# When is it okay or at least helpful to use energy arguments in orbital mechanics?

When faced with energy-based questions about spaceflight we often comment something like orbits are about momentum, not energy and use delta-v math in answers.

But in this answer to How big a nuke would be needed to break Phobos out of orbit? it was thought necessary to consider energy. Comments show a disagreement of a factor of 10 in considering kinetic vs potential energy, but we're still looking into that.

There is the concept of C3, and searching @MarkAdler & C3 finds several answers about it including this one for example:

It can be calculated at any distance from Earth as your specific energy (energy per unit mass), times two:

$$C_3 = v_\infty^2 = v^2-{2\mu\over r}$$

and Elon Musk's famous wrong 3C tweet of 12 km^/2^2 (see this answer)

Question: When is it okay or at least helpful to use energy arguments in orbital mechanics?

If we have an object of mass $$m$$ and energy $$E$$ for example, in what cases can something quantitative be said about how $$E$$ can be used to affect $$m$$'s orbit?

Energy is always conserved, but different oberservers will disagree about how much energy there is, and what forms it takes. Also you have to be sure to include the whole system. Let's return to the Phobos example from the linked question, but be a bit more careful.

Suppose what we actually do is use our teraton nukes to split Phobos into two equal halves of mass roughly $$5\times 10^{15} kg$$ each, and push them apart at a mutual velocity of 1.8 km/s. Now we can analyse these events from the viewpoint of a couple of different observers.

1. An observer in spaceship which starts out close to, and at rest relative to, Phobos. They see zero initial kinetic energy and $$0.5\times 10^{16}\times 900^2 \sim 4\times 10^{21}J$$ of final kinetic energy, which nicely matches the energy released by a 1 teraton explosion.

2. An observer at rest relative to the centre of mass of Mars.They see an initial KE of $$0.5\times 10^{16}\times 2100^2 \sim 2.2\times 10^{22} J$$ from the orbital velocity of Phobos. Afterwards, they see one part escaping at 3 km/s carrying $$2.5\times 10^{15}\times 3000^2 \sim 2.25\times 10^{22} J$$ and the other about to crash into Mars at 1.2km/s carrying $$2.5\times 10^{15}\times 1200^2 \sim 3.2\times 10^{21} J$$. Thus they can see about $$2.6\times 10^{22}J$$ of KE a gain over the starting conditions of about $$4\times 10^{21}$$ again matching the nukes.

So you can use energy calculations, but you need to keep track of the reaction mass and both the initial and final KE of all the various elements. Incidentally, if you wanted to launch Phobos relatively intact using a much smaller amount of reaction mass and a higher exhaust velocity you'd need much more energy.

• I wouldn't want to be observer 1. – Organic Marble Dec 7 '19 at 11:37
• And observer 3, on the surface of Phobos, will disagree pointing out the importance of being in an inertial frame. – asdfex Dec 7 '19 at 17:14
• The largest nuclear explosion we've ever produced was the Tsar bomb which was 50 megatons so just for a reference point more easily grasped how many of those are we talking about here? – Pelinore Dec 7 '19 at 20:10
• 1 teraton is a million megatons, so 20,000 Tsar Bombas. Another way of looking ay it is about 20 tons of antimatter annihilating with 20 tons of matter. – Steve Linton Dec 7 '19 at 20:49
• Might this be applied incrementally (one smaller thrust at a time) to increase orbital velocity over time until it breaks orbit (I presume this would be less likely to cause the whole moon to fracture)? – Pelinore Dec 7 '19 at 21:53