Question

A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing experiments, one of their tasks is to demonstrate the concept of the escape speed by throwing rocks straight up at various initial speeds. With what minimum initial speed ????esc will the rocks need to be thrown in order for them never to "fall" back to the asteroid? Assume that the asteroid is approximately spherical, with an average density ????=2.02×106 g/m3 and volume ????=3.09×1012 m3 . Recall that the universal gravitational constant is ????=6.67×10−11 N·m2/kg2 .

Answer 1

Answer:

9.60 m/s

Explanation:

The escape speed of an object from the surface of a planet/asteroid is given by:

$v=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the mass of the planet/asteroid

R is the radius of the planet/asteroid

In this problem we have

$\rho = 2.02\cdot 10^6 g/m^3$ is the density of the asteroid

$V=3.09\cdot 10^{12}m^3$ is the volume

So the mass of the asteroid is

$M=\rho V=(2.02\cdot 10^6 g/m^3)(3.09\cdot 10^{12} m^3)=6.24\cdot 10^{18} g=6.24\cdot 10^{15} kg$

The asteroid is approximately spherical, so its volume can be written as

$V=\frac{4}{3}\pi R^3$

where R is the radius. Solving for R,

$R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.09\cdot 10^{12} m^3)}{4\pi}}=9036 m$

Substituting M and R inside the formula of the escape speed, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(6.24\cdot 10^{15})}{(9036)}}=9.60 m/s$