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Question: The gravitational potential due to a solid sphere A. Increases as we move away from the surface of...

The gravitational potential due to a solid sphere
A. Increases as we move away from the surface of the sphere
B. Reduces as we move away from the surface of the sphere
C. Remain constant as we move away from the surface of the sphere
D. Becomes negative as we move away from the surface of the sphere

Explanation

Solution

Hint: First write down the definition of the gravitational potential. Then after finding an expression for this potential, we will see the effect of distance on it. Therefore, just looking at the mathematical expression we can answer this question.

Formula used:
V=−GMrV=-\dfrac{GM}{r}

Complete step-by-step solution:
We should know that; gravitational potential of a point is defined as the work done in bringing a unit
mass from infinity to the point of interest. So basically, gravitational potential has same dimension as
energy or work done.
Now, let the mass of the sphere be M. So, the force it will exert on a unit mass is given as,
F=GMr2F=\dfrac{GM}{r^2}
Here ‘R’ is the distance of the unit mass from the center of the sphere. This is obtained by
Newton’s law of gravitation, which states that the attraction force between two objects is
proportional to the product of their mass and inversely proportional to the square of their
distance.
Now the work done to bring this unit must from infinity to a distance R is given by,
V=∫∞RGMr2.dr=−GMRV=\int\limits_{\infty}^{R}\dfrac{GM}{r^2}.dr=-\dfrac{GM}{R}
This is the expression for the gravitational potential at a distance R from the center of the
sphere. So now if we move away from the surface of the sphere, the value of R increases.
As a result, the value of the fraction GMR\dfrac{GM}{R} decreases. But due to the negative sign in
the expression of V, gravitational potential increases as the value of R increases.
Hence option A is the correct answer.
Additional information:
The expression for the acceleration due to gravity is given by,
g=GMR2g=\dfrac{GM}{R^2}
Here, R is the radius of Earth. The value of G is 6.674×10−11Nm2/kg26.674\times 10^{-11} Nm^2/kg^2.

Note: Make sure you put the negative sign in the expression for gravitational potential.
Gravitational potential is always negative and increases with distance. At infinity, its value
reaches to zero. It can’t be positive. Students might forget the negative sign and get option B as the correct answer. But this is wrong.