NEET-XII-Physics
11: Gravitation
- #22Find the height over the Earth’s surface at which the weight of a body becomes half of its value at the surface.Ans : Assume that at height h, the weight of the body becomes half.
Weight of the body at the surface = mg
Weight of the body at height h above the Earth's surface = mg', where g' is the acceleration due to gravity at height h
`` \,\mathrm{\,Now\,},g\text{'}=\frac{1}{2}g``
`` \therefore \left(\frac{1}{2}\right)\frac{\,\mathrm{\,G\,}M}{{R}^{2}}=\frac{\,\mathrm{\,G\,}M}{{\left(R+h\right)}^{2}}\left[\because g=\frac{\,\mathrm{\,G\,}\,\mathrm{\,M\,}}{{\,\mathrm{\,R\,}}^{2}}\right]``
`` \Rightarrow 2{R}^{2}={\left(R+h\right)}^{2}``
`` \Rightarrow \sqrt{2}R=R+h``
`` \Rightarrow h=\left(\sqrt{2}-1\right)R``
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