Physics - Gravitation Question with Solution | TestHub

The orbital velocity of a satellite revolving just above the Earth's surface is given by

$v=\sqrt{\frac{GM}{R}}$

For a satellite at an altitude $h=\frac{3}{4}R$, the distance from the center of the Earth is

$r=R+h=R+\frac{3}{4}R=\frac{7}{4}R$

The orbital velocity at this altitude is

$v^{\prime }=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{\frac{7}{4}R}}$

$v^{\prime }=\sqrt{\frac{4}{7}}\sqrt{\frac{GM}{R}}=\frac{2}{\sqrt{7}}v$

Answer: $\frac{2}{\sqrt{7}}v$