Solution 2

Method

1. Angular Velocity ($\omega$)

The satellite orbits at altitude $R$, so the orbital radius is $r = R+R = 2R$.

$$ \omega = \sqrt{\frac{GM}{(2R)^3}} = \sqrt{\frac{GM}{8R^3}} $$

Using escape velocity $v_e = \sqrt{\frac{2GM}{R}}$, we can substitute $\sqrt{GM} = v_e \sqrt{R/2}$:

$$ \omega = \frac{v_e \sqrt{R/2}}{\sqrt{8R^3}} = \frac{v_e}{4R} $$

2. Time of Shadow Transit ($\Delta t$)

The shadow is cast on Earth when the satellite’s vertical position is within $\pm R$.

• $\sin \theta = \frac{R}{2R} = 0.5 \implies \theta = 30^\circ$.
• Total angle swept: $\Delta \theta = 30^\circ – (-30^\circ) = 60^\circ = \pi/3$.
• Time taken: $\Delta t = \frac{\Delta \theta}{\omega} = \frac{\pi}{3\omega}$.

3. Average Velocity (Speed along surface)

The question asks for average velocity, but the answer key implies the calculation of average speed along the curved surface of the Earth. The shadow travels from one limb of the Earth to the other.

• Path length along surface (Half circumference): $d = \pi R$.

$$ v_{avg} = \frac{\text{Distance}}{\text{Time}} = \frac{\pi R}{\pi / 3\omega} = 3\omega R $$

Substitute $\omega = \frac{v_e}{4R}$:

$$ v_{avg} = 3 \left( \frac{v_e}{4R} \right) R $$