Question

Calculate 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 $10$R, where R is the radius of the earth and is equal to $6.4\times {10}^6$m.
(Given: Mass of the earth $=6\times {10}^{24}$kg, gravitational constant $G=6.7\times {10}^{-11}N{m}^{2}k{g}^{-2}$)

• A. $2.07\times {10}^{4}m{s}^{-1}$.
• B. $1.07\times {10}^{4}m{s}^{-1}$.
• C. $7\times {10}^{4}m{s}^{-1}$.
• D. $4\times {10}^{4}m{s}^{-1}$.

Answer 1

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

The gravitational potential on earth surface $U\left(R\right)=\frac{-{GM}_m}{R}$

Now at a distance from the earth surface is $11R$ from center. Thus gravitational potential $=U\left(11R\right)=\frac{-{GM}_m}{11R}$

Change in $PE=U\left(11R\right)-U\left(R\right)$

$=\frac{-{GM}_m}{11R}-\left(\frac{-{GM}_m}{R}\right)=\frac{{10GM}_m}{11R}$

The difference is equal to the kinetic energy $=\frac{1}{2}{mV}^2$ (initial velocity)

$\frac{1}{2}{mV}^2=\frac{{GM}_m}{11R}$

$V=\sqrt{\frac{{20GM}_m}{11R}}$

Putting value,

$V=\sqrt{\frac{20\times 6.7\times {10}^{-11}\times 6\times {10}^{24}}{11\times 64\times {10}^6}}$

$=1.07\times {10}^{4}m/s$