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Q.
Average density of the earth:
NTA AbhyasNTA Abhyas 2022
Solution:
Acceleration due to gravity, $g=\frac{G M}{R^{2}},$ where $M=\left(\frac{4}{3} \pi R^{3}\right)\rho $
$\therefore \, g=\frac{4 G}{3}\frac{\pi R^{3}}{R^{2}}\rho $
$\Longrightarrow g=\left(\frac{4 G \pi R}{3}\right)\rho \, \left(\rho = a v e r a g e \, d e n s i t y\right)$
$\Longrightarrow g \propto \rho $ or $\rho \propto g$