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Q. If the density of the earth is doubled keeping radius constant, find the new acceleration due to gravity? $ (g = 9.8\, m/s^2) $

MHT CETMHT CET 2010

Solution:

Acceleration due to gravity $g=\frac{4}{3} \pi \rho G R$
or $\,\,\,g \propto d R$ $\,\,\,\left[\because\right.$ density $\left.d=\frac{4}{3} \pi \rho\right]$
$\therefore \frac{g_{1}}{g_{2}}=\frac{d_{1}}{d_{2}} $
$ \frac{g_{1}}{g_{2}}=\frac{d}{2 d} \,\,\,\,\left(\because d_{2}=2 d\right)$
$g_{2}=g_{1} \times 2=9.8 \times 2$
$g_{2}=19.6\, m / s ^{2}$