# Ellipse Area & Perimeter Calculator

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,

Equivalently

So cosA=(e+cosf) /(1+ecosf)

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/r= 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.

The ratio is 1 when cosf=-e or

cosf1=cos(180-f) =e and

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 ]

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

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) = ba(1 +[ f'(x)]2)dx. For sine wave, it is 0√(1 +cos2 x )dx =4√2 π/20√(1 +(1/2)sin2 x )dx

we do it for a cos curve  with  f(x) =4√2 π/20√(1 -(1/2)sin2 x )dx

This is elliptical integral of second kind of the form E(m) =π/20√(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)]

cos ADB=(b2-a2 )/(b2+a2 ) =-1/(2/e2   - 1)

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

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)

L =2GMm2 *ab/(a+b)

Putting this in T2, we get

T2= (2)[(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 [r2ψ1/∂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 [r2ψ1/∂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+(x22)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.

∫ 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 +2B2u/x y +C 2u/y2 +D u/x +Eu/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 is 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

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    of the system :

= T - V  where T=KE  and V=PE

T=(GMm /2a) * (1+e2+2ecosf)/(1-e2)

V=(-GMm /a) * (1+ecosf)/(1-e2)

=(-TE)* [(3+e2) /(1-e2) ]*[1 +  4ecosf /(3+e2) ]

()max =(-TE)* [(3+e2) /(1-e2) ]*[1 +  4e/(3+e2) ]

()min =(-TE)* [(3+e2) /(1-e2) ]*[1 -  4e/(3+e2) ]

()av =(-TE)* [(3+e2) /(1-e2) ]

()e=0 =(-TE)*3

* with total energy of the system remaining the same

(a) = constant if e=0

(b) = 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'/ =x *[(3+e2) /(1-e2) ]*[1 +  4ecosf /(3+e2) ] / [(3+e'2) /(1-e'2) ]*[1 +  4e'cosf /(3+e'2) ]

if 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 =(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 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

 Another Approach :    GIVEN Major axis(2a) Any value for arbitrary angle α in degree (measure of eccentricity) Angle θ in degree of any Point P in the perimeter  so that x=acosθ   and y=bsinθ with centre as origin Find sinα cosα sinθ cosθ tanθ tanα tan2α  (n1 to be less than tan2α , limiting value is equal to tan2α) Choose n1 √(sin2α -n1cos2α) ; Eccentricity (e) =sinα Flattening/Ellipticity =1-cosα semi major axis(a) semi minor axis(b)=acosα Semi-latus rectum l =b2/a=acos2α c=ae=a*sinα k=2/(1-e2) =2sec2α ke=2tanαsecα Perihilion distance from focus F1= F1A=dp=a(1-sinα ) Apehilion distance from focus F1=F1B= da=a(1+sinα ) dp/da=(1-sinα )/(1+sinα )=(secα -tanα)2 ; radius vector from focus F1 ;rf1=F1P=a*(1-sinα cosθ) radius vector from other focus F2 ;rf2=F2P=a*(1+sinα cosθ) radius vector from origin  rf0=OP=a*√(cos2θ+sin2θcos2α) Slope of rf0 with major axis:(tanθ*cosα) Angle of rf0 with major axis=tan-1 (y-coordinate of P/x-coordinate of P)=tan-1(tanθ*cosα) Angle POF2=180-angrf0 Vertcal distance  from 1st focus F1 on tangent at P=h=acosα *√(1-sinα cosθ) /√(1+sinα cosθ) x-coordinate of P =acosθ y-coordinate of P =bsinθ=asinθcosα slope of the tangent at P with center as origin=mtp=dy/dx=-cosαcotθ slope of rf1 with center as origin=m1r=sinθcosα /(cosθ-sinα) Angle of rf1 with major axis                                                            or slope of rf2 with center as origin=m2r=sinθcosα /(cosθ+sinα) Angle F2PF1 between rf2  and rf1 in degree=tan-1 [(m1-m2)/(1+m1m2)] Angle PF2F1 between rf2 and 2ae in degree= cos-1  [(cosθ+sinα) /(1+sinα cosθ)] = use formula of triangle, a2 +b2-c2=2abcosθ Angle PF1F2 between rf1 and 2ae in degree= Angle OPF1=sin-1[ sinα*bsinθ/((1-sinα cosθ)*rf0)]=sin-1[c*sinPOF1/rf1] Angle OF1P=rf0*sin angrf0/rf1                                                           or Angle PF1A=180*1-OF1P*1=true anomaly                                               or Angle PF2O Angle F2PO Perimeter of triangle PF2F1=2a(1+e)=2a(1+sinα) Area of triangle PF2F1=a2cosα*sinαsinθ =b2 *√[(rf1*rf2/b2)-1] Area of triangle OPF1=a2cosα*sinαsinθ/2 Area of ellipse=πa2cosα Area of triangle PF2F1/Area of ellipse =sinαsinθ /π Angle between tangent at P and rf1=tan-1[(mtp-m1r)/(1+mtp*m1r)]=tan-1(-cotα/sinθ) Angle between tangent at P and rf2=tan-1[(mtp-m2r)/(1+mtp*m2r)]=tan-1(cotα/sinθ) F2K=Perpendicular from F2 on first radius vector rf1=-asin2α*sinθ /√(1+cos2θ-2sinαcosθ); PK=√[(2nd radius vector)2-F2K2] Δ1=F2K+PK-l R(E) : Elliptical Rotation Matrix η in degree ------* *    ------- or det : x'=re11a*x+re12*y y'=re21*x+re22a*y

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)

(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.