Question

A Mars satellite moving in an orbit of radius 9.4 × 10³ km takes 27540 s to complete one revolution. Calculate the mass of Mars.

Answer 1

Time period of revolution of the satellite around the Mars is give by
$T=2\mathrm{\pi }\sqrt{\frac{r^3}{\mathrm{G}M}}$,
where M is the mass of the Mars and r is the distance of the satellite from the centre of the planet.

$$\begin{gathered} \mathrm{Now},27540=2\times 3.14\sqrt{\frac{{\left(9.4\times {10}^3\times {10}^3\right)}^3}{6.67\times {10}^{-11}\times \mathrm{M}}} \\ \Rightarrow {\left(27540\right)}^2={\left(6.28\right)}^2\times \frac{{\left(9.4\times {10}^5\right)}^3}{6.67\times {10}^{-11}\times \mathrm{M}} \\ \Rightarrow M=\frac{{\left(6.28\right)}^2\times {\left(9.4\right)}^3\times {10}^{18}}{6.67\times {10}^{-11}\times {\left(27540\right)}^2} \\ \Rightarrow M=6.5\times {10}^{23}\mathrm{kg} \end{gathered}$$