# Astronomers can calculate the mass of a planet by using the orbital parameters of one of its...

## Question:

Astronomers can calculate the mass of a planet by using the orbital parameters of one of its moons. Consider a hypothetical planet, which has a hypothetical moon that orbits the planet in a roughly circular orbit with a radius of 3.95*10{eq}^{12} {/eq} m. The moon takes 12 Earth days to orbit the planet. What is the mass of the planet?

## Circular Orbits:

When an object shows a circular path around the massive object in the sky, the gravitational force supports the centripetal force of the system. The orbital velocity is related to the mass of the massive object and the orbital radius of the circular path. In addition, the period squared is proportional to the cube of the orbital radius.

## Answer and Explanation:

According to Kepler's third law, the relationship between the orbital period T and the orbital radius r is given by

{eq}T^2=\dfrac{4\pi^2 r^3}{GM} \\ \rm Here:\\ \,\,\,\, \, \bullet \, G(=6.67\times10^{-11} N m^2/kg^2)\text{: gravitational constant}\\ \,\,\,\, \, \bullet \,M \text{: mass of the planet}\\ \,\,\,\, \, \bullet \, T(=12\, \rm days) \text{: orbital period of Io}\\ \,\,\,\, \, \bullet \,r(=3.95\times 10^{12}\, m) \text{: radius of the circular orbit} {/eq}

Please note, 1 day has 24 hours, and there are 3600 seconds in an hour. So, the mass of planet is

{eq}\begin{align} M&=\dfrac{4\pi^2r^3}{GT^2}\\\\ &=\dfrac{4\pi^2\times (3.95\times 10^{12})^3}{6.67\times10^{-11}\times (12\times 24\times 3600)^2}\\\\ &=\boxed{3.39\times 10^{37}\, kg} \end{align} {/eq}