NEET-XII-Physics
11: Gravitation
- #31A Mars satellite moving in an orbit of radius 9.4 × 103 km takes 27540 s to complete one revolution. Calculate the mass of Mars.Ans : 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.
`` \,\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\,}``
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