An Astronaut Stands on the Surface of a Spherical Asteroid: Escape and Rotation Limits

Answer

(a) Largest asteroid radius

The astronaut's Earth jump speed follows from near-surface energy conservation:

$$\frac12 m v_j^2=m g_E h \quad\Rightarrow\quad v_j^2=2g_Eh.$$

For a uniform spherical asteroid with Earth-like mean density $\rho_E$,

$$M=\frac43\pi\rho_E r^3, \qquad v_{\mathrm{esc}}^2=\frac{2GM}{r}=\frac{8\pi G\rho_E}{3}r^2.$$

Earth's surface gravity satisfies $g_E=4\pi G\rho_E R_E/3$, so

$$v_{\mathrm{esc}}^2=2g_E\frac{r^2}{R_E}.$$

At the largest allowable radius, the jump speed just equals escape speed:

$$2g_Eh=2g_E\frac{r^2}{R_E} \quad\Rightarrow\quad r_{\max}=\sqrt{hR_E}.$$

Using the stated values,

$$r_{\max}=\sqrt{(0.5)(6.4\times10^6)} =1.79\times10^3\,\mathrm{m} \approx \boxed{1.8\,\mathrm{km}}.$$

(b) Rotation limit

At the equator, contact is just maintained when gravity supplies exactly the required centripetal acceleration:

$$\omega_{\max}^2r=g_{\mathrm{ast}}, \qquad g_{\mathrm{ast}}=\frac{GM}{r^2}=g_E\frac{r}{R_E}.$$

Therefore

$$\omega_{\max}=\sqrt{\frac{g_E}{R_E}} =\sqrt{\frac{9.8}{6.4\times10^6}} \approx \boxed{1.24\times10^{-3}\,\mathrm{rad/s}}.$$

The corresponding minimum rotation period is

$$T_{\min}=\frac{2\pi}{\omega_{\max}} \approx 5.08\times10^3\,\mathrm{s} \approx \boxed{84.6\,\mathrm{min}}.$$

Faster rotation would make a freely standing astronaut lose contact at the equator. For the maximum-radius asteroid, the equatorial speed at this limit is about $2.21\,\mathrm{m/s}$.