gravitational potential energy

Gravitational potential energy

Gravitational potential energy is the potential energy an object has due to its position in a gravitational field. It is the mechanical work done by the gravitational force to bring the mass from a chosen reference point (often an "infinite distance" from the mass generating the field) to some other point in the field, which is equal to the change in the kinetic energies of the objects as they fall towards each other. Gravitational potential energy increases when two objects are brought further apart and is converted to kinetic energy as they are allowed to fall towards each other.

Definition

For two pairwise interacting point particles, the gravitational potential energy \[U\] is the work that an outside agent must do in order to bring the masses together, which is therefore, exactly opposite the work done by the gravitational field on the masses. Mathematically, \[\Delta U=+W_{\text{external}}=-W_{\text{field}}=-\int \mathbf{F}\cdot d\mathbf{s}\] (see conservative forces), where \[\mathbf{F}\] is the gravitational force and \[d\mathbf{s}\] is the displacement vector.

Derivation

Based on Newton's law of universal gravitation, gravitational force, \[\mathbf{F}=-G\frac{Mm}{r^{2}}\hat{\mathbf{r}}\] where \[r=\left\lVert \hat{\mathbf{r}} \right\rVert\]. Let \[d\mathbf{s}=dr\,\hat{\mathbf{r}}\]. Now we calculate the total work done by the gravitational force in bringing point mass \[m\] from \[\infty\] to a final distance \[r\] from point mass \[M\], i.e. \[W=\int \mathbf{F}\cdot d\mathbf{s}\].

First, we expand the dot product,

\begin{align*} \mathbf{F}\cdot d\mathbf{s}&=\left( -G \frac{Mm}{r^{2}}\hat{\mathbf{r}} \right)\cdot \left( dr\, \hat{\mathbf{r}} \right)\\ &=-G \frac{Mm}{r^{2}}\,dr\times \left( \hat{\mathbf{r}}\cdot \hat{\mathbf{r}} \right)\\ &=-G \frac{Mm}{r^{2}}\,dr\\ \end{align*}

then integrate from \[\infty\] to \[r\],

\begin{align*} \int^{r}_{\infty}\mathbf{F}\cdot d\mathbf{s}&=\int_{\infty}^{r}-G \frac{Mm}{\left( r^{\prime} \right)^{2}}\,dr^{\prime}\\ &=\biggl[ \frac{GMm}{r^{\prime}} \biggr]_{\infty}^{r}\\ &=\frac{GMm}{r}\\ \end{align*}

We used \[r^{\prime}\] (a dummy variable) to avoid confusion with the variable \[r\].

Now, combine this with the definition of gravitational potential energy, \[\Delta U=-\frac{GMm}{r}\].

Simplified derivation for calculations on the Earth's surface

A more commonly used version will be \[\Delta U=mgh\].

Based on our previous derivation, \[\Delta U=\frac{GMm}{r}\]. We can then find the change in potential energy moving from the surface to a height \[h\] above the surface with,

\begin{align*} \Delta U&=\frac{GMm}{r}-\frac{GMm}{r+h}\\ &=\frac{GMm}{r}\left( 1-\frac{1}{1+\frac{h}{r}} \right)\\ &\approx\frac{GMmr}{r}\left( 1- \left( 1-\frac{h}{R} \right) \right)\quad\text{by binomial approximation}\\ &\approx \frac{GMmh}{r^{2}}\\ &\approx m \frac{GM}{r^{2}}h\\ \end{align*}

Then substituting the gravitational field, \[g=\frac{GM}{r^{2}}\] we get \[\Delta U\approx mgh\].

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