Question

A science fiction story I once read had the premise that a small, heavy piece of rock could orbit a planet at a height of 1.5 m. Because of this, the natives would always duck down when walking across the small moon's path, a behavior that seemed inexplicable to the explorers who had recently arrived but who didn't know about the moon. If air resistance weren't a problem, and such a situation existed on earth, (a) how fast would the moon travel and (b) how long would the orbital period be

Answer 1

Answer:

$v=7905.8m/s$

$t=506.3s$

Explanation:

The force of gravity between the Earth, of mass $M=5.97\times10^{24}Kg$ and a rock of mass m orbiting at a distance that would be the radius of the Earth R=6371000m (1.5m is insignificant) would be:

$F=\frac{GMm}{R^2}$

Where $G=6.67\times10^{-11}Nm^2/Kg^2$ is the gravitational constant.

Under this force, the rock experiments a (centripetal) acceleration given by:

$F=ma=m\frac{v^2}{R}$

Putting all together:

$m\frac{v^2}{R}=\frac{GMm}{R^2}$

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

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

Which for our values is:

$v=\sqrt{\frac{(6.67\times10^{-11}Nm^2/Kg^2)(5.97\times10^{24}Kg)}{(6371000m)}}=7905.8m/s$

And since the circumference of the orbit would be $C=2\pi R$, the time taken to travel it would be:

$t=\frac{C}{v}=\frac{2\pi R}{v}=\frac{2\pi (6371000m)}{7905.8m/s}=506.3s$