Discussion Forum



1)

The ratio of binding energy of a satellite at rest on earth's surface to the binding energy of a satellite of the same mass  revolving around the earth at a height h above the earth's surface is (R= radius of the earth)


A) $\frac{2(R+h)}{R}$

B) $\frac{(R+h)}{2}$

C) $\frac{(R+h)}{R}$

D) $\frac{(R)}{R+h}$

Answer:

Option A

Explanation:

 Binding energy on the surface of the earth is 

$E_{1}=\frac{GMm}{R}$         .............(i)

 Binding energy of revolving  satellite  at a height h is 

$E_{2}=\frac{GMm}{2(R+h)}$......(ii)

  From Eqs. (i) and (ii) , we get

$\therefore$     $\frac{E_{1}}{E_{2}}=\frac{2(R+h)}{R}$