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If the density of the earth is doubled keeping radius constant, find the new acceleration due to gravity? $\left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right)$
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Verified Answer
The correct answer is:
$19.6 \mathrm{~m} / \mathrm{s}^{2}$
Acceleration due to gravity $g=\frac{4}{3} \pi \rho G R$
or $g \propto d R$
$$
\left[\because \text { density } d=\frac{4}{3} \pi \rho\right]
$$
$$
\begin{aligned}
\therefore \quad & \frac{g_{1}}{g_{2}}=\frac{d_{1}}{d_{2}} \\
\frac{g_{1}}{g_{2}} &=\frac{d}{2 d} \quad\left(\because d_{2}=2 d\right) \\
& g_{2}=g_{1} \times 2=9.8 \times 2 \\
& g_{2}=19.6 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
$$
or $g \propto d R$
$$
\left[\because \text { density } d=\frac{4}{3} \pi \rho\right]
$$
$$
\begin{aligned}
\therefore \quad & \frac{g_{1}}{g_{2}}=\frac{d_{1}}{d_{2}} \\
\frac{g_{1}}{g_{2}} &=\frac{d}{2 d} \quad\left(\because d_{2}=2 d\right) \\
& g_{2}=g_{1} \times 2=9.8 \times 2 \\
& g_{2}=19.6 \mathrm{~m} / \mathrm{s}^{2}
\end{aligned}
$$
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