Total energy in orbit

In the example of the previous section, we found how much work it took to raise an object to a certain altitude. To put a satellite in orbit, we can’t follow that same method because the satellite would just fall down. For a satellite to remain in (circular) orbit, we need to have Fg=Fc\displaystyle F_{\text{g}} = F_{\text{c}}. The satellite will have a certain amount of total energy at all times:

Etot=Eg+Ek\displaystyle E_{\text{tot}} = E_{\text{g}} + E_{\text{k}}.

For a stable orbit, we must have

Etot=12Gm1m2r\displaystyle E_{\text{tot}} = - \frac{1}{2} G \frac{m_{1} m_{2}}{r}.


What amount of work does it take to put a 745 kg satellite into orbit 1108 km above the surface of the Earth?

The satellite begins on Earth, so r1=rE\displaystyle r_{1} = r_{\text{E}}. When it reaches the altitude of 1108 km, it will be at a radius of

r2=rE+1108km=7.488×106m\displaystyle r_{2} = r_{\text{E}} + 1108 \, \text{km} = 7.488 \times 10^{6} \, \text{m}.

Now since W=ΔEtot\displaystyle W = \Delta{} E_{\text{tot}}, we should find the change in energy. For the initial value we simply use Eg\displaystyle E_{\text{g}} because the satellite begins at rest, but for the final value we use the special formula for stable orbit energy:

ΔEtot=ΔEtot,2ΔEtot,1=(12Gm1m2r2)(Gm1m2r1)\displaystyle \Delta{} E_{\text{tot}} = \Delta{} E_{\text{tot,2}} - \Delta{} E_{\text{tot,1}} = \left ( - \frac{1}{2} G \frac{m_{1} m_{2}}{r_{2}} \right ) - \left ( - G \frac{m_{1} m_{2}}{r_{1}} \right ).

Substituting all the known quantities and evaluating gives us the answer: it would take 2.68 × 1010 J of work to put the satellite in orbit.