Answer to Question #280700 in Astronomy | Astrophysics for Susan

Question

A satellite was launched in 1983. It followed an elliptical orbit that varied from about 2.00 x 10^3 km to 5.00 x 10^3 km above earth. What was its orbital speed at its closest approach to earth?

Answer 1

Explanations & Calculations

• If the satellite was launched at the highest margin: $\small 5000\,km$ in 1983, then it was meant to stay in that orbit (ideally) moving at the corresponding speed of that orbit.
• That was,

$\qquad\qquad \begin{aligned} \small G\frac{M_Em}{r_1^2}&=\small m\frac{v_1^2}{r_1}\\ \small v_1&=\small \sqrt{\frac{GM_E}{r_1}} \end{aligned}$

• When it approaches the lowest orbit, its speed is meant to increase according to the law of conservation of angular momentum ($\small I=mvr$ ).
• Then, it can be calculated as follows.

$\qquad\qquad \begin{aligned} \small mv_1r_1&=\small mv_2r_2\\ \small v_2&=\small v_1.\frac{r_1}{r_2}\\ &=\small \sqrt{\frac{GM_E}{r_1}}.\Big(\frac{r_1}{r_2}\Big) \end{aligned}$

• Here G and $\small M_E$ are the Universal gravitational constant and the mass of the earth respectively.