A geostationary satellite is revolving around the earth. To make it escape from gravitational field of earth, its velocity must be increased

The correct option is B

$41.4\%$

Explanation for the correct option:

Step 1: Formula used:

Escape speed for satellite

$\begin{array}{l}{v}_e=\sqrt{\frac{2G{M}_e}{r}}\end{array}$…………..$\left(1\right)$

where, $v_e$ is escape speed

$G$ is Universal gravitational constant

$M_e$ is mass of the earth

Also, we know that

Centripetal force on satellite $=$ Gravitational force between satellite and earth

$\Rightarrow \begin{array}{l}\frac{m{v}_0^2}{r}=\frac{G{M}_{e}m}{r^2}\end{array}$ where, $r$ is distance between earth and satellite and $m$ is mass of the satellite.

$\Rightarrow \begin{array}{l}{v}_0=\sqrt{\frac{G{M}_e}{r}}\end{array}$…………..$\left(2\right)$

Step 2: Finding the increase in the velocity

Substituting equation $\left(2\right)$ in $\left(1\right)$.

$\begin{array}{rcl}{v}_e& =& \sqrt{2}{v}_0\\ & \Rightarrow & v_e=1.414v_0\\ & \Rightarrow & \frac{v_e}{v_0}=1.414\end{array}$

Fractional increase $=\frac{(v_e–v_0)}{v_0}$

$$\begin{gathered} =\frac{1.414–1}{1} \\ =0.414 \\ =41.4\% \end{gathered}$$

Hence, option B is correct.