Satellites in Orbit found the speed of a circular orbit, and Gravitational Potential Energy and Escape Speed gave the potential energy $U = -GMm/r$. Put them together and orbital energy has a surprise in it. Explore it in the Orbit Lab.
Why it works
In a circular orbit gravity supplies the centripetal force: $\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$, so $mv^2 = \dfrac{GMm}{r}$ and $$K = \tfrac{1}{2}mv^2 = \dfrac{GMm}{2r} = -\tfrac{1}{2}U.$$ Adding, $E = K + U = -\dfrac{GMm}{2r}$. A larger orbit has a less negative $E$ (more energy) but a smaller $K$ (less speed): its $U$ rises by twice as much as its $K$ falls.
Key idea
For a circular orbit of radius $r$: $$E = -\dfrac{GMm}{2r}, \qquad K = -E, \qquad U = 2E.$$ Moving to a higher orbit takes energy, $\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$, yet the satellite ends up slower, $v = \sqrt{GM/r}$.
Worked example: A satellite's energy
A $1000\ \text{kg}$ satellite orbits $400\ \text{km}$ above Earth ($GM = 3.986 \times 10^{14}\ \text{m}^3/\text{s}^2$, $R = 6.371 \times 10^{6}\ \text{m}$). Find its total energy and speed.
- $r = 6.771 \times 10^{6}\ \text{m}$.
- $$E = -\dfrac{GMm}{2r} = -\dfrac{(3.986 \times 10^{14})(1000)}{2(6.771 \times 10^{6})} = -2.94 \times 10^{10}\ \text{J} = -29.4\ \text{GJ}$$
- $v = \sqrt{GM/r} = 7.67\ \text{km/s}$.
Answer: $E \approx -29.4\ \text{GJ}$; $v \approx 7.67\ \text{km/s}$.
Worked example: Raising the orbit
How much energy does it take to raise that satellite to an $800\ \text{km}$ circular orbit, and what is its new speed?
- $r_2 = 7.171 \times 10^{6}\ \text{m}$; $$\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) = 1.64 \times 10^{9}\ \text{J} = 1.64\ \text{GJ}$$
- New speed: $\sqrt{GM/r_2} = 7.46\ \text{km/s}$, slower than before.
Answer: $1.64\ \text{GJ}$; $7.46\ \text{km/s}$.
Common mistake
Thinking a satellite in a higher orbit must move faster, because it took more energy to get there. The energy went into climbing out of the gravity well, which raises $U$ a lot; part of it even came out of the kinetic energy. Check with $v = \sqrt{GM/r}$: the Moon, far out, moves about 1 km/s; the space station, close in, about 7.7 km/s.
A strange consequence: air drag on a low satellite takes energy away, so $E$ becomes more negative and the orbit shrinks, and on the smaller orbit the satellite moves faster. Drag speeds it up, because gravity does more positive work during the slow spiral inward than drag does negative work.
The total energy is negative for every bound orbit and zero exactly at escape. For an ellipse, $E = -\dfrac{GMm}{2a}$ with $a$ the semi-major axis; you'll derive that in Speed on an Elliptical Orbit.