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Question: On the surface of the earth acceleration due to gravity will be \(g\) and gravitational potential wi...

On the surface of the earth acceleration due to gravity will be gg and gravitational potential will be VV.

| COLUMN 1| | COLUMN 2
---|---|---|---
a| At heighth=Rh=R, value of gg| p| Decreases by a factor 14\dfrac{1}{4}
b| At depthh=R2h=\dfrac{R}{2}, value of gg | q| Decreases by a factor 12\dfrac{1}{2}
c| At heighth=Rh=R, value of VV | r| Increases by a factor 118\dfrac{11}{8}
d| At depthh=R2h=\dfrac{R}{2}, value of VV | s| Increases by a factor 22
| | t| None

A.(ap),(bq),(cs),(dt) B.(aq),(bp),(ct),(ds) C.(at),(bs),(cp),(dq) D.none of these \begin{aligned} & A.\left( a-p \right),\left( b-q \right),\left( c-s \right),\left( d-t \right) \\\ & B.\left( a-q \right),\left( b-p \right),\left( c-t \right),\left( d-s \right) \\\ & C.\left( a-t \right),\left( b-s \right),\left( c-p \right),\left( d-q \right) \\\ & D.\text{none of these} \\\ \end{aligned}

Explanation

Solution

The value of acceleration due to gravity at the surface can be found by taking the ratio of the product of the gravitational constant and the mass of the earth to the square of the radius of the earth. The acceleration due to gravity at a specific height from the surface of earth will be found by taking the ratio of the product of the gravitational constant and the mass of the earth to the square of the height on the surface.

Complete step by step answer:
Value of gg at a distance rr from the centre of the earth =GMr2=\dfrac{GM}{{{r}^{2}}}
The value of gg at the surface =GMR2=\dfrac{GM}{{{R}^{2}}}
The value of gg at a height RR from the surface =GM(2R)2=\dfrac{GM}{{{\left( 2R \right)}^{2}}}
Therefore the acceleration due to gravity decreased by a factor of 14\dfrac{1}{4}.
Dependence of the acceleration of gravity gg on the depth hh be,
gh=4π3Gρh{{g}_{h}}=\dfrac{4\pi }{3}G\rho h
Hence it will decrease by a factor of 12\dfrac{1}{2} ​ at h=R2h=\dfrac{R}{2}.
The gravitational potential varies with distance rr from the centre of the earth as GMr\dfrac{-GM}{r}.
Hence at the surface will be having a value be GMR\dfrac{-GM}{R}
At a height RR, it will be having a value as, GM2R\dfrac{-GM}{2R}
Therefore it will be increased by a factor 22, as the potential is negative.

So, the correct answer is “Option A”.

Note: The gravitational potential at a point will be equivalent to the work per unit mass that is required to move a body to that point from a fixed reference point. It will be analogous to the electric potential where the mass is playing the role of the charge. In short the gravitational potential at a point will be the gravitational potential energy per unit time.