Answer to Question #88068 in Mechanics | Relativity for Vaishali suri

Question

A satellite moves in an elliptical orbit with the Earth at one focus. At the perigee its
speed is v and its distance from the centre of the Earth is R . If the eccentricity of its
orbit is 0.5, calculate its speed at the apogee.

Answer 1

Let $v_a$ denote the speed of the satellite at the apogee to be found, and let its distance from the center of the Earth at the apogee be $R_a$. The conservation of energy of the satellite gives the equation

$$\frac{m v_a^2}{2} - \frac{G m M}{R_a} = \frac{m v^2}{2} - \frac{G m M}{R} \, ,$$

where $m$ and $M$ are the masses of the satellite and of the Earth, respectively, and $G$ is Newton's gravitational constant. Solving this equation with respect to $v_a$, we have

$$v_a = \sqrt{ v^2 - \frac{G M}{R} \left( 1 - \frac{R}{R_a} \right) } \, .$$

The ratio of the distances in the perigee and apogee is related to the eccentricity $\epsilon$ as

$$\frac{R}{R_a} = \frac{1 - \epsilon}{1 + \epsilon} \, .$$

Since $\epsilon = 1/2$, we obtain $R / R_a = 1/3$ and

$$v_a = \sqrt{ v^2 - \frac{2 G M}{3 R} } \, .$$

Answer: $v_a = \sqrt{ v^2 - \frac{2 G M}{3 R} } \, .$