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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.
Solution
Time period of revolution of the satellite around the Mars is give by \[T = 2\pi\sqrt{\frac{r^3}{GM}}\] where M is the mass of the Mars and r is the distance of the satellite from the centre of the planet.
\[\text { Now }, 27540 = 2 \times 3 . 14\sqrt{\frac{\left( 9 . 4 \times {10}^3 \times {10}^3 \right)^3}{6 . 67 \times {10}^{- 11} \times M}}\]
\[ \Rightarrow \left( 27540 \right)^2 = \left( 6 . 28 \right)^2 \times \frac{\left( 9 . 4 \times {10}^5 \right)^3}{6 . 67 \times {10}^{- 11} \times M}\]
\[ \Rightarrow M = \frac{\left( 6 . 28 \right)^2 \times \left( 9 . 4 \right)^3 \times {10}^{18}}{6 . 67 \times {10}^{- 11} \times \left( 27540 \right)^2}\]
\[ \Rightarrow M = 6 . 5 \times {10}^{23} kg\]
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