Question

The velocity with which a body must be thrown vertically upward from the surface of the earth so that it may reach a height of $10R$, where $R$ is the radius of the earth and is equal to $6.4\times {10}^{6}m$ is
(Given: Mass of the earth = $6\times {10}^{24}kg$, gravitational constant $G=6.7\times {10}^{-11}N{m}^2{kg}^{-2}$)

• A. $1.70\times {10}^{4}m{s}^{-1}$
• B. $1.07\times {10}^{4}m{s}^{-1}$
• C. $2.07\times {10}^{4}m{s}^{-1}$
• D. None of these

Answer 1

The correct option is B $1.07\times {10}^{4}m{s}^{-1}$

Kinetic energy of the body will be used for increasing its potential energy.

$\frac{1}{2}m{v}^2=-\frac{GMm}{R+h}-(-\frac{GMm}{R})\Rightarrow v=\sqrt{-\frac{2GM}{11R}+\frac{2GMm}{R}}$

$v=\sqrt{\frac{2\times 6.67\times {10}^{-11}\times 6\times {10}^{24}}{6.4\times {10}^6}\times \frac{10}{11}}=1.07\times {10}^{4}m/s$