Question

The orbital velocity of a satellite is $ V_0 $, at a height $ h = R $ (where $ R $ is the radius of the Earth) from the surface of the Earth. What is the relationship between $ V_0 $ and the escape velocity $ V_e $?

• A. \( V_0 = \frac{V_e}{2} \)
• B. \( V_0 = \frac{V_e}{\sqrt{2}} \)
• C. \( V_0 = \frac{V_e}{4} \)
• D. \( V_0 = \frac{V_e}{3} \)

Hint:

The orbital velocity at a certain height and the escape velocity are related through the gravitational potential energy of the satellite at that height.

Answer 1

The Correct Option is B

The escape velocity \( V_e \) at the surface of the Earth is given by: \[ V_e = \sqrt{\frac{2GM}{R}} \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth. The orbital velocity \( V_0 \) of a satellite at a height \( h = R \) from the surface of the Earth (i.e., at a distance \( 2R \) from the center of the Earth) is given by: \[ V_0 = \sqrt{\frac{GM}{2R}} \] Now, to find the relationship between \( V_0 \) and \( V_e \), we take the ratio: \[ \frac{V_0}{V_e} = \frac{\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{2GM}{R}}} = \frac{1}{\sqrt{2}} \] 
Thus, the relationship is: \[ V_0 = \frac{V_e}{\sqrt{2}} \] Hence, the correct answer is \( \frac{V_e}{\sqrt{2}} \).