Question

The orbital velocity of an artificial satellite in a circular orbit just above the earth's surface is 'V'. For the satellite orbiting at an altitude of half the earth's radius, the orbital velocity is

• A. $\sqrt{\frac{3}{2}} V$
• B. $\sqrt{\frac{2}{3}} V$
• C. $\frac{3}{2} V$
• D. $\frac{2}{3} V$

Hint:

Orbital speed scales inversely with the square root of the distance from the center ($v \propto 1/\sqrt{r}$). Since the total distance increases by a factor of $\frac{3}{2}$ (from $R \rightarrow 1.5R$), the speed must change by the reciprocal square root factor, which is exactly $\sqrt{\frac{2}{3}}$!

Answer 1

The Correct Option is B

The orbital velocity $v_{\text{orb}}$ of a satellite circulating around a planet of mass $M$ at a distance $r$ from the planetary center is: $$v_{\text{orb}} = \sqrt{\frac{GM}{r}}$$ 1. Just above the Earth's surface, the radial distance equals the planet's radius ($r_1 = R$): $$V = \sqrt{\frac{GM}{R}}$$ 2. At an altitude of half the Earth's radius ($h = \frac{R}{2}$), the total distance from the center is $r_2 = R + h = R + \frac{R}{2} = \frac{3R}{2}$: $$V' = \sqrt{\frac{GM}{\frac{3R}{2}}} = \sqrt{\frac{2GM}{3R}}$$ Setting up the ratio between the new velocity and our baseline surface profile: $$\frac{V'}{V} = \frac{\sqrt{\frac{2GM}{3R}}}{\sqrt{\frac{GM}{R}}} = \sqrt{\frac{2}{3}} \implies V' = \sqrt{\frac{2}{3}} V$$
Final Answer:
The orbital velocity at that altitude is $\sqrt{\frac{2}{3}} V$, which corresponds to option (B).