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# If R is the radius of the earth and g is the acceleration due to gravity on the earth's surface, then mean density of the earth isA) $3gR/4\pi G$B) $4gG/3\pi R$C) $3g/4\pi RG$D) $\pi g/12RG$

Last updated date: 19th Apr 2024
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Hint Use the formula of density and replace the value of Mass of the earth by $M = \dfrac{{g{R^2}}}{G}$ and the value of Volume of the earth $V = \dfrac{4}{3}\pi {R^3}$ and solve to find the answer.

Step by step Solution
The acceleration due to gravity on the earth’s surface is given by the formula,
$\begin{gathered} g = \dfrac{{GM}}{{{R^2}}} \\ \Rightarrow M = \dfrac{{g{R^2}}}{G} \\ \end{gathered}$
Now, the volume of the earth is given by,
$V = \dfrac{4}{3}\pi {R^3}$, where R is the radius of the earth.
Also the density is given by the formula,
mass = Density $\times$Volume.
Hence density = Mass/volume.
$\Rightarrow d = \dfrac{M}{V}$
Putting the values of mass (M) and the volume (V) of the earth we have,
$\begin{gathered} \Rightarrow d = \dfrac{{\dfrac{{g{R^2}}}{G}}}{{\dfrac{4}{3}\pi {R^3}}} \\ \Rightarrow d = \dfrac{{3g}}{{4\pi RG}} \\ \end{gathered}$
Hence the density is given by $\Rightarrow d = \dfrac{{3g}}{{4\pi RG}}$.

And the correct option for the following is option C.

Note
The value of acceleration due to gravity (g) depends on the following factors.
1) Acceleration due to gravity changes with height above the surface of the earth.
2) The Acceleration due to gravity changes with the depth of the earth , Here d is taken as the distance from the centre of the earth.
3) the Acceleration due to gravity changes with the rotation of the earth.
4) The Acceleration due to gravity changes with the shape and size of the earth.