More gravitational potential energy

Near the surface of the Earth we can use $\mathrm{\Delta }{E}_{\text{g}}=mg\mathrm{\Delta }h\text{,}$ 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 $E_{\text{g}}$ itself, not just $\mathrm{\Delta }{E}_{\text{g}}\text{.}$ To do this, we will use $r=\infty$ as a reference point. At an infinite radius, we decide that $E_{\text{g}}=0\text{.}$ 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, $E_{\text{g}}$ must always be negative: it starts out at zero when $r=\infty$ and only goes down from there.

The general equation for gravitation potential energy is

$E_{\text{g}}=-G\frac{m_1{m}_2}{r}\text{.}$

Example

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

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

$r_1=r_{\text{E}}$ and $r_2=r_{\text{E}}+\text{altitude}\text{.}$

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

$\mathrm{\Delta }{E}_{\text{g}}=G{m}_{\text{E}}{m}_{\text{obj}}(\frac{1}{r_1}-\frac{1}{r_2})\text{.}$

After plugging in the values and evaluating, we find that the amount of work required is 4.21 × 10⁸ J.