Question

A sky laboratory of mass $2\times {10}^{3}Kg$ is raised from a circular orbit of radius 2 R to a circular orbit of radius 3 R. The work done is approximately

• A. $1\times {10}^{16}Joule$
• B. $2\times {10}^{10}$
• C. $1\times {10}^{6}Joule$
• D. $3\times {10}^{10}Joule$

Answer 1

The correct option is B $2\times {10}^{10}$

Total energy at $2R,T_2=\frac{-GMm}{2R}$
similarly at $3R,T_3=\frac{-GMm}{3R}$
Net change = final - initial = $\frac{-GMm}{3R}-\frac{-GMm}{2R}=\frac{GMm}{6R}$
Substituting the given values we get ,
Net change = $\frac{6.67\times {10}^{-11}\times 2\times {10}^3\times 5.97\times {10}^{24}}{6\times {10}^6}=2\times {10}^{10}$