Question

A man can jump up to height $\frac{4}{3}$ m on the surface of a planet of radius 12000 km. Assume another planet having same density as first where man escapes through jumping. The radius of the second planet is ______ km.

Answer 1

4

Answer 2

The man's jumping ability is characterized by an initial velocity $v_0$. On the first planet (radius $R_1$, density $\rho$), this velocity allows him to reach a height $h_1$. Using energy conservation, $\frac{1}{2}mv_0^2 = \frac{GM_1mh_1}{R_1(R_1+h_1)}$. Since $h_1 \ll R_1$, this simplifies to $\frac{1}{2}mv_0^2 \approx mgh_1$, where $g = \frac{GM_1}{R_1^2}$. Substituting $M_1 = \rho \frac{4}{3}\pi R_1^3$, we get $g = \frac{4}{3}\pi G \rho R_1$, so $v_0^2 \approx 2gh_1 = \frac{8}{3}\pi G \rho R_1 h_1$.

On the second planet (radius $R_2$, density $\rho$), the man escapes through jumping, meaning his initial velocity $v_0$ is at least the escape velocity $v_e$. For escape, $v_e^2 = \frac{2GM_2}{R_2}$. Substituting $M_2 = \rho \frac{4}{3}\pi R_2^3$, we get $v_e^2 = \frac{8}{3}\pi G \rho R_2^2$. Assuming the man can just escape, $v_0^2 = v_e^2$. Equating the expressions for $v_0^2$ and $v_e^2$: $\frac{8}{3}\pi G \rho R_1 h_1 = \frac{8}{3}\pi G \rho R_2^2$ $R_1 h_1 = R_2^2$.

Given $R_1 = 12000$ km and $h_1 = \frac{4}{3}$ m. To maintain consistency in units for $R_2$ in km, we convert $h_1$ to km: $h_1 = \frac{4}{3} \times 10^{-3}$ km. $R_2^2 = (12000 \text{ km}) \times (\frac{4}{3} \times 10^{-3} \text{ km})$ $R_2^2 = 16 \text{ km}^2$ $R_2 = 4 \text{ km}$.