In the section CIRCULAR ORBITS we learnt that circular orbits are only possible for particular speeds that can be calculated using the formula:
$$\begin{aligned} \mathrm{v = \sqrt\frac {GM}{r}} \end{aligned}$$
Where M is the larger mass of the system (which can be the Sun in planet orbit calculation or the Earth in a satellite calculation) and m is the smaller mass (e.g. planet orbiting the Sun or satellite).
Based on that we can write the expression for Ek (kinetic energy) of a circular orbit:
$$\begin{aligned} \mathrm{E_k = \frac {mv^2}{2}= \frac {GmM}{2r}} \end{aligned}$$
Ep is given by:
$$\begin{aligned} \mathrm{E_p = - \frac {GmM}{r}} \end{aligned}$$
So the total energy of the circular orbit is :
$$\begin{aligned} \mathrm{E_T= \frac {GmM}{2r} - \frac {GmM}{r}= - \frac {GmM}{2r} = E_p / 2} \end{aligned}$$
Negative Et means that it is a bound system, as discussed in the potential energy section.
Example calculation
A satellite of mass orbits a planet of mass in a circular orbit of radius r. How much energy is needed to move it to an orbit with radius 2r? What about radius 4r?
Answer:
The total energy of a circular orbit of radius r (as shown above) is:
$$\begin{aligned}
\mathrm{E_T= - \frac {GmM}{2r} }
\end{aligned}$$
If we change r by 2r it becomes:
$$\begin{aligned} \mathrm{E_{T2}= - \frac {GmM}{4r} } \end{aligned}$$
The energy difference between the 2 is
$$\begin{aligned} \mathrm{\Delta E = - \frac {GmM}{2r} (\frac {1}{2}- 1) = \frac {GmM}{4r} } \end{aligned}$$
In the case of the orbit of radius 4r the calculation is:
$$\begin{aligned} \mathrm{\Delta E = - \frac {GmM}{2r} (\frac {1}{4}- 1) = \frac {3GmM}{8r} } \end{aligned}$$