I'm trying to understand work obtained in the context of gravitational potential energy.

It's stated as the **work obtained** when an object is moved from infinity to a point

$$U = -GMm/r.$$

# [Physics] What’s the difference between work done and work obtained?

conventionsnewtonian-gravitynewtonian-mechanicspotential energywork

## Best Answer

The underlying intuition is that when you drop an object from a given height h, the object will accelerate towards the ground thus converting the potential energy into kinetic energy. During this process we have "done nothing", the object is accelerated without our doing thus we obtain work and reduce the potential of the object. If we lifted the object a certain height we would need to invest energy and thus do work.

A more specific example: We choose our potential to be such that it vanishes for $h = 0\; m$. For convenience $g = 10\; \frac{m}{s^2}$. If we have a ball of mass $m = 2\;kg$ at an initial height of $h = 10 \;m$ the potential energy of the ball is given by $E_{pot}= m\cdot g \cdot h = 2 \cdot 10 \cdot 10 \;[J] = 200 \;[J]$. When the ball drops it does this on its own and converts the potential energy. If we would like to reverse this process we would need to invest energy (meaning we have to do work). (Note that this relies on the approximation that the (gravitational-)force exerted on the mass is constant and thus only valid for small enough h)

A more general derivation: The potential energy for gravity is $$U(r) =m V(r)$$ where $m$ is the mass of the object and $V$ the gravitational potential. For a unit test mass $m_t =1$ we have that $$V(r) = \frac{W}{m_t} = \frac{1}{m_t} \int_{\infty}^r \frac{Gm_tM}{\bar{r}^2}d\bar{r} = -\frac{1}{m_t}\frac{Gm_tM}{r} = -\frac{GM}{r} $$ Using the formula for the potential energy we obtain that $$U = - \frac{GMm}{r}$$ just as the formula desired. (The example above should give you and intuition for the minus. Throughout the whole problem spherical symmetry was assumed.)