Gravitation

JEE Main level, single correct, medium

111 of 223

Question 111 of 223

An astronaut takes a ball of mass mm from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km318.5 \mathrm{~km}. From earth's surface to the orbit, the change in total mechanical energy of the ball is xGMem21Rex \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}. The value of xx is (take Re=6370 km)\left.\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km}\right) :

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Answer: (D) 11

h=318.5≈(Re20)T⋅Ei=−GMemReT⋅Ef=−GMem2(Re+h)=−GMem2(Re+Re20)⇒T⋅Ef=−10GMem21Re\begin{aligned} & \mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right) \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{i}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}} \\ & \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\mathrm{h}\right)}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{e}}+\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)} \\ & \Rightarrow \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}\end{aligned}
Change in total mechanical energy
=TEf−TEi=GMemRe⁡[1−1021]=11GMem21Re⁡\begin{aligned} & =\mathrm{TE}_{\mathrm{f}}-\mathrm{TE}_{\mathrm{i}} \\ & =\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\operatorname{Re}}\left[1-\frac{10}{21}\right]=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \operatorname{Re}}\end{aligned}

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