Question

A moon revolves around a planet with a speed of 9.0×10^3mls.the distance from the moon to the centre of the planets is 5.4×10^6m.what is period of the moon?

Answer 1

Let's first find the mass of the planet about which the moon revolves:

$$F_c=F_g$$ $$\dfrac{GM_{m}M_p}{R^2}=\dfrac{M_mv^2}{R},$$ $$M_p=\dfrac{v^2R}{G}=\dfrac{(9.0\cdot10^3\ \dfrac{m}{s})^2\cdot5.4\cdot10^6\ m}{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}}=6.56\cdot10^{24}\ kg.$$

Finally, we can find the period of the moon from the Newton's form of Kepler's third law:

$$\dfrac{T^2}{R^3}=\dfrac{4\pi^2}{GM_p},$$ $$T=\sqrt{\dfrac{4\pi^2R^3}{GM_p}},$$ $$T=\sqrt{\dfrac{4\pi^2\cdot(5.4\cdot10^6\ m)^3}{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot6.56\cdot10^{24}\ kg}}=3767\ s=1.05\ hr.$$