The correct formula for the height h of a sat | vedclass

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Question

(A) For a satellite of mass $m$ orbiting at a height $h$ above the Earth's surface,the gravitational force provides the necessary centripetal force:$\

Vedclass · Accepted Answer

$\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R$

Vedclass · Answer

(A) For a satellite of mass $m$ orbiting at a height $h$ above the Earth's surface,the gravitational force provides the necessary centripetal force:$\frac{GMm}{(R+h)^2} = \frac{mv^2}{(R+h)}$$\Rightarrow \frac{GM}{(R+h)} = v^2 \quad \dots(1)$Since $v = (R+h)\omega = (R+h)\frac{2\pi}{T}$,we have $v^2 = (R+h)^2 \frac{4\pi^2}{T^2} \quad \dots(2)$Also,the acceleration due to gravity at the Earth's surface is $g = \frac{GM}{R^2}$,so $GM = gR^2 \quad \dots(3)$Substituting $(2)$ and $(3)$ into $(1)$:$\frac{gR^2}{(R+h)} = (R+h)^2 \frac{4\pi^2}{T^2}$$\Rightarrow (R+h)^3 = \frac{gR^2 T^2}{4\pi^2}$$\Rightarrow R+h = \left(\frac{gR^2 T^2}{4\pi^2}\right)^{1/3}$$\Rightarrow h = \left(\frac{T^2 R^2 g}{4\pi^2}\right)^{1/3} - R$

The correct formula for the height $h$ of a satellite from the Earth's surface is:

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