## Ellipse
a=semi major axis; c=ae= b Taking
e
(1+x) (1-e So
a and
a 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)(64-3h ^{2})/(64-16h) ;
where h=(a-b)^{2}/(a+b)^{2} ;Euler Formula : - P =2π* √[(a^{2}+b^{2})/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)
; 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.
PP1=a -√(a Let
e
e sin sinr2=[1+√(1 - e
sinr2=[1-√(1 - e we take the second value. Latus Rectum = 2b.b/a Radius of Minor Auxiliary Circle = b
Radius of Equivalent Circle = √(ab) Homo focal ellipses are those sharing common foci. Circle is a degenerate ellipse. A st.line y=mx+c touches the ellipse when x c =
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. e 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+
If x' >1 ,b > ae, e <
If x' < 1, then b < ae,
e > 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.
f(A)=a
f '(A)=(ae f "(A) =( ae At inflection point ,
f "(A) = 0 or [cos2A - e simplifying, we get
Beyond this point, the sign of curvature changes . f " (A) > 0, concave f " (A) < 0, convex
Put value of y in equation of ellipse.we get the following quadratic equation- x
+ b^{2})
-2a^{2}^{3}x + (a^{4} -ab^{2})
=0^{2}solving, x=a(a + b^{2})
^{2}hence x=a or a(a ) /(a^{2}
+ b^{2}) =a cos A'^{2}
y=0 or
2ab + b^{2})
= b sinA'^{2} cosA'= 1 or
(a ) /(a^{2}
+ b^{2})^{2} sinA' = 0 or 2ab/
(a )^{2}Area swept by Ellipse : The total area swept in an elliptical orbit of semi major
axis a is √(1-e
the arithmetic mean=AM= semi major axis= the geometric
mean=GM=semi minor axis=√ the harmonic mean=HM=semi latus rectum=([1/da
+1/dp]*(1/2)) the quadratic mean=QM= root mean square =√( (a.a +b.b )/2 ) used in averaging radius AM*
HM=(GM) 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-e radf0=a radfo+radf1=2a radf0-radf1=2ae*(e+cosf) /(1+ecosf)
Equivalently radf1=a(1-ecosA) radf0= radfo+radf1=2a radfo-radf1=2aecosA
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
. x 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)
(xn)
*Let P be any point on the perimeter of the ellipse. Let the
radius vector of P
Then
*Let P be any point on the perimeter of the ellipse. Let the
radius vector of P from
Then
^{2}x1/a^{2}y1).
slope of radius vector of P from center is mr=tanA. Angle
between the two is tan^{-1} [(mt-mr) /(1+mt*mr)]
d =(A*0 +B*0+C) /
)
where co-ordinate of focus is (0,0)^{2}
tan A=-√(1-e tanA=2tanA/2 / (1-tan tan A/2 =-1/√(1-e Since tan A/2=√[(1-e)/(1+e)]tan f/2 we get tan f/2=[√(2-e tan f= * tanf1=tan(180-f)=[(-b when slope is 45 degree, cotk=1 and tanf1=((1-e * Orbit Equation for
Ellipse : h
-2a/r =-1 where r is the radius vector with focus as origin^{2}* 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-e +2ecosf)^{2}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
perihelion : a(1-e)
vertex of latus rectum : a(1-e
another point:a(1-e
vertex of minor axis:a(1-e 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-e
at another point
rf1=a(1-e
at another point rf1=a(1-e ..... .....
at nth point
rf1=a(1-e 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, a Putting this formula, the angle K between 2 radius vectors with n th integer is given by
cosK = [a
when n--> ∞ , e
;
Sum of cos of angles(Σ
f(α)=2sinα
-2sin
Differentiating, 2cosα
-2sin2α
=0 or cosα
=cos(90-2α) or 3α=90 and α=30 hence
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+2cos
If we take
Σ cosθi =y and
5/4=(n+1)/n and cos
(n+1)/n - cosnγ / n and keep a fixed value of
n, the equation becomes
α
= cos^{2}α can be both positive and negative,
γ
and value of sin
for +Ve values of
for -Ve values of
=cos of angle formed at minor axis extended to length a +2* cos of
angles formed at focus =((1-e ))
+2e/^{2}√(1+e) =[(1-e^{2})
+2e ^{2}√(1+e)]/(1+e^{2})
^{2}When e=0, sum=1; when e=1, sum = We have to maximize f(e)
=[(1-e √(1+e)]/(1+e^{2})
or[ cos^{2}α +2sinα^{2}√(1+sinα)]/(1+sin^{2}α)^{2}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,
y A]^{2}
y )tan^{2}A)]^{2}
Similarly x )tan^{2}A)]^{2}
x
x A)]^{2}
x=a√[ )tan^{2}A)]^{2}
y=a√[ )tan^{2}A)]^{2}
cos F1P1F2 =(F1P1 Putting the values, we get
A)]^{2}
But to maximize the sum , we have to
differentiate f(e)=[2√(2-e sum (maximum)=3/2 or the triangle becomes equilateral. or √(b to become the semi minor axis. if Length of semi minor axis
is b, length of major axis is √3b and eccentricity is 1- a 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-( b cos of the angle formed at any point on the ellipse perimeter by the lines emanating from 2 ends of major axis : BP AP AP AP
AP 2AP*BP =2sin
cosAPB =(2a
m1=alope of radius vector1=tanf =bsinA / (acosA=ae)
m2=(-x1/y1)*(b
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' 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
x 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
radf1=a(1-e when radf1=1,
cosf=[a(1-e
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) =
we do it for a cos curve with
f(x) =4√2
This is elliptical integral of second kind of the
form E(m) =
* Area of ellipse =A =πab. If a circle of radius b is
drawn inside, area of circle=C =πb 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' /
Elliptical Rotation Matrix 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+e)]
; This angle is same as the angle between the latus rectum
and 2nd focus.^{2}
we use tan X formula, and tan2X =2tan X / (1-tan
perimeter of F2DF1 / perimeter of F2CF1=(1+e)/[1+ angle ADB=2angle ODB
cos ADB=( b
) =-1/(2/^{2}+a^{2} e
- 1)^{2}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- When e=0, CD=0 e=1 CD=a
DB
+a^{2}(1-e^{2})
=a^{2}(2-e^{2})^{2}
CF2 +c^{2})
=a^{2}(1+e^{2})
^{2}
DB
when e^{2}
=1/2 and DB=CF2 when e=1/^{2}√2
DB when e^{2}
>1/2 and DB>CF2 when e>1/^{2}√2
DB when e^{2}
<1/2 and DB<CF2 when e<1/^{2}√2A 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 At any other point z,
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.
√[1+e^{2}+2ecosB1]/(1+e)]
*v(x)= √[1+e^{2}+2ecosB1]/(1+e)]
*v(max) where B1 refers to
true anomaly.
sinB/sinA=radc/radf ;
dA/dt =L/2m or A=LT / 2m or T=2mA/L or T A^{2}/L^{2}
= 4m^{2}π^{2}a^{2}b^{2}/L^{2}
= 4m^{2}π^{2}a^{2}(1-e^{4})
/L^{2}^{2}L =mv(x)a(1-e)
L v^{2}(x)a^{2}(1-e)^{2}
=GMm^{2}a(1-e^{2})^{2}
Hence T /GM)a^{2}
; (Kepler's 3rd law)^{3}
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
T π^{2}a^{2}b^{2}/L^{2}
;^{2}
L is constant as per law of conservation of angular momentum
and L(x)=mr(x)v(x)sin 90=ma
L
*ab/(a+b)^{2}
Putting this in
T
T )[(1-e^{2})
+ ^{2}
√(1-e)]
*(1/GM) * a^{2}^{3} ;
Hence T ^{3} .
When sun is at the focus, the constant of proportionality
becomes 4π
When
[(1-e
√(1-e)]
=2 , both constants of proportionality become same . Solving
the equations,
^{2}we get either e=0 when the ellipse becomes a circle or
or e
= 1-b^{2}/a^{2}
=1-4 =-3^{2}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)/(
Ekr ---kinetic energy at P =(1/2)mv 1+e)/(1+e)^{2}+2ecosf
=(GMm/2a)*(^{2}1+e)/(^{2}+2ecosf1-e)
^{2}Epr+Ekr=-GMm/2a =constant
x= Epr/Ekr=(Epp/Ekp) *(1+ecosf)/(1+e)
/( 1+e)^{2}+2ecosf
if dx/df=0 for ratio to be extremum, [ 2esinf /( 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/e
Epr/Ekr - 1 =( 1+e)^{2}+2ecosf
or,
Epr/Ekr = 1+( 1+e)
neglecting the minus sign^{2}+2ecosfWhen e=0, Epr=2Ekr magnitude wise and because potential energy is negative, total energy is negative e=1 Epr=Ekr, total energy is zero.
Total energy =E=mv 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=mL Rearranging, we get
where a=GM/(-2E/m) =GM/(-2E'), since E=-GMm/2a ;E'=E/m
b=(L/m) /
semi- latus rectum=l=J
Thus 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 =
Area of triangle OPB =
tanOPF2=e/(1-e tanBPF2=1/(1+e)
tan OPB =(tanOPF2 +tanBPF2)/(1-tanOPF2*tanBPF2) =(1+e) /(1-e
When tan OPB =90 degree ,e we get e=0.754877666 and e=-0.87745883 + 0.7448617666i e=-0.87745883 - 0.7448617666i
For OPB =90 degree , l
e=
if 2E'J
if -1< 2E'J
if 2E'J
if 2E'J 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.
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)J
b =J*
3)GM*l=J
l = J
4) e =
e =
5) c =ae =
focus2focus=2ae=(1/E')
6)dp =a(1-e) = (-GM/2E')[1 -
7)da =a(1+e) = (-GM/2E')[1
+
For computational purpose , we assign L= a constant arbitray value
At aphelion, E=L
with condition that GM >L
GM * semi latus rectum =(L/m)
{GM(g) / semi latus rectum(cm)}
*
a(1-e^{2})=1/semi
latus rectum=GM/J^{2}
;
So GM/J^{2}
=0 OR d^{2}u/df^{2} +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
=3GMu^{2}/c^{2}
=3GM/l^{2}cwhere c is the velocity of light
in empty space with solution
^{2}*(1+e^{2}cos^{2}f
+2ecosf)....(b)
u=(1+ecosf) / l + * 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 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 [r θ*∂ψ2/∂θ] =0^{2}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 [r
[1/ψ2]*(1/sinθ)∂/∂θ[sin 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 d
+ u - GM/J^{2}
=0 where U=1/rf=(1+ecosf) / a(^{2}1-e)
and
d^{2}u/df^{2}
+ u =1/a(^{2}1-e) = 1/semi latus rectum . But
J =GM*semi latus rectum^{2}and rf is radius vector and f=real anomaly and in spherical coordinate is akin to φ which is polar coordinate.
Einstein
d
+ u - GM/J^{2} =3GMu^{2}/c^{2}^{2}
*We write Einstein Field equation as
r
Λ
is cosmological constant. At low energy and low space time
curvature, the equation reduces to Poisson Equation. (k=8πG/c 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
x ^{2}y/dx + x dy/dx+(x^{2}-α^{2})y=0
for an arbitrary complex number
α . The most important cases are when α is integer or half
integer.^{2}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-e cos^{2}f)^{2}
radf1c-radf1=a(1-e cos^{2}f)^{2}
(radf1c-radf1) /(radf1c+radf1) = ecosf
Let u=1/(radf1c+radf1)=(1-e f)/2a(1-e^{2})^{2}
∫ udf =1/2a(1-e /4]^{2}
d
+4u =2/a^{2}
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
∂
+2B∂^{2}u/∂x
∂y +C
∂^{2}u/∂y^{2}
+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 B^{2}
-AC < 0.^{2}Example-- Laplace Equation which is
f(x,y) =
∆u=∇ u/∂x^{2}
+ ∂^{2}u/∂y^{2}=0^{2}
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,
electrostatics, gravitation, and fluid
dynamics.
In the study of heat
conduction,
the Laplace equation is the steady-state heat
equation. 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
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 The superposition principle follows directly from the fact that Laplace's equation is linear in the potential
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) = t -3t)k
^{2}
then v=2ti +2j
+(6t-3)k and
||v|| =
+9 -36t )=^{2}√(40t
-36t+13) ;^{2}
a=2i +6k
and
||a|| =
aT
=(v.a) / ||v|| = (40t-18) /
aN
=(vxa) / ||v|| = ]
= ^{2}
√[√40
- (40t-18)
/
(40t^{2}
-36t+13)]^{2}
ℒ = T - V where T=KE and V=PE
T=(GMm /2a) *
( 1-e)
^{2}
V=(-GMm
/a) * (
ℒ
=(-TE)* [( 1-e)
]*[1 + 4ecosf /(^{2}3+e)
]^{2}
(ℒ)max
=(-TE)*
[( 1-e)
]*[1
+ 4e/(^{2}3+e)
]^{2}
(ℒ)min
=(-TE)* [( 1-e)
]*[1
- 4e/(^{2}3+e)
]^{2}
(ℒ)av
=(-TE)* [( 1-e)
]^{2}
(ℒ)e=0
=(-TE)* * 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)*
[( 1-e)
]*[1 + 4ecosf /(^{2}3+e)
]^{2}
ℒ'/ℒ
=x *[( 1-e)
]*[1 + 4ecosf /(^{2}3+e)
] / [(^{2}3+e)
/(^{'2}1-e)
]*[1 + 4e'cosf /(^{'2}3+e)
]
^{'2}if ℒ'/ℒ =1, then
x=
[( 1-e)
]*[1 + 4e'cosf /(^{'2}3+e)
] / [(^{'2}3+e)
/(^{2}1-e)
]*[1 + 4ecosf /(^{2}3+e)
]
^{2}
Lagrangian has 2 components , one constant and the other variable.
Lagrange
Equn in polar coordinates is d/dt (∂ℒ
/
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
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.
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-e
mL
or L 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 = r
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
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) d If the force is radial (rd (d (rd or (r Area swept by a polar curve =Area=dA=(1/2)r The area is dependent on only time interval. acceleration due to gravity=g=a=- and a=(d (rd also a =(-∂φ /∂r) -∂φ /∂r =d
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=(d 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
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 - 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."
* 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 * 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')
/ Relativistic Longitudinal Doppler Effect is given by n1= n When β << 1, β cosθ =cosθ' +β and n1=n(1+βcos θ') =n(1+βcos θ) This is the classical Doppler effect at the When we get n1=n We know in elliptic orbit V(a)= V(p)= On substituting e by β and Gravitational Red Shift: Ep +Ek=TE=constant For a photon m=E/c -GMm / r + hv = hv(at infinity) or
hv(at infinity) - hv =-GMhv/rc 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. |