Question

A small moon orbits its planet in a circular orbit at a speed of $7.5km/s$. It takes $28hours$ to complete one full orbit. What is the mass of the planet?

Short answer

The mass of the planet is${10}^{26}kg.$

Step-by-step solution

Given information

Time period = $28hours=100800s,$speed of orbiting =$7.5km/s.$

Calculation

According to the Kepler's 3rd law, the time period is given by $T^2=\frac{4{\pi }^2{R}^3}{GM}$

Where, T is time period, R is the radius of the orbit, G is the universal gravitational constant and M is the required mass of the planet.

Also,$v=\frac{2\pi R}{T}$

Here, R is unknown and is given by the below formula. v is the speed, R is the radius and T is the time period.

Now solving we get,

$$\begin{gathered} v=\frac{2\pi R}{T} \\ R=\frac{v\times T}{2\pi } \\ R=\frac{(7500m/s)(28\times 60\times 60s)}{2\pi } \\ R=1.203\times {10}^8\mathrm{m} \end{gathered}$$

Now substituting all the values in Kepler's third law equation we get,

$$\begin{gathered} T^2=\frac{4{\pi }^2{\mathbf{r}}^3}{GM} \\ {T}^2=\frac{4{\pi }^2{R}^3}{GM} \\ M=\frac{4{\pi }^2{R}^3}{G{T}^2} \\ M=\frac{4{\pi }^2{\left(1.203\times {10}^{8}m\right)}^3}{(6.67\times {10}^{-11}N{m}^2/k{g}^2){\left(1.008\times {10}^{5}kg\right)}^2} \\ M=8.83\times {10}^{25}\mathrm{kg} \\ \simeq 1\times {10}^{26}\mathrm{kg} \end{gathered}$$

Final answer

The mass of the planet is${10}^{26}kg.$