Question

The orbital velocity of a body at higher $h$ above the surface of the Earth is $36$% of that near the surface of the Earth of radius $R$. If the escape velocity at the surface of Earth is $11.2km{s}^{-1}$, then its value at the height $h$ will be:-

• A. $11.2km{s}^{-1}$
• B. $\sqrt{\frac{h}{R}}\times 11.2km{s}^{-1}$
• C. $\frac{9}{25}\times 11.2km{s}^{-1}$
• D. $\sqrt{\frac{R}{h}}\times 11.2km{s}^{-1}$

Answer 1

The correct option is C $\frac{9}{25}\times 11.2km{s}^{-1}$

$V_e=\sqrt{2}{V}_0$

If $V_0$ becomes $36$%, $V_e$ will also be $36$%

$\begin{array}{c}\therefore V{e}^{\prime }=\frac{36}{100}\times 11.2km/s\\ =\frac{9}{25}\times 11.2km/s\end{array}$

Hence, Option $C$ is correct.