Question
Download Solution PDFIf the radius of the earth were to shrink by 1%, the density remaining constant, the acceleration due to gravity on the surface of the earth will be :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The acceleration due to gravity is acceleration by which the earth attracts anything falling on earth by virtue of gravity is called acceleration due to gravity.
It is given by
\(g = \frac{GM}{r^2}\) -- (1)
G is Universal Gravitational Constant, M is the mass of the Earth, and r is the radius of the earth.
Calculation:
Given:
- Now, if the radius is reduced to 1% which means volume has been changed, without any change in density. New mass will be \(M'=ρ. \frac{4}{3} πR^{' 3} \)
- Then the new gravitational constant is g'.
Then ,
\(g=\frac{4Gρ π R^3}{3R^2} \\ g= \frac{4Gρ π R}{3}\\ g \propto R\)
\(\frac{g^{'}}{g}=\frac{ \frac{4Gρ π (R-0.01R)}{3}}{ \frac{4Gρ π R}{3}}\\ \frac{g^{'}}{g}=0.99 \ \ \)
Hence, the new gravity is less than that of old one.
If we reduce the radius by 1%, then g will also be reduced.
The correct option is (4).
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