# Actual meaning of “Gravitational Potential”?

In a gravitational field, the gravitational force acting on a body (of mass m) at a point x metres away from the attracting body (of mass M) is $\frac{GMm}{x^2}$. Integrating this force from a point at infinity to x gives $-\frac{GMm}{x}$, or the work done by the gravitational field on the mass in moving the mass from a point at infinity to x. Then, the gravitational potential at point x is $-\frac{GM}{x}$.

However, the definition that my school gives is : "Gravitational potential at a point in a gravitational field is the amount of work done by an external agent in moving one unit of mass from a point at infinity to the point in the gravitational field, without any acceleration". The lecturers explained that as a body moves closer to the attracting body, it accelerates and gains kinetic energy. In order to maintain a constant velocity, negative work must be done by an external agent to remove this extra kinetic energy.

Now I am very confused, because both seem valid to me. What does "gravitational potential" actually mean?

Both definitions are fine as long as you're careful with signs.

1. Here's a derivation of your gravitational potential energy using the idea of the negative work done "by the field." (Note the initial negative sign.)

$$U(r)=-W_\text{by field}=-\int\vec{F}_\text{field}\cdot d\vec{s}=-\int_{r=\infty}^{r=x}\underbrace{\frac{-GMm\,\hat{r}}{r^2}}_\text{Toward origin}\cdot d\vec{s}=\int_{r=\infty}^{r=x}\frac{GMm}{r^2}dr=-\frac{GMm}{x}$$

The negative inside the integral indicates that the force by the field is toward the origin, in the $-\hat{r}$ direction. Now, divide by $m$ and you have your expression.

2. Here's a derivation using the work done by the external agent.

If the object doesn't accelerate, the net force on the object is zero. Thus, the force exerted by the external agent is equal in magnitude and opposite in direction to that by the field. We indicate this by explicitly writing that this force is away from the origin:

$$U(r)=+W_\text{ext agent}=+\int\vec{F}_\text{ext}\cdot d\vec{s}=+\int_{r=\infty}^{r=x}\underbrace{\frac{+GMm\,\hat{r}}{r^2}}_\text{Away from orig.}\cdot d\vec{s}$$

We can stop there since we already have the same expression as our first derivation.

You don't do any negative work.

All that gravitational potential says is that an object at a higher altitude (say, a ball in your hand) is at a relatively higher potential than the same object when it's at the Earth's surface. Mind you, I meant "relatively" higher potential, because it's still negative - gravitational potential is negative everywhere, except an infinite distance).

The work done is the difference between the two, which is always positive. And, it's consistent with our principles, that any system at rest, is at its lowest possible energy state. So, you take the ball in your hand, you're doing some work. When you drop it, the ball falls down (i.e) gets transferred to the lowest energy state (which is ground)...

In other words, the object is in a bound state, when it's at a lower potential. To get the object out, you do a positive work, which is the kinetic energy given to the object. The "infinity" comes into the expression because potential energy is always chosen relative to a zero reference position. In case of gravity, zero turns out to be the highest potential energy state (i.e) an infinite distance. For such a case, we introduce the negative sign so that when an object is taken to a farther distance from a massive body, we can still think that it's at a higher potential energy.

I hope this explains you everything about P.E= $-\frac{GMm}{r}$ !