JEE Mains · Physics · STD 11 - 7. gravitation
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 ______ \(\times 10^{10} \mathrm{~J}\).
(Mass of earth \(=6 \times 10^{24} \mathrm{~kg}\), Radius of earth \(=\) \(6.4 \times 10^6 \mathrm{~m}\), Gravitational constant \(=\)\(\left.6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\right)\)
- A 5
- B 4
- C 3
- D 2
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
\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)} \\ &…
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