JEE Main 2025 Apr 2 Shift 2, Physics Q25: Satellite, Planetary Motion And Kepler’s Law

JEE Main2025Apr 2, Shift 2Physics

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 ________ $\times 1 0^{10}$ J.

(Mass of earth $= 6 \times 1 0^{24}$ kg, Radius of earth $= 6.4 \times 1 0^{6}$ m, Gravitational constant $= 6.67 \times 1 0^{- 11}$ Nm $^{2}$ kg $^{- 2}$ )

Your answer

Solution

For a circular orbit at radius $r = R_{E} + h$ , the orbital speed satisfies $\frac{m v ^{2}}{r} = \frac{G M _{E} m}{r ^{2}}$ , giving kinetic energy

$KE = \frac{1}{2} m v^{2} = \frac{G M _{E} m}{2 r} = \frac{G M _{E} m}{2 ( R _{E} + h )}$ .

Substituting numbers with $R_{E} + h = 6.4 \times 1 0^{6} + 2.7 \times 1 0^{5} = 6.67 \times 1 0^{6}$ m:

$KE = \frac{( 6.67 \times 1 0 ^{- 11} ) ( 6 \times 1 0 ^{24} ) ( 1000 )}{2 ( 6.67 \times 1 0 ^{6} )}$ .

Numerator $= 6.67 \times 6 \times 1 0^{16} = 40.02 \times 1 0^{16} J\cdotm$ .

Denominator $= 13.34 \times 1 0^{6} m$ .

$KE \approx 3 \times 1 0^{10} J$ .

$∴$ the coefficient is $3$ .

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JEE Main 2025 Apr 2 Shift 2, Physics Q25: Satellite, Planetary Motion And Kepler’s Law

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