# An earth’s satellite of mass m revolves in a circular orbit at a height h from the surface g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by Option 1) $\frac{gR^{2}}{R+h}$ Option 2) $gR$ Option 3) $\frac{gR}{R+h}$ Option 4) $\sqrt{\left (\frac{ gR^{2}}{R+h} \right )}$

As we learnt in

Variation in 'g' with height -

$g'=g\left ( \frac{R}{R+h} \right )^{2}$

$g'\rightarrow$ gravity at height $h$ from surface of earth.

$R\rightarrow$ Radius of earth

$h\rightarrow$ height above surface

- wherein

$g'\alpha\, \frac{1}{r^{2}}$

$r=R+h$

At height h, effect of gravity is ${g}'=\frac{g}{\left ( 1+\frac{h}{R} \right )^{2}}$

$\Rightarrow m{g}'=\frac{mv^{2}}{R+h}$

$\Rightarrow \frac{g}{\left ( 1+\frac{h}{R} \right )^{2}}=\frac{v^{2}}{R+h}\ \: \: or\ \: \: \frac{R^{2}g}{R+h}=v^{2}$

$\Rightarrow v=\sqrt{\frac{gR^{2}}{R+h}}$

Option 1)

$\frac{gR^{2}}{R+h}$

This is incorrect option

Option 2)

$gR$

This is incorrect option

Option 3)

$\frac{gR}{R+h}$

This is incorrect option

Option 4)

$\sqrt{\left (\frac{ gR^{2}}{R+h} \right )}$

This is correct option

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