# More gravitational potential energy

Near the surface of the Earth we can use
$\displaystyle \Delta{} E_{\text{g}} = m g{\Delta{}} h$,
but this equation breaks down in situations where the radius of the gravitational attraction is not constant. At this point we are going to want to talk about
$\displaystyle E_{\text{g}}$
itself, not just
$\displaystyle \Delta{} E_{\text{g}}$.
To do this, we will use $\displaystyle r = \infty$ as a reference point. At an infinite radius, we decide that
$\displaystyle E_{\text{g}} = 0$.
Why, you ask? Because we need *some* reference point, and we use this one by convention.

When radius decreases, the object cannot do as much work during its descent, so gravitation potential decreases. Therefore, $\displaystyle E_{\text{g}}$ must always be negative: it starts out at zero when $\displaystyle r = \infty$ and only goes down from there.

The general equation for gravitation potential energy is

$\displaystyle E_{\text{g}} = - G \frac{m_{1} m_{2}}{r}$.

## Example

How much work is needed to lift a 101 kg object to an altitude of 456 km?

Since $\displaystyle W = \Delta{} E_{\text{g}}$, we want to find the change in gravitational potential. The changing quantity is radius:

$\displaystyle r_{1} = r_{\text{E}} \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad r_{2} = r_{\text{E}} + \text{altitude}$.

Now, we substitute our new equation into $\displaystyle \Delta{} E_{\text{g}} = E_{\text{g,2}} - E_{\text{g,1}}$, giving us

$\displaystyle \Delta{} E_{\text{g}} = G m_{\text{E}} m_{\text{obj}} \left ( \frac{1}{r_{1}} - \frac{1}{r_{2}} \right )$.

After plugging in the values and evaluating, we find that the amount of work required is 4.21 × 10^{8} J.