Physics · Gravitation

JEE Main 2025 — 2 April, Evening Shift — Question 68

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 $\qquad$

$\times 10^{10} \mathrm{~J}$ . (Mass of earth $=6 \times 10^{24} \mathrm{~kg}$ ,

Radius of earth $=6.4 \times 10^{6} \mathrm{~m}$ , Gravitational constant

$=6.67 \times 10^{-11}$ $\mathrm{Nm}^{2} \mathrm{~kg}^{-2}$ )

Answer: 3

Numerical answer — enter this value.

$K=-\frac{U}{2}=\frac{G M m}{2 r}$

\begin{aligned} \\ = & \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 10^{3}}{2 \times 6670 \times 10^{3}} \\ & =3 \times 10^{10} \mathrm{~J} \end{aligned}

Answer key and solution verified before publishing.

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