NEET-XII-Physics
11: Gravitation
- #35The radius of a planet is R1 and a satellite revolves round it in a circle of radius R2. The time period of revolution is T. Find the acceleration due to the gravitation of the planet at its surface.Ans : The time period of revolution of the satellite around a planet in terms of the radius of the planet and radius of the orbit of the satellite is given by `` T=2\,\mathrm{\,\pi \,}\sqrt{\frac{{R}_{2}^{2}}{g{R}_{1}^{2}}}``, where g is the acceleration due to gravity at the surface of the planet.
`` \,\mathrm{\,Now\,},{T}^{2}=4{\,\mathrm{\,\pi \,}}^{2}\frac{{R}_{2}^{2}}{g{R}_{1}^{2}}``
`` \Rightarrow g=\frac{4{\,\mathrm{\,\pi \,}}^{2}}{{T}^{2}}\frac{{R}_{2}^{2}}{{R}_{1}^{2}}``
∴ Acceleration due to gravity of the planet = `` \frac{4{\,\mathrm{\,\pi \,}}^{2}}{{T}^{2}}\frac{{R}_{2}^{2}}{{R}_{1}^{2}}``
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