Satellites: Orbits and Energy — Question 8

Question 8

Slow orbital energy loss can lead to a higher eventual circular speed

A satellite evolves slowly between nearly circular orbits under weak drag. Its initial kinetic energy is $K_0$ and initial radius $r_0$. The net orbital energy removed is $.01K_0$. Compare circular endpoints; do not treat the entire change as one instantaneous impulse.

1. Find the final circular radius and speed relative to their initial values.

2. Explain why this does not mean the drag force accelerates the satellite forward.

Question 8 - Solution

Slow orbital energy loss can lead to a higher eventual circular speed

(a) Derivation and calculation

For circular endpoints, $E_0=-K_0$. Removing $.01K_0$ gives $$E_f=E_0-.01K_0=-1.01K_0.$$ Since $E=-\mu m/(2r)$, $$\boxed{r_f/r_0=1/1.01\simeq.990099},$$ $$\boxed{v_f/v_0=\sqrt{r_0/r_f}=\sqrt{1.01}\simeq1.00499}.$$

(b) Result and interpretation

The final circular kinetic and potential energies are $$K_f=1.01K_0,\qquad U_f=-2.02K_0.$$ Hence $$\Delta K=+.01K_0,\quad\Delta U=-.02K_0,\quad \Delta E=-.01K_0.$$ Drag does negative work directly. During orbital contraction, the larger decrease in potential energy allows a higher eventual circular speed.

Physical check

This compares settled nearly circular states; it does not claim that the immediate drag acceleration points forward.