Question

An astronaut takes a ball of mass m from earth to space. He throws the ball into a circular orbit about earth at an altitude of \(318.5 \mathrm{~km}\). From earth's surface to the orbit, the change in total mechanical energy of the ball is \( x \frac{\mathrm{GM}_{e} \mathrm{~m}}{21 \mathrm{R}_{e}}\). The value of x is (take \(\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km}\)):

(1) 11

(2) 9

(3) 12

(4) 10

Answer 1

Correct option is : (1) 11 

\(\because \text { Total mechanical energy }=\frac{\mathrm{PE}}{2} \quad\left(\because \frac{\mathrm{R}_{\mathrm{e}}}{20}=318.5\right) \) 

\( \text { ME on surface of earth }=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}}(\mathrm{KE} \text { on surface }=0) \)  

\( \text { ME at an altitude }=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)}=-\frac{20 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2 \times 21 \mathrm{R}_{\mathrm{e}}}\) 

\(=\frac{-10 \mathrm{Gm}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}\)  

\(\text { Change in Total M.E. }=\mathrm{E}_{\mathrm{f}}-\mathrm{E}_{\mathrm{i}}\)  

\( =-\frac{10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}+\frac{\mathrm{GM_e} \mathrm{m}}{\mathrm{Re}} \) 

\(=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}\ +\ 21 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}\)  

\(\mathrm{x}=11\)