A satellite of mass 1000 kg revolves around Earth in a circular orbit at a height of 270 km above the surface. Take M(Earth) = 6×10²⁴ kg, R(Earth) = 6.4×10⁶ m and G = 6.67×10⁻¹¹ N m² kg⁻². If its kinetic energy is N×10¹⁰ J, find N.
Physics · Gravitation · sattelite and keplers law
JEE Mains 2025 2 April Evening Shift

Question

A satellite of mass 1000 kg revolves around Earth in a circular orbit at a height of 270 km above the surface. Take M(Earth) = 6×10²⁴ kg, R(Earth) = 6.4×10⁶ m and G = 6.67×10⁻¹¹ N m² kg⁻². If its kinetic energy is N×10¹⁰ J, find N.

Physics · Gravitation · sattelite and keplers law

JEE Mains 2025 2 April Evening Shift

G1 Physics · Accepted Answer

For a satellite in a circular orbit, K = GMm/[2r], where r = R + h. Given R = 6.4×10⁶ m, h = 270 km = 2.70×10⁵ m. Thus r = 6.4×10⁶ + 0.27×10⁶ = 6.67×10⁶ m. Now K = [(6.67×10⁻¹¹)(6×10²⁴)(1000)]/[2(6.67×10⁶)]. Cancel 6.67: K = [6×10¹⁶]/2 = 3×10¹⁰ J. Therefore N = 3.