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The acceleration due to gravity g and density of the earth P are related by which of the following relations? (Here G is the gravitational constant and R is the radius of the earth).
(A) \[p = \dfrac{{4\pi }}{{3GRd}}\]
(B) \[p = \dfrac{{3g}}{{4\pi GR}}\]
(C) \[p = \dfrac{{3G}}{{4\pi GR}}\]
 (D) \[p = \dfrac{{4\pi GR}}{{3G}}\]

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Last updated date: 20th Jun 2024
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Answer
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Hint Write down the equation for acceleration due to gravity. Rearrange the equation and find the formula for mass. Then calculate the volume of earth. Finally substitute, mass and volume in the equation which states that the ratio of mass to volume gives us density.

Complete step-by-step solution
Acceleration due to gravity on earth is given as,
 \[g = \dfrac{{GM}}{{{R^2}}}\,\,\, \ldots (1)\]
Where, g: Acceleration due to gravity
G: Gravitational constant
M: Mass of earth
R: Radius of earth

Now we know, density is given as,
 \[density = \dfrac{{mass}}{{volume}}\]
 \[\therefore p = \dfrac{M}{V}\,\,\, \ldots (2)\]
Volume of Earth is
 \[V = \dfrac{4}{3}\pi {R^3}\,\, \ldots (3)\]
Now, rearranging equation.(1) gives us,
 \[{\text{M = }}\dfrac{{{\text{g}}{{\text{R}}^{\text{2}}}}}{{\text{G}}}\,\,\,{\ldots (4)}\]
By substituting equation.(3) and equation.(4) in equation.(2) we get,
$p=\dfrac{{{\text{g}}{{\text{R}}^{\text{2}}}}}{ \dfrac{4}{3}\pi {R^3}}$
 \[\therefore {\text{p = }}\dfrac{{{\text{3g}}}}{{{\text{4}\pi \text{GR}}}}\]

So, the relation between density and acceleration due to gravity is given as \[p = \dfrac{{3g}}{{4\pi GR}}\].
Acceleration due to gravity is proportional to mean density of earth and its radius.

Hence, the correct answer is option (B).

Additional Information

Equation of gravitational constant and acceleration due to gravity can be calculated as shown below:
Suppose, a body of mass m is placed on the surface of earth and assumes the shape of the earth to be round. If mass of the earth is M and radius of earth is R, then by Newton’s law of gravitation we get,
 \[{\text{F = G}}\dfrac{{{\text{Mm}}}}{{{{\text{R}}^{\text{2}}}}}\,\,{ \ldots (1)}\]
Now, from Newton’s Second law of motion,
 \[{\text{F = mg}}\,\,{ \ldots (2)}\]
From equation.(1) and equation.(2) we get,
  \[{\text{mg = G}}\dfrac{{{\text{Mm}}}}{{{{\text{R}}^{\text{2}}}}}\]
 \[\therefore {\text{g = }}\dfrac{{{\text{GM}}}}{{{{\text{R}}^{\text{2}}}}}\,\,\]

Note This expression can also be used at and within the earth at distance R from the centre of the earth. Generally, we use P (rho) as the symbol for density but in this question it is denoted by symbol p. So don’t get confused by the initialization. Solve the question using the initials mentioned in the question.