NEET-XII-Physics
11: Gravitation
- #6Find the acceleration due to gravity of the moon at a point 1000 km above the moon’s surface. The mass of the moon is 7.4 × 1022 kg and its radius is 1740 km.Ans : The acceleration due to gravity at a point at height h from the surface of the moon is given by
`` g=\frac{\,\mathrm{\,G\,}M}{{r}^{2}}``,
where M is the mass of the moon; r is the distance of point from the centre of the moon and G is universal gravitational constant.
`` \therefore g=\frac{GM}{{\left(R\mathit{+}h\right)}^{2}}``
`` \Rightarrow g=\frac{6.67\times {10}^{-11}\times 7.4\times {10}^{22}}{{\left(1740+1000\right)}^{2}\times {10}^{6}}``
`` \Rightarrow g=\frac{6.67\times 7.4\times {10}^{11}}{\left(1740+{1000}^{2}\times {10}^{6}\right)}``
`` \Rightarrow g=\frac{6.67\times 7.4\times {10}^{11}}{2740\times 2740\times {10}^{6}}``
`` \Rightarrow g=0.65\,\mathrm{\,m\,}/{\,\mathrm{\,s\,}}^{2}``
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