# What is the reason for work done by gravitational force being negative?

When a body is moved from $$r=∞$$ to say, $$r=R$$, work done by the gravitational force is $$-(GMm)/R$$. Why is it negative even though the Force and the displacement of the body is in the same direction. I understand the potential energy aspect but I don't get it if solely looked at the work done aspect.

PS I have seen other similar questions but they deal with the $$mgh$$ formula not the original one.

• FWIW, the work by gravity in your example is positive: Force and displacement are in the same direction as you say. Commented Jan 15, 2020 at 9:48

The reason here is a calculation error. The formula for work done is

$$W_\gamma = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$

in moving along the path $$\gamma$$ under the influence of the force $$\mathbf{F}$$, here gravity, $$\mathbf{F}_G$$. What you need to take account of though is the dot product, and the nature of the differential element: $$d\mathbf{r}$$ is a differential vector, and we have a vector dot product. This means that the work depends on the direction of both the force and the object's motion.

In general, work will be positive when the particle is moving with the force, and negative when the particle is moving against the force. Positive work means the force is contributing energy to the particle, negative means it is stealing energy from the particle (which means there must be some other source - which could be the particle's kinetic energy, or it could also be another source of force, like a rocket attached to it burning some fuel to leave the gravity well). So if you're falling into the gravity well, as you are suggesting, it should be positive, and indeed it is: the work in moving from $$r = \infty$$ to $$r = R$$ is

$$W_\gamma = \frac{GMm}{R}$$

for a gravitator of mass $$M$$ and craft of mass $$m$$ coming in.

Your formula, I believe, was that for the gravitational potential energy, which is related, but not same. The potential energy is

$$U_\mathrm{relative\ to\ \infty} = -\frac{GMm}{R}$$

and the key is in the subscript. The potential energy represents stored energy, and as the object is drawn into the gravity well, it should decrease, hence the negative sign. The relationship is that potential energy is converted into positive work, and negative work is converted to increased potential energy.

The reason is that the work done by a conservative force is $$W = -\Delta U$$ and since $$W \gt 0$$ therefore $$\Delta U \lt 0$$. Also $$\Delta U = U_f-U_i$$

If a body falls radially under gravity from $$r=R_1$$ to $$r=R_2 < R_1$$ then the work done by gravity is indeed positive, and is equal to $$W_g = GMm(\frac 1 {R_2} - \frac 1 {R_1})$$.

To account for the source of this energy, we ascribe a scalar potential $$U(r)$$ to the gravitational field such that $$W_g = U(R_1) - U(R_2)$$. $$U(r)$$ is defined up to a constant, and by convention we take $$U(\infty) = 0$$, which leaves $$U(r) = -\frac{GMm}{r}$$.

For the body to get back to $$r=∞$$ it would need to make up that delta. As such in going to $$r=R$$ it has lost potential energy, hence negative.

A good example might be to imagine a stationary body at the top of a cliff compared to the same stationary body at the bottom - the body at the bottom has a negative energy delta by being closer to the centre of gravity.

The work done by the gravitational force is not negative. It is positive.

It's clear to see that from the definition of work, as you say. And it's also clear to see by the fact that an object increases speed and kinetic energy when subjected to gravity. If the kinetic energy is increasing, the work done on it is positive.