A satellite is orbiting just above the surface of the planet of density ' ρ ' with periodic time ' T '. The quantity T²ρ is equal to ( G = universal gravitational constant) (a) (4π²)/(G) (b) (3π²)/(G) (c) 3π/G (d) π/G

Answer: (c). Just above the surface: T² = (4π² R³)/(GM) with M = 4/3π R³ρ. T² = (4π² R³)/(G · 4/3π R³ ρ) = (3π)/(Gρ) T²ρ = 3π/G So the answer is (c) 3π/G.

A satellite is orbiting just above the surface of the planet of… | MHT CET 2024 (16 May, Shift 1)

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Satellites - Orbital Velocity & Period

A satellite is orbiting just above the surface of the planet of density ' $\rho$ ' with periodic time ' $T$ '. The quantity $T^2\rho$ is equal to ( $G =$ universal gravitational constant)

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Answer: (c)

Just above the surface: $T^2 = \frac{4\pi^2 R^3}{GM}$ with $M = \frac{4}{3}\pi R^3\rho$ .
$T^2 = \frac{4\pi^2 R^3}{G \cdot \frac{4}{3}\pi R^3 \rho} = \frac{3\pi}{G\rho}$
$T^2\rho = \frac{3\pi}{G}$
So the answer is (c) $\frac{3\pi}{G}$ .

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