Gravitational potential energy

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Gravitational potential energy, commonly noted $U$, refers to the energy incorporated in a system as a result of the gravitational attraction between its elements.

Zero reference point

Generally, we define the zero reference point:

U(r=\infty)=0

Since there is no gravitational interaction when objects are at an infinite distance from each other.

Then, the gravitational potential energy of a system measures the amount of work done by gravitational forces to move all masses of the system away from each other until the distance between them is infinite.

Basic case: 2 point masses

For two point masses $m$, $M$, we can calculate the gravitational potential energy of the system mostly easily by constructing a cartesian coordinate system with one of the point masses, $M$, at the origin.

Then, if $m$ has position vector $ \mathbf r_1$:

$$\mathbf F_g(m)=-G\frac{Mm}{r_1^2}\hat{\mathbf r}_1$$

The potential energy of the system is:

$$U=-\int_{\infty}^{r_1}\left(-G\frac{Mm}{r^2}\right) \hat{\mathbf r} \cdot dr=-G\frac{Mm}{r}$$

$$\hat{\mathbf r} \cdot dr = dr$$

Many point masses

Recall the gravitational potential energy between 2 point masses is

U=-G\frac{Mm}{r}

Applying the formula to all pairs of point masses in the system:

U=-G\sum_{i<j}\frac{m_i m_j}{\mathbf{r_i}-\mathbf{r_j}}

We can similarly derive a formula for the gravitational potential energy of a point mass external to the system:

U(\mathbf{r})=-Gm\sum_i \frac{M_i}{\mathbf{r}-\mathbf{r}_i}

Continuous mass distribution

Any generic mass system can be modelled by a continuous mass distribution, with $\rho=0$ where there is no mass.

If the density at point with position vector $\mathbf{r}$ is $\rho(\mathbf{r})$:

U=-\frac12 G\int\int\frac{\rho(\mathbf{r})\rho(\mathbf{r}_1)}{|\mathbf{r}-\mathbf{r}_1|}d^3r\,d^3r_1

$dV=dr^3$: infinitesimal volume

Gravitational potential

If a test mass $m$ has gravitational potential energy $U(\mathbf{r})$ at a point, then the gravitational potential at that point is

$$\Phi(\mathbf{r})=\frac{U(\mathbf{r})}{m}$$

For another test mass $m_1$, the gravitational potential energy at the same point is then simply:

U(\mathbf{r})=\Phi(\mathbf{r})m_1