Near the surface of the Earth we can use `Delta E_"g" = mgDelta 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 `E_"g"` itself, not just `Delta E_"g"`. To do this, we will use `r=oo` as a reference point. At an infinite radius, we decide that `E_"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, `E_"g"` must always be negative: it starts out at zero when `r=oo` and only goes down from there.

The general equation for gravitation potential energy is

`E_"g" = -G(m_1m_2)/r`.

Always remember that gravitational potential energy is divided by `r`, not by `r^2` like gravitational force is. It’s easy to confuse the two equations.

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

Since `W = Delta E_"g"`, we want to find the change in gravitational potential. The changing quantity is radius:

`r_1 = r_"E" qquad and qquad r_2 = r_"E" + "altitude"`.

Now, we substitute our new equation into `Delta E_"g" = E_"g,2" - E_"g,1"`, giving us

`Delta E_"g" = Gm_"E"m_"obj"(1/r_1 - 1/r_2)`.

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