min = e
Exa- suppose maximum of the ratio is
Comparison between Gravitational Force & Coulomb Force:
similarities---
1.
both obey inverse square law.
2.
both can exist in vacuum.
3.Both are conservative, central forces
difference:
1.Gravitational force is only attractive while the other one
can be attractive or repulsive.
2.Material medium does not affect gravitational force but it
affects Coulomb force through permittivity of the medium.
3.Gravitational force does not mean presence of coulomb
force but since charge cannot exist without mass, coulomb
force exists along with gravitational force,
4.Gravitational force is much weaker than the coulomb force.
Is
coulomb force instantaneous ?
the
law says the same implicitly which is not correct. If one of
the charges moves, the electric field close to it moves with
it but thereafter the change travels as a wave at the speed
of light.
Lorentz Force:
When 2 charged particles move relative to each other, The
force each particle experiences is a combination of electric
and magnetic force which is called Lorentz force.
Conservation of Energy :
In planetary motion in elliptical orbits, total energy Which
is a sum of potential energy and kinetic energy) is
conserved at every point of the ellipse. The force is
central , conservative and hence only radial acceleration is
non zero but the tangential acceleration is zero. We also
know that
(PE/TE) +(KE/-TE) =1 or 2a/radf1 -
radf0/radf1=1. we can take TE=-radf1, PE=-2a,KE=radf0 and
find that PE is conserved , TE & KE adjust accordingly. This
is a case where system is not insulated. Here there are 3
parameters a, radf1,radf2 .
(PE/TE) +(KE/-TE) =1 or 2(1+ecosf)/(1-e2)
- (1+e2 +2ecosf) /
(1-e2) =1; We take
total energy asTE=-(1-e2)
, KE= (1+e2 +2ecosf)
,PE=-2(1+ecosf) . Here there are 2 parameters e, f. In a
closed system, total energy is conserved, But in an open
system, one can change TE along with either PE or KE so that
the other of PE and KE is conserved
Description |
value in SI / MKS unit |
Mass of sun M |
1.9891 * 1030 |
Mass of earth m |
5.97219 * 1024 |
Radius of the Earth r1 |
6.378*106 |
Radius of the Sun |
6.96340*108 |
Moment of Inertia of Earth =I=(2/5)mr12 |
1.523576*1037 |
Time period of rotation of Earth=Tr |
86400s |
mass of electron |
9.10938356*10-31 |
charge of electron e |
1.602 *10-19 |
e/m ratio of electron |
1.758820 *1011 |
proton to electron mass ratio= |
1836.15267343(11) |
m/M |
3 * 10-6 |
Spin Angular Velocity
of earth
ωs
=2π
/Tr |
7.27*10-5 |
Spin or Rotational Angular Momentum,earth= S1=Iωs |
1.105*1033 |
Orbital Angular Momentum of earth L |
2.663 *
1040 |
Angular velocity=ω=L/(mr2sinφ1)=2πab/r2T |
|
radial acceleration=a(r) =v2
/r |
|
Tangential Acceleration=a(T)
=αr |
|
1Angstrom Unit |
1.5*1011
m |
Universal Gravitational Constant G |
6.67 * 10-11 |
velocity of light in vacuum =c |
3*108 |
GM |
1.32673*1020 |
Mean Radius of earth from the sun=R |
150*109 |
GM/R |
0.884487*109 |
GM/Rc2 |
0.98276*10-2 |
4GM/Rc2
in arcsec(divide by 3.6*103
) |
4*0.273*10-5 |
Semi Major axis,earth (a) |
149.6 * 109 |
Perihelion distance of earth =d(p) |
147.1 * 109 |
Apehelion distance of earth =d(a) |
152.16 * 109 |
eccentricity e of earth |
0.01671 |
Eccentricity of Halley comet |
0.967 |
time period of revolution earth=T |
365.26days=3.1556 *107
s |
dθ /dt = 360/T in degree( for circle ) |
1.14 *10-5 |
speed at aphelion |
29290 |
speed at perihelion |
30290 =0.1 c |
average speed |
29790=0.0993c |
Av. perimeter of orbit |
9.4 *
1011 |
Hill Radius a(1-e)(m/M)1/3 |
|
Semi Latus rectum of Earth
a(1-e2) |
14.955823*1010 |
specific angular momentum l of earth |
4.45447*1015 |
l2
of earth=GMa(1-e2) =GM*Semi latus
rectum |
19.84234*1030 |
GM /semi latus rectum = GM/a(1-e2) |
0.8871*109 |
Orbit of Earth in elliptic motion about the sun
rotates through an angle Ψ =(6π/e2)*(GM
/semi latus rectum) |
0.59860*1014
radian |
Kinetic energy K = U/2 |
2.6672 *
1033 |
Potential energy -U |
-5.29201 * 1033 |
Total energy E= - U/2 |
-2.62439 * 1033 |
2K+U =0 (virial theorem) |
|
mL2
/m2h2 = GMm/r
or L2 =GMm2 (h2
/r) .Hence
(h2 /r) is a constant
The revolution of earth around the sun takes place in
anti-clockwise direction when viewed from above the north
pole.
dA/dt for a circular orbit = r2ω/2
or dA/dt=(1/2)r2
dθ /dt= πab/T
dθ /dt = 360/T where T is the time period
a=[d(a)+d(p)] /2
b=√[d(a)*d(p)]
Moment of Inertia of an elliptical mass M=I(z) =M(a2+b2)/4
δψ/δt =perehelion precession rate (arc sec/year).
Arc sec is the unit of very small angular measurements and
is used in astronomy. 1 arcsec =
π/ 648000 radian =1/3600 of a degree
serial(i) |
planet name |
m/M |
T (years) |
R(AU) |
δψ/δt(theory) |
δψ/δt(observed) |
1 |
Mercury |
0.166*10-6 |
0.241 |
0.387 |
5.50 |
5.75 |
2 |
Venus |
2.45*10-6 |
0.615 |
0.723 |
10.75 |
2.04 |
3 |
Earth |
3.04*10-6 |
1 |
1 |
11.87 |
11.45 |
4 |
Mars |
0.323*10-6 |
1.881 |
1.52 |
17.60 |
16.28 |
5 |
Jupiter |
955*10-6 |
11.86 |
5.20 |
07.42 |
06.55 |
6 |
Saturn |
286*10-6 |
29.46 |
9.54 |
18.36 |
19.50 |
7 |
Uranus |
43.6*10-6 |
84.01 |
19.19 |
02.72 |
03.34 |
8 |
Pluto |
51.8*10-6 |
164.8 |
30.07 |
00.65 |
00.36 |
|
|
|
|
|
|
|
Expressing displacement, velocity, acceleration
(2-dimensional) in polar coordinates :
Let r=r(t) and θ=θ(t)
x= rcosθ + rsinθ
dx/dt =
dr/dt * (cosθ +sinθ) +rdθ/dt
* (cosθ -sinθ) ( We need to explore the consequence of
v having only radial part or only transverse part)
d2x/dt2=(d2r/dt2-r(dθ
/dt)2)(cosθ
+sinθ) +(rd2θ
/dt2
+2dr/dt * dθ /dt)(cosθ -sinθ)
If the force is radial
(rd2θ
/dt2
+2dr/dt * dθ /dt) =0......(transverse component of acceleration)
(d2r/dt2-r(dθ
/dt)2)
= acceleration due to central force (radial component of acceleration)
(rd2θ
/dt2
+2dr/dt * dθ /dt) =0 or d/dt(r2dθ/dt)
=0
or (r2dθ/dt)
=constant =c or r2ω
=c = L/m where L is Angular Momentum and c is specific angular momentum.
This is conservation of angular momentum.
Area swept by a polar curve =Area=dA=(1/2)r2dθ
so A =∫t0(1/2)r2dθ/dt
*dt =(1/2)c∫t0dt =(1/2)c[t-t0]
The area is dependent on only time interval.
acceleration due to gravity=g=a=-∇φ(r,θ)
=(-∂φ /∂r)r +1/r(-∂φ /∂θ)θwhere
φ is the gravitational potential, r & θ
in bold are unit vectors. We assume φ(r,θ) =φ(r)
*φ(θ) by separation of variables.
and a=(d2r/dt2-r(dθ
/dt)2) r
+(rd2θ
/dt2
+2dr/dt * dθ /dt)θ where r,θ in bold represent
unit vectors.
(rd2θ
/dt2
+2dr/dt * dθ /dt) =(1/r)d[r2*dθ /dt]/dt
=(1/r)dJ/dt where J=r2*dθ /dt
=specific angular momentum of revolving body
also a =(-∂φ /∂r)r +1/r(-∂φ /∂θ)θ,
it follows
-∂φ /∂r =d2r/dt2 - r(dθ
/dt)2
1/r(-∂φ /∂θ) =rd2θ
/dt2
+ 2dr/dt * dθ /dt=(1/r)dJ/dt since J=r2
dθ /dt and dJ/dt = rd2θ
/dt2+2r*dr/dt
*dθ /dt
Hence dJ/dt=-∂φ/∂θ (When J=constant, transverse
component of acceleration vanishes. otherwise not)
When the motion of the body is such that gravitational
potential is azimuthally symmetric and not spherically symmetric , φ= φ(r,θ)
,then dJ/dt ≠ 0, hence Orbital Angular Momentum is not conserved.
If J of a planet around the sun is not conserved, then
the sum of the orbital and spin angular momentum must be a conserved
quantity which implies at different r positions, the excess orbital
angular momentum is transferred to the spin angular momentum
or the spin of a planet about its own axis must vary and in the instant
case increase (resulting in increasing time period T) leading to variation
in length of the day depending upon the radial position. This is due to
interchange of spin and orbital angular momentum so that total momentum
is unchanged since g remains the same.
a=(d2r/dt2-r(dθ
/dt)2) r
+(rd2θ
/dt2
+2dr/dt * dθ /dt)θ where r,θ in bold represent
unit vectors.
There are 3 situations ---
(I) Both radial and transverse part of velocity are non
zero and
(a) only radial part of acceleration
is non zero .........forces are central and conservative
(b) only transverse part is non zero
(c) both radial and transverse part
are non zero
(ii) Radial part of velocity are non zero and
(a) only radial part of acceleration
is non zero
(b) only transverse part is non zero
(c) both radial and transverse part
are non zero
(iiI) Transverse part of velocity are non zero and
(a) only radial part of acceleration
is non zero
(b) only transverse part is non zero
(c) both radial and transverse part
are non zero
Force Law in Rotating Frame of Reference:
In all inertial frames of reference, the Newtonian force
law is F=ma
Transferring this equation to a rotating frame of
reference where ω is the angular velocity of the rotating frame,
r' is the position vector of the object w.r.t the rotating frame, v' is
the velocity of the
object w.r.t. the rotating frame of reference , the
force equation becomes ---
F'= F - m(dω/dt x r')
-2m(ω x v') -mω x (ω
x r') =ma'
The blue part is called Euler force, red part is called
Coriolis force and green part is called centrifugal force. These are
notreal forces but arise out of transformation to a rotating frame and
have the
structure of force , we call them Pseudo force or
fictitious force.
These forces all disappear when
(a) ω =0
or
(b) m=0
When ω is parallel to v', Coriolis force becomes zero.
When Velocity is in the direction of rotation, Coriolis
force is assumed to be outward from above.
Coriolis force travelling in circular trajectories is
called "inertial circle."
Doppler Effect :
* A moving source with velocity v along x-axis of Lab
frame (L) emits light with frequency n on its rest frame(R).
* However, in the L frame, 2 successive crests are
emitted at a time interval Δt=t2-t1=γT=γ/n (time dilation
effect of SR)
* For an observer located at fixed point Q on the
x-axis of L frame , time interval between arrival of 2 crests is (1-vcosθ/c)*γ/n
where
θ is the angle between the light and x-axis of L frame
* Hence light frequency measured by an observer is
n1=n√(1-β2)
/(1-βcosθ)
* If
θ' is the angle between the light and x'-axis which is
parallel to x-axis lying in R frame
cosθ=(cos
θ' + β) /(1+βcos
θ')
cosθ' =(cos
θ - β) /(1- βcos
θ)
Hence n1=n(1+βcosθ')
/√(1-β2)
Relativistic Longitudinal Doppler Effect is given by
n1= n√[(1+β)/(1-β)]
when θ=0 degree
When β << 1, β2
approximates 0
cosθ =cosθ'
+β and n1=n(1+βcos
θ') =n(1+βcos
θ)
This is the classical Doppler effect at the first
order of β , showing frequency change due to a relative motion
between the source and receiver.
When β --> 1, but cos
θ=0, we put in n1=n√(1-β2)
/(1-βcosθ)
we get n1=n√(1-β2)
which is time dilation effect of special relativity (SR)
We know in elliptic orbit
V(a)= √(GM/a) √[(1-e)/(1+e)]
V(p)=√(GM/a) √[(1+e)/(1-e)]
On substituting e by β and √(GM/a)
by n, we get n1=n√[(1+β)/(1-β)]
or n1=n√[(1-β)/(1+β)]
assuming wavelength to be unity(V=nλ =n) which is relativistic
Doppler Effect.
Gravitational Red Shift:
Ep +Ek=TE=constant
For a photon m=E/c2=hv/c2
=
-GMm / r + hv = hv(at infinity) or
hv(at infinity) - hv =-GMhv/rc2
; or ( v(at infinity) - v )/v = -GM/rc2
Red shift of spectral lines due to action of
gravitational field is drawn from Newtonian Mechanics. The shift is due
to loss of energy of the photons.
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