Answer to Question #87120 in Mechanics | Relativity for Chanchal

Question

A satellite moves in an elliptical orbit with the earth at one focus . At the perigee it's speed is v and it's distance from the centre of the earth is r . If the eccentricity of orbit is 0.5, calculate it's speed at the apogee

Answer 1

satellite velocity on elliptical orbit is

$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$

but for apogee this this equation is

$$r=a(1+e)$$

$$v_A^2=\frac{GM}{a} \times\frac{(1-e)}{(1+e)}$$

for perigee

$$r=a(1-e)$$

$$v_P^2=\frac{GM}{a} \times\frac{(1+e)}{(1-e)}$$

Thus

$$\frac{v_A^2}{v_P^2}=\frac{\frac{GM}{a} \times\frac{(1-e)}{(1+e)}}{\frac{GM}{a} \times\frac{(1+e)}{(1-e)}}$$

we have

$$v_A=\frac{(1-e)}{(1+e)}\times v_P$$

finally using initial conditions, we have

$$v_A=\frac{1}{3}\times v$$

satellite velocity at apogee is three times less than perigee