# Measuring work done by gravity over non-constant gravitational acceleration

A question from an example from a MIT Classical Mechanics Lecture on Work.

Here's the given definition for gravitational potential energy (~32:00): "The gravitational potential energy at a point $P$ is the work that I, Walter Lewin, [the lecturer] have to do to bring that mass from $\infty$ to that point $P$. My force is always the same as gravity with a minus sign. "

This diagram is given at ~40:00:

It is unclear to me why W. Lewin's force $F_{WL}$ is pointing towards $\infty$ (+ $r$ direction) if he is measuring the force to bring the object from $\infty \to P$.

Furthermore, the work required by WL is given as $\int_{\infty}^{R} \frac{mMG}{r^2} dr$. Since $\infty \to$r is the $-r$ direction, shouldn't this integral have a minus sign?

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The integral itself doesn't need the minus sign -- if you integrate from $a$ to $b$, that is the same as integrating from $b$ to $a$ with a $-$ sign. –  tpg2114 Jan 13 '13 at 6:06

The force is pointing in the $+r$ direction because it cancels out the force of gravity. So if the object is not accelerating, the force applied has to be the negative of the force of gravity.

He is not measuring "The Force" to bring an object from infinity to P, because there is no singular force. There's an amount of work done, yes, but it doesn't make much sense to phrase it as "the force to move an object from A to B".

The integral does have a negative value. Lets evaluate it.

$$\int^{R}_{\infty} \frac{1}{r^2} dr=\left.-\frac{1}{r} \right|_{r=\infty}^{r=R} =-\frac{1}{R}-(-\frac{1}{\infty})=-\frac{1}{R}$$ (with a factor of $GmM$ tacked on)

So, as you can see, the integral evaluated to a negative value.

Maybe you're having a problem with the definition/workings of definite integrals? For example, one might ask, "If we're summing up an infinite number of infinitesimal quantities $\frac{GMm}{r^2}dr$ which are all positive, how do we end up with a negative value?" The answer can be viewed as saying $dr$ takes into account the direction which you integrate in. So: it's already handled for you.

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Gravitation would be driving the mass from $\infty$ to your point $P$, so according to the definition you gave, no work needs to be done by Mr. Lewin to accomplish that. It's actually more like "the work that I have to do to bring that mass from $\infty$ to that point $P$, and have it stay there", with the highlighted part being equivalent to "moving at a constant, infinitesimally small speed". So your negative work is actually done in preventing the object from accelerating towards the point it is being pulled to.

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