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As I know gravitational potential energy of an object relative to $U=0$ point is defined as $U(h)=mgh$ and this came from work-energy theorem.

But In my book, there is another definition for it.

Consider that we have 2 objects that can interract with each other by gravitational force. Set the first object steady. Now, there is $\vec{r_{0}}$ distance between these two objects. Move the second object $d\vec{r}$ in direction $\vec{r_{0}}$/$\hat{r_{0}}$. Let's call $\vec{r}$ to new distance.

Work done by us (agent),

$$\vec{F}_{g}=-G\frac{M_{1}M_{2}}{r^2}\hat{r}$$

$$\vec{F}_{ag}=-\vec{F}_{g}=G\frac{M_{1}M_{2}}{r^2}\hat{r}$$

$$W(\text{done by us})=W_{ag}=\int_{r_{0}}^{r}\vec{F}_{ag}.d\vec{r}$$

$$W_{ag}=\int_{r_{0}}^{r}{G\frac{M_{1}M_{2}}{r^2}\hat{r}.dr\hat{r}}$$

$$W_{ag}=GM_{1}M_{2}\int_{r_{0}}^{r}{\frac{1}{r^2}dr}$$

$$W_{ag}=GM_{1}M_{2}(-\frac{1}{r}+\frac{1}{r_{0}})$$

$$W_{ag}=-G\frac{M_{1}M_{2}}{r}+G\frac{M_{1}M_{2}}{r_{0}}$$

$$r_{0}\to\infty\Rightarrow G\frac{M_{1}M_{2}}{r_{0}}=0$$

$$U(r)=-G\frac{M_{1}M_{2}}{r}$$

This is it. What is the difference?

Qmechanic
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M. Çağlar TUFAN
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1 Answers1

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$U = mgh$ is valid when the gravitational field is uniform; $U = -\tfrac{GMm}{r}$ is valid for point masses.

If one object moves only a small distance relative to the total distance between two masses, we can Taylor expand the point mass potential:

$$\Delta U \approx \frac{G M m}{r^2} \Delta r = mg\Delta r$$

where $g = \tfrac{GM}{r^2}$. This is why the former expression is valid at, for example, the surface of the earth over human-scale distances.

DavidH
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