This week the science blogosphere has been spending a lot of time on space exploration and the science of understanding the universe outside our own planet. At risk of being the buzzkill distracting from all the cool space travel history and heated debates about NASA’s future, I think it will be interesting to do a little bit of the math behind basic rocketry. As a bonus, we’ll do it in the Lagrangian formulation of classical mechanics. (And maybe to draw in a few Google hits from lost students, I’ll note that this is pretty much Goldstein 1.13) Non-mathy readers don’t have to worry about the math, it’s certainly possible to get the general idea without worrying about the details of how it works.
The advantage of the Lagrangian formulation is that it avoids having to deal with vectors and forces directly. We really won’t benefit from this here because we’ll launch our mathematical rocket straight up, in 1-d motion. Still, good practice. The quantity that’s actually called the Lagrangian is the difference between the kinetic and potential energy, so we’d better write those down first. First the kinetic energy:
Then the potential:
The Lagrangian L = T – U. A few quick things to note: first, the mass m is not constant. It’s a rocket, and mass is constantly being blasted out the back end. Second, I’m calling the vertical position x because why not? Now that we have the Lagrangian, we need Lagrange’s equation to plug it into. His equation looks like this for our case:
Normally there would be a 0 in the place of that Q. That’s in the case where all the forces can be expressed in terms of the potential. But that’s rather difficult here because the thrust of the exhaust is not really something where a potential can easily be written down. Instead we’re using the generalized force Q, which in this case happens to be just the thrust itself. Via Newton’s second law, the thrust is just the rate of change of the momentum of the rocket. And that’s just the velocity of the rocket exhaust times the burn rate. Call that velocity v, and remember that v is not the velocity of the rocket. We’re representing that with x-dot.
Do the partial derivatives and the above equation becomes this:
Do the total derivative and simplify:
This is pretty bad for one main reason: v is an annoyingly involved quantity. It’s the velocity of the rocket exhaust, with respect to the ground. What we want is the nice constant speed of the exhaust with respect to the rocket. Call that v’. The relationship between them is simple. Exhaust velocity w/r/t the ground is the negative of the exhaust speed w/r/t the rocket (negative because it’s pointing down) plus the rocket w/r/t the ground. Substitute into the above and switch dm/dt into dot notation:
And simplify, canceling the m-dot x-dot products:
Ok, time for a brief break to let everyone catch their breaths. Do you think I’ll get in trouble for posting cheesecake? Probably, but I don’t care!
All right. The last equation we just derived is a differential equation. It’s easy to solve if you’ve enough math experience to follow all the previous stuff, so I won’t bore you. Also this is bad form but COMMAND DECISION: we’re declaring the previous definition of v to be dead and buried; henceforth v will be the velocity of the rocket. The solution for the velocity is:
The final velocity will clearly occur at the time the rocket runs out of fuel and reaches its final mass. With that in mind,
So finally we need to write the time the rocket motor burns, assuming a constant burn rate. We can then stuff that back into the previous equation. The time is just the mass burned over the burn rate:
Which after all that gives us the final rocket velocity:
Escape velocity is about 11.7 km/s. For a rocket with exhaust speed of 4 km/s and a burn time of 1000 s, you need more than 200 kilograms of fuel for each kilogram of payload. Which is why it takes one of these:
To get one of these to the moon:
Whew! I think that’s enough. It’s not called rocket science for no reason.