An asteroid of mass $1.4 \times 10^{14} \mathrm{~kg}$ is observed to have an escape speed equal to a typical person's maximum jumping speed, $3.0 \mathrm{~m} / \mathrm{s}$. What is its radius?

Question

Quizplus · Accepted Answer

The escape speed from a spherical object of radius $R$ and mass $M$ is given by:$$v_{esc}=\sqrt{\frac{2GM}{R}}$$where $G$ is the gravitational constant. Solving for $R$, we get:$$R=\frac{2GM}{v_{esc}^2}$$Plugging in the given values, we get:$$R=\frac{2(6.674\times10^{-11}\mathrm{~N}\cdot\mathrm{m}^2/\mathrm{kg}^2)(1.4\times10^{14}\mathrm{~kg})}{(3.0\mathrm{~m}/\mathrm{s})^2}=2.1\mathrm{~km}$$

An Asteroid of Mass 1.4×1014 kg1.4 \times 10^{14} \mathrm{~kg}1.4×1014 kg Is Observed to Have an Escape Speed Equal to a to Have

An Asteroid of Mass $1.4 \times 10^{14} \mathrm{~kg}$ Is Observed to Have an Escape Speed Equal to a to Have