# Spacecraft Slingshot

1. Jan 31, 2012

### Opus_723

1. The problem statement, all variables and given/known data

Spacecrafts can gain in mechanical energy as they encounter a planet. This may appear as a violation of the conservation of mechanical energy, but it is not. The gained energy is at the expense of the orbital energy of the planet. The easiest way to see how this works in principle is to treat the problem as a one-dimensional collision. Lett he spacecraftha ve a mass m and just before the encounter a velocity v, the planet a mass M and velocity V. Both velocities are relative to the sun and they are in opposite directions. Thus the angle between v and V is 180◦. Assume that the spacecraft rounds the planet and departs in the opposite direction. Thus, after the encounter the velocity of the spacecraft is in the same direction as V.

What is the speed of the spacecraft after the encounter in terms of m, M and the speed of the spacecraft before the encounter and the speed of the planet before the encounter?

2. Relevant equations

mv + MV = mv$_{f}$+MV$_{f}$

$\frac{mv^{2}}{2}$ + $\frac{MV^{2}}{2}$ = $\frac{mv^{2}_{f}}{2}$ + $\frac{MV^{2}_{f}}{2}$

3. The attempt at a solution

I'm pretty sure this is an elastic collision (at least, if you assume the objects are infinitely separated before and after the collision, so that they have no potential energy). So I can use conservation of momentum and conservation of kinetic energy. It talks about the two initial velocities being in opposite directions, but I think you can use the equations above and just roll the signs into the variables. Since I can express v$_{f}$ in terms of m, M, and V but not V$_{f}$, I figured I'd solve one equation for V$_{f}$ and plug it into the other.

I used the above kinetic energy equation to solve for V$_{f}$=$\sqrt{\frac{mv^{2}+MV^{2}-mv^{2}_{f}}{M}}$

And then I plug that into the momentum equation.

But when I try to solve that, I end up with a crazy quadratic equation, which seems wrong to me. It seems to me that there should only be one solution. Am I just getting screwed up in the algebra somewhere, or is this the wrong approach? Is there an easier way to do this?

2. Jan 31, 2012

### Staff: Mentor

Another equation to consider using when doing conservation of momentum problems with perfectly elastic collisions is one that relates the relative speeds of the approaching, and later separating, bodies. In particular, the relative speed at which the bodies separate from each other after the collision is the negative of the relative speed of approach.

So if the two speeds before collision are v1 and v2, and their speeds after the collision are u1 and u2, then

v1 - v2 = -(u1 - u2)

This property of elastic collisions can take the place of the conservation of energy equation, and has the benefit of not involving the squares of velocities.