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. (Mass of earth = 6 × 10²⁴ kg, Radius of earth = 6.4 × 10⁶ m, Gravitational constant = 6.67 × 10⁻¹¹ N m² kg⁻²) Numerical answer type. Work it out, then open the answer.

Answer: 3. r = R + h = 6.4×10⁶ + 0.27×10⁶ = 6.67×10⁶ m. K = GMm/2r = 6.67×10⁻¹¹× 6×10²⁴× 10002× 6.67×10⁶. K = 3×10¹⁰ J. So the answer is 3.

A satellite of mass 1000 kg is launched to revolve around the earth… | JEE Main 2025 (2 Apr, Shift 2)

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Satellite Energy & Weightlessness

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____________ $\times 10^{10}\text{ J}$ . (Mass of earth $= 6 \times 10^{24}\text{ kg}$ , Radius of earth $= 6.4 \times 10^{6}\text{ m}$ , Gravitational constant $= 6.67 \times 10^{-11}\text{ N m}^{2}\text{ kg}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 3

$r = R + h = 6.4\times10^6 + 0.27\times10^6 = 6.67\times10^6$ m.
$K = \frac{GMm}{2r} = \frac{6.67\times10^{-11}\times 6\times10^{24}\times 1000}{2\times 6.67\times10^6}$ .
$K = 3\times10^{10}$ J.
So the answer is 3.

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