# 4.7 Gravitational potential  (Page 2/3)

 Page 2 / 3

$\text{Δ}U=-\underset{{r}_{1}}{\overset{{r}_{2}}{\int }}{F}_{G}dr$

Clearly, change in gravitational potential is equal to the negative of work by gravitational force as a particle of unit mass is brought from one point to another in a gravitational field. Mathematically, :

$⇒\text{Δ}V=\frac{\text{Δ}U}{m}=-\underset{{r}_{1}}{\overset{{r}_{2}}{\int }}Edr$

We can easily determine change in potential as a particle is moved from one point to another in a gravitational field. In order to find the change in potential difference in a gravitational field due to a point mass, we consider a point mass “M”, situated at the origin of reference. Considering motion in the reference direction of “r”, the change in potential between two points at a distance “r” and “r+dr” is :

$⇒\text{Δ}V=-\underset{{r}_{1}}{\overset{{r}_{2}}{\int }}\frac{GMdr}{{r}^{2}}$

$⇒\text{Δ}V=-GM\left[-\frac{1}{r}\underset{{r}_{1}}{\overset{{r}_{2}}{\right]}}$

$⇒\text{Δ}V=GM\left[\frac{1}{{r}_{1}}-\frac{1}{{r}_{2}}\right]$

In the expression, the ratio “ $\frac{1}{{r}_{1}}$ ” is smaller than “ $\frac{1}{{r}_{2}}$ ”. Hence, change in gravitational potential is positive as we move from a point closer to the mass responsible for gravitational field to a point away from it.

## Example

Problem 1: A particle of mass 2 kg is brought from one point to another. The increase in kinetic energy of the mass is 4 J, whereas work done by the external force is -10 J. Find potential difference between two points.

Solution : So far we have considered work by external force as equal to change in potential energy. However, if we recall, then this interpretation of work is restricted to the condition that work is done slowly in such a manner that no kinetic energy is imparted to the particle. Here, this is not the case. In general, we know from the conservation of mechanical energy that work by external force is equal to change in mechanical energy:

${W}_{F}=\Delta {E}_{mech}=\Delta K+\Delta U$

Putting values,

$⇒-10=4+\Delta U$

$⇒\Delta U=-10-4=-14\phantom{\rule{1em}{0ex}}J$

As the change in potential energy is negative, it means that final potential energy is less than initial potential energy. It means that final potential energy is more negative than the initial.

Potential change is equal to potential energy change per unit mass. The change in potential energy per unit mass i.e. change in potential is :

$⇒\Delta V=\frac{\Delta U}{m}=-\frac{14}{2}=-7\phantom{\rule{1em}{0ex}}J$

## Absolute gravitational potential in a field due to point mass

The expression for change in gravitational potential is used to find the expression for the potential at a point by putting suitable values. When,

${V}_{1}=0$

${V}_{2}=V\left(\mathrm{say}\right)$

${r}_{1}=\infty$

${r}_{2}=r\left(\mathrm{say}\right)$

$⇒v=-\frac{GM}{r}$

This is the expression for determining potential at a point in the gravitational field of a particle of mass “M”. We see here that gravitational potential is a negative quantity. As we move away from the particle, 1/r becomes a smaller fraction. Therefore, gravitational potential increases being a smaller negative quantity. The magnitude of potential, however, becomes smaller. The maximum value of potential is zero for r = ∞.

This relation has an important deduction. We know that particle of unit mass will move towards the particle responsible for the gravitational field, if no other force exists. This fact underlies the natural tendency of a particle to move from a higher gravitational potential (less negative) to lower gravitational potential (more negative). This deduction, though interpreted in the present context, is not specific to gravitational field, but is a general characteristic of all force fields. This aspect is more emphasized in the electromagnetic field.

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