Answer to Question #104560 in Analytic Geometry for mm

Question #104560
Suppose a planet describes an ellipse with sun S as its focus, whose major axis is 2a and minor axis is 2b . Let P(x, y) be the position of the planet after time t after starting from rest from perihelion position A , referred to S as origin. Let θ be the eccentric angle of the position P .
Show that
h t = ab (θ − esin θ)
1
Expert's answer
2020-03-11T11:41:53-0400

Let P(x,y)P(x, y) be the position of the planet after time t after starting from rest from perihelion position A , referred to S as origin. Then


x=acosθae, y=bsinθx=a\cos \theta-ae,\ y=b\sin \theta

By Kepler's second law of motion, we have: the radius vector drawn from the centre of the sun to the planet sweeps out equal area in equal time. The rate of description of area is constant:


h=2(rate of description of area)h=2(rate\ of\ description \ of \ area )

h2=12r2dνdt{h\over 2}={1 \over 2}r^2 {d\nu \over dt}

Hence


h=xdydtydxdth=x{dy \over dt}-y{dx\over dt}

dxdt=asinθdθdt, dydt=bcosθdθdt{dx\over dt}=-a\sin\theta{d\theta\over dt}, \ {dy\over dt}=b\cos\theta{d\theta\over dt}

h=(acosθae)bcosθdθdt+bsinθ(asinθdθdt)h=(a\cos \theta-ae)b\cos\theta{d\theta\over dt}+b\sin \theta (a\sin\theta{d\theta\over dt})

h=ab(cos2θ+sin2θecosθ)dθdth=ab(\cos^2\theta+\sin^2\theta-e\cos\theta){d\theta\over dt}

h=ab(1ecosθ)dθdth=ab(1-e\cos\theta){d\theta\over dt}

hdt=ab(1ecosθ)dθhdt=ab(1-e\cos\theta)d\theta

Integrate


ht=ab(θesinθ)+c1ht=ab(\theta-e\sin \theta)+c_1

Initially when t=0,t=0, the planet is is at A whose eccentric angle θ=0.\theta=0.

Find c1c_1


h(0)=ab(0esin(0))+c1=>c1=0h(0)=ab(0-e\sin (0))+c_1=>c_1=0

Therefore


ht=ab(θesinθ)ht=ab(\theta-e\sin \theta)


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