More gravitational potential energy

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

The general equation for gravitation potential energy is

Eg=Gm1m2r\displaystyle E_{\text{g}} = - G \frac{m_{1} m_{2}}{r}.


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

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

r1=rEandr2=rE+altitude\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 ΔEg=Eg,2Eg,1\displaystyle \Delta{} E_{\text{g}} = E_{\text{g,2}} - E_{\text{g,1}}, giving us

ΔEg=GmEmobj(1r11r2)\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 × 108 J.