KVPY Sample Paper KVPY Stream-SX Model Paper-17

The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth \[=6.4\times {{10}^{3}}\text{km}\]) is \[{{\text{E}}_{\text{1}}}\]and kinetic energy required for the satellite to be in a circular orbit at this height is \[{{\text{E}}_{\text{2}}}\] are The value of h for which \[{{\text{E}}_{\text{1}}}\] and \[{{\text{E}}_{\text{2}}}\]are equal is:

A) \[1.6\times {{10}^{3}}\,km\]

B) \[3.2\times {{10}^{3}}km\]

C) \[6.4\times {{10}^{3}}km\]

D) \[1.28\times {{10}^{4}}km\]

Correct Answer: B

Solution :

\[{{E}_{1}}=-\frac{GMm}{R+h}-\left( -\frac{GMm}{R} \right)\]

\[{{E}_{2}}=\frac{1}{2}m{{\left( \frac{\sqrt{GM}}{R+h} \right)}^{2}}\]\[=\frac{GMm}{2\,(R+h)}\]

\[{{E}_{1}}={{E}_{2}};\] \[h=\frac{R}{2}.\]