NEET-XII-Physics
11: Gravitation
- #12If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what per cent?Ans : If we consider the Earth to be a perfect sphere, then the acceleration due to gravity at its surface is given by `` g=G\frac{M}{{R}^{2}}``.
Here, M is the mass of Earth; R is the radius of the Earth and G is universal gravitational constant.
If the radius of the earth is decreased by 1%, then the new radius becomes
`` R\text{'}=R-\frac{R}{100}=\frac{99}{100}R``
`` \Rightarrow R\text{'}=0.99R``
New acceleration due to gravity will be given by
`` g\text{'}=G\frac{M}{R{\text{'}}^{2}}=G\frac{M}{(0.99R{)}^{2}}``
`` \Rightarrow g\text{'}=1.02\times \left(G\frac{M}{{R}^{2}}\right)=1.02g``
Hence, the value of the acceleration due to gravity increases when the radius is decreased.
Percentage increase in the acceleration due to gravity is given by
`` \frac{g\text{'}-g}{g}\times 100``
`` =\frac{0.02g}{g}\times 100``
`` =2\%``
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