A satellite is orbiting just above the surface of the earth with period T . If d is the density of the earth and G is the universal constant of gravit - Physics MCQ NEET 2023 2026 | ExamTest.live

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A satellite is orbiting just above the surface of the earth with period T . If d is the density of the earth and G is the universal constant of gravitation, the quantity \frac{3 π}{G d} represents:

Solution & Step-by-step Explanation

The time period of a satellite orbiting close to the surface (r \approx R) is: T = 2 π \frac{R ^{3}}{GM} Mass M = Density \times Volume = d \times \frac{4}{3} π R^{3} . T^{2} = 4 π^{2} [\frac{R ^{3}}{G ( d \cdot \frac{4}{3} π R ^{3} )}] T^{2} = 4 π^{2} [\frac{3}{4 π G d}] = \frac{3 π}{G d} Thus, \frac{3 π}{G d} = T^{2} .

Try it yourself before checking the explanation above.

A satellite is orbiting just above the surface of the earth with period T . If d is the density of the earth and G is the universal constant of gravitation, the quantity \frac{3 π}{G d} represents:

T

T^{2}

T

T^{3}

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