Question

A man can jump over $b=4m$ wide trench on earth. If mean density of an imaginary planet is twice that of the earth, then the maximum possible radius $r$ so that he may escape from it by jumping. Given radius of earth $=$ $6400km$ then find the value of $r^2$ in $k{m}^2$?

• A. $6.4$
• B. $6$
• C. $7$
• D. None of these

Answer 1

The correct option is A $6.4$

Velocity required to jump over a trench at an angle of ${45}^{\circ }$ is $v_0=\sqrt{bg}$.

Kinetic energy required to jump over this trench$=\frac{1}{2}m{v}_0^2=\frac{1}{2}mbg$

To jump out of the imaginary planet $=\frac{G{M}_{p}m}{r}=\frac{1}{2}mbg$

$⟹\frac{G\left(\rho \frac{4}{3}\pi r^3\right)m}{r}=\frac{1}{2}mb(G\frac{\rho \left(\frac{4}{3}\pi R_e^3\right)}{R_e^2}$)

$⟹r^2=\frac{1}{4}b{R}_e=6.4k{m}^2$