Question

the period of the moon around the earth is 28 days and the radius of its orbit is 4*10 to the power of 5 km, if G =6.67*10 to the power of -11.Find the mass of the earth?

Answer 1

The period of the moon is calculated from this equation: $T=\frac{2\pi r}{v}.$Where r is the radius of the orbit and v is the velocity.

Since the moon moves in a circular motion, therefore, the centripetal force is: $F_c=\frac{mv^2}{r}$.

In this example, the gravitational force of the Earth acts as the centripetal force. therefore, $F_g=F_c$.

Therefore, $\frac{G*m_mm_e}{r^2}=\frac{m_mv^2}{r}$. Where $m_m$ is the mass of the moon and $m_e$ is the mass of earth.

$\frac{G*m_mm_e}{r^2}=\frac{m_mv^2}{r}$

$\frac{Gm_e}{r}=v^2\\ v=\sqrt{\frac{Gm_e}{r}}$.

Substitute by $v$ in the period equation:

$T=\frac{2\pi r}{\sqrt{\frac{Gm_e}{r} } }\\T=2\pi \sqrt{\frac{r^3}{Gm_e} }$.

The required is $m_e$. But first, we need to convert the period from days to seconds.

Therefore, T = 28 days*24 hours* 60 minutes*60 seconds = 2419200 seconds.

Therefore, $2419200=2\pi \sqrt{\frac{(4*10^5*10^3)^3}{6.67*10^{-11}*m_e} }\\385027.6=\sqrt{\frac{9.59*10^{35}}{m_e}} \\(385027.6)^2=\frac{9.59*10^{35}}{m_e}$

Therefore, $m_e=\frac{9.59*10^{35}}{(385027.6)^2}=6.4*10^{24}\:kg$