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 )}

 

Answers (1)

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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