Gravitation

JEE Main level, numerical answer, medium

109 of 223

Question 109 of 223

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ______ ×1010 J\times 10^{10} \mathrm{~J}.
(Mass of earth =6×1024 kg=6 \times 10^{24} \mathrm{~kg}, Radius of earth == 6.4×106 m6.4 \times 10^6 \mathrm{~m}, Gravitational constant ==
6.67×10−11Nm2 kg−2)\left.6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\right)

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Answer: 3

KE=12mv2=12 mGMer=GMem2r=GMem2(RE+h)=6.67×10−11×6×1024×6.4×1062(6.4×106+2.7×105)=3×1010 J\begin{aligned} & \mathrm{KE}=\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2 \mathrm{r}}=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{2\left(\mathrm{R}_{\mathrm{E}}+\mathrm{h}\right)} \\ & =\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 6.4 \times 10^6}{2\left(6.4 \times 10^6+2.7 \times 10^5\right)}=3 \times 10^{10} \mathrm{~J}\end{aligned}

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