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The gravitational potential energy of a mass at a point in a field is defined as the work done by an external agent in bringing that mass from infinity to that point, without a change in kinetic energy.

So, assume that we are applying a force away from the body's gravitational field (to keep acceleration or KE zero) and the displacement is towards the body. The potential energy at infinity, $U_\infty = 0$.

$dw = F \cdot dr$

$\int_{W_\infty}^{W_r} dw = \int_\infty^r F\,dx\, cos\theta$

$\int_{W_\infty}^{W_r} dw = \int_\infty^r F\,dx\, cos180^\circ$

$\int_{W_\infty}^{W_r} dw = -\int_\infty^r \frac{GMm}{x^2}\,dx\qquad$(since, $cos180^\circ = -1$)

$\int_{W_\infty}^{W_r} dw = -GMm\, [\frac{-1}{x}]_\infty^r$

$\int_{W_\infty}^{W_r} dw = -GMm\, [\frac{-1}{r} + \frac{1}{\infty}]$

$U_r - U_\infty = \frac{GMm}{r} \qquad$ (since the work done by external force = potential energy and $\frac{1}{\infty} \rightarrow0$)

$U_r = \frac{GMm}{r}$

But the correct expression for it is $U_r = \frac{-GMm}{r}$

Why is this wrong?

I know this can also be derived by bringing the mass from $r$ to $\infty$, but why this doesn't work?

Qmechanic
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1 Answers1

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It comes down to the definition of potential energy. Remember that potential energy is thought of as energy that is "available" to do work: if you have already expended that energy (bringing a particle from point A to point B under the influence of the external field) it is no longer available, and you need to subtract it from the (arbitrary) reference energy (e.g. 0) that you started with.

TLDR
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