Question

An earth satellite of mass $m$ revolves in a circular orbit at a height $h$ from the surface of the earth. $R$ is the radius of the earth and $g$ is accelaration due to gravity at the surface of the earth. The velocity of the satellite in orbit is given by

• A. $g{R}^2/(R+h)$
• B. $gR$
• C. $gR/(R-h)$
• D. $\sqrt{\left[g{R}^2/\right(R+h\left)\right]}$

Answer 1

The correct option is D $\sqrt{\left[g{R}^2/\right(R+h\left)\right]}$

According to Newton gravitational force
$\frac{G{M}_{e}m}{R_e^2}=mg$ & $\frac{G{M}_{e}m}{(R_e+h{)}^2}=m{g}^{\prime }$
$\Rightarrow g^{\prime }=g\frac{1}{(1+h/R_e{)}^2}=\frac{g{R}_e^2}{(R_e+h{)}^2}$
where $M_e$ is mass of earth, $G$ is gravitational constant, $R_e$ is radius & earth, $h$ is height of satellite above the surface of earth, $g$ is value at the surface of earth, $g$' is value at height $h$ above the surface of earth.
so $\frac{m{v}^2}{(R_e+h)}=m{g}^{\prime }$

$\frac{m{v}^2}{(R_e+h)}=\frac{mg{R}_e^2}{(R_e+h{)}^2}$

$\Rightarrow v=\sqrt{\frac{g{R}_e^2}{R_e+h}}$