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:

Option 1 - <math> <mroot> <mrow> <mrow> <mi mathvariant="normal">T</mi> </mrow> </mrow> <mrow></mrow> </mroot> </math>

Option 2 - T

Option 3 - <math> <msup> <mrow> <mrow> <mi mathvariant="normal">T</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math>

Option 4 - <math> <msup> <mrow> <mrow> <mi mathvariant="normal">T</mi> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math>

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Raj Pandey | Contributor-Level 9

5 months ago

Correct Option - 3

Detailed Solution:

Time period of satellite

$T = 2 π \sqrt[]{\frac{R^{3}}{G M}}$

$= 2 π \sqrt[]{\frac{R^{3}}{G d \frac{4}{3} π R^{3}}}$

$\Rightarrow T = \sqrt[]{\frac{3 π}{G d}}$ $\Rightarrow \frac{3 π}{G d} = T^{2}$

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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 3 π G d represents:

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