Answer to Question #134006 in Mechanics | Relativity for night

Question

Show that the period of a particle that moves in a circular orbit close to the surface of a sphere
depends only upon G and the average density ρ of the sphere. Find what this period would be
for any sphere having an average density equal to that of water.

Answer 1

Newton's second law gives us

$ma = F_G$

$\displaystyle m \omega^2 r = \frac{GmM}{r^2}$

$\displaystyle (\frac{2\pi}{T})^2 r = \frac{GM}{r^2}$

$\displaystyle T^2 = \frac{4 \pi^2}{GM} r^3$

If object is a sphere, then $M = \rho \cdot \frac{4 \pi}{3} R^3$ . Particle moves close to the surface, so$r= R+h \approx R$

$\displaystyle T^2 = \frac{3\cdot4 \pi^2}{G \rho 4 \pi R^3} R^3 = \frac{3 \pi}{G \rho}$

$\displaystyle T = \sqrt{\frac{3\pi}{G\rho}}= \sqrt{\frac{3 \cdot \pi}{6.67 \cdot 10^{-11} \cdot 10^3}}=11\,887 \; s = 198 \, m \; 7\,s = 3\, h \; 18\,m\;7\,s$

Answer: $3\, h \; 18\,m\;7\,s$