Question

The orbital velocity $v_o$ of a satellite at height R from the surface in terms of escape velocity $v_e$ from the earth is:

• A. $\frac{v_{e}}{2}$
• B. $\frac{v_{e}}{4}$
• C. $\frac{v_{e}}{\sqrt{2}}$
• D. $v_{e}$
• E. $\sqrt{2}v_{e}$

Hint:

Remember to always measure distance from the center of the earth (Radius + Height) for orbital mechanics.

Answer 1

The Correct Option is A

Step 1: Concept
Escape velocity $v_e = \sqrt{\frac{2GM}{R}}$. Orbital velocity $v_o = \sqrt{\frac{GM}{r}}$.

Step 2: Analysis
At height $R$ from the surface, distance from centre $r = R + R = 2R$. $v_o = \sqrt{\frac{GM}{2R}}$.

Step 3: Calculation
$v_e = \sqrt{\frac{2GM}{R}} \implies \sqrt{\frac{GM}{R}} = \frac{v_e}{\sqrt{2}}$. $v_o = \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}} = \frac{1}{\sqrt{2}} \cdot \frac{v_e}{\sqrt{2}} = \frac{v_e}{2}$. Final Answer: (A)