The density of a newly discovered plant is twice that of earth. The acceleration due to gravity at the…
The density of a newly discovered plant is twice that of earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth $R$, the radius of the planet would be:
$2 R$
$4 R$
$\frac{1}{4} R$
$\frac{1}{2} R$
Solution
We know that:
$\begin{aligned}
g=\frac{G M}{R^2} & =\frac{G \frac{4}{3} \pi R^3 \rho}{R^2} \\
& =\frac{4}{3} \pi G \rho R
\end{aligned}$
New according to the question
$\begin{aligned}
g_{\text {planet }} & =g_{\text {earth }} \\
\Rightarrow \quad R_{\text {planet }} & =\frac{R_{\text {earth }}}{2}
\end{aligned}$
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