Answer to Question #104560 in Analytic Geometry for mm

Question

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)

Answer 1

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. Then

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

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:

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

$${h\over 2}={1 \over 2}r^2 {d\nu \over dt}$$

Hence

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

$${dx\over dt}=-a\sin\theta{d\theta\over dt}, \ {dy\over dt}=b\cos\theta{d\theta\over dt}$$

$$h=(a\cos \theta-ae)b\cos\theta{d\theta\over dt}+b\sin \theta (a\sin\theta{d\theta\over dt})$$

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

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

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

Integrate

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

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

Find $c_1$

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

Therefore

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