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Question: The gravitational field intensity at a point \( 10000km \) from the centre of the earth is \( 4.8Nk{...

The gravitational field intensity at a point 10000km10000km from the centre of the earth is 4.8Nkg−14.8Nk{g^{ - 1}} . Calculate the gravitational potential at the same point.
(A) −4.8×107Jkg−1- 4.8 \times {10^7}Jk{g^{ - 1}}
(B) −2.4×107Jkg−1- 2.4 \times {10^7}Jk{g^{ - 1}}
(C) 4.8×106Jkg−14.8 \times {10^6}Jk{g^{ - 1}}
(D) 3.6×106Jkg−13.6 \times {10^6}Jk{g^{ - 1}}

Explanation

Solution

Hint
To solve this problem, first compare the formulae for gravitational field intensity and potential. The relation connecting the potential and gravitational field intensity can be used to solve the problem.
Formulae used:
E=GMR2E = \dfrac{{GM}}{{{R^2}}}
V=−GMRV = - \dfrac{{GM}}{R}
where EE is the gravitational field intensity, VV is the gravitational potential, GG is the gravitational constant, MM is the mass of earth, and RR is the distance to the point from the centre of the earth.

Complete step by step answer
The gravitational field intensity EE at a point is the gravitational force experienced by a unit mass placed at the same point. The gravitational potential at a point in the gravitational field is the work done while moving the unit mass to the point from infinity (where the potential is zero). Let us consider that the point of interest is ‘A’ and it is given that the gravitational field intensity EE at ‘A’ is E=4.8Nkg−1E = 4.8Nk{g^{ - 1}} .
The distance from the centre of the Earth to A is given as
R=10000km=10000000m=107mR = 10000km = 10000000m = {10^7}m
But the gravitational intensity is given by:
E=GMR2E = \dfrac{{GM}}{{{R^2}}}
At the same time, we know that the gravitational potential,
V=−GMRV = - \dfrac{{GM}}{R}
To compare the above two equations and to estimate the gravitational potential at A, we divide both the equations, i.e.,
VE=−R\dfrac{V}{E} = - R
∴V=−E×R\therefore V = - E \times R
Thus, we can calculate the gravitational potential at the point A by substituting the values of EE and RR .
Substituting and solving we get,
V=−4.8Nkg−1×107m=−4.8×107Nm.kg−1V = - 4.8Nk{g^{ - 1}} \times {10^7}m = - 4.8 \times {10^7}Nm.k{g^{ - 1}}
Since the unit Nm(newton−meter)Nm\left( {newton - meter} \right) is equivalent to the unit of energy, that is, J(joules)J\left( {joules} \right) , we can also write the final answer as:
V=−4.8×107Jkg−1V = - 4.8 \times {10^7}Jk{g^{ - 1}}
Therefore, the correct answer is option (A).

Note
It is nice to think why the gravitational potential is taken negative, and there are many ways to explain this. If we discuss it in a simple way, the gravitational potential is assumed to be zero at the infinite distance from the centre of the earth. So, when the distance between the body and the earth is infinity, then the potential is zero. Based on that reference, when we try to bring a body from infinity to earth’s gravitational field, the potential drops from zero to a negative quantity. When it comes closer and closer to the earth, the potential becomes more and more negative.