# Elastic collision with the sun

1. Jun 22, 2008

### levi2613

1. The problem statement, all variables and given/known data
Which of the following is a true statement concerning a comet passing near the sun? (Mass of comet and mass loss are negligible)

1. The momentum of the comet is nearly conserved because the interaction with the sun is similar to an elastic collision.
2. The momentum of the comet is nearly conserved because the interaction with the sun is similar to an inelastic collision.
3. The mechanical energy (kinetic plus potential) of the comet is nearly conserved because the interaction with the sun is similar to an elastic collision.
4. The mechanical energy (kinetic plus potential) of the comet is nearly conserved because the interaction with the sun is similar to an inelastic collision.

3. The attempt at a solution
The correct answer is C. I understand that the momentum of just the comet is not conserved, but rather the system's as a whole is, so that knocks out A and B. And since it is clearly not inelastic, it can't be D either.

My question: NASA often uses this "slingshot" effect to accelerate objects without the use of fuel. If the object comes around the other side with more velocity than when it started, its KE must increase... where does that come from? The only potential energy I see is from the gravitational force of the sun. So I don't see how at equidistant points from the sun, one before the "collision" and one after, the KE could increase but the PE would be the same.

Thank you so much!

2. Jun 22, 2008

### Nick89

The slingshot effect you mention I believe comes from the fact that the planet they use to slingshot around is not standing still. The movement of the planet gives you the extra KE needed.

In fact you could calculate this...

Suppose you have a large planet (Jupiter for example) with mass M, moving with speed V, to the left.
Then suppose you have a spacecraft of mass m, moving with speed v, to the right. (So the planet and spacecraft are moving towards eachother).
After the spacecraft has passed jupiter (so it is now moving to the left aswell, it 'bounced' off the planet) it has a speed w (to the left) and the planet has a speed W.
The relation between their momentum and KE would be:
$$mv - MV = P = -mw - MW$$
$$1/2 mv^2 + 1/2MV^2 = KE = 1/2mw^2 + 1/2MW^2$$

Working this out further:
$$m(w^2 - v^2) = M(V^2 - W^2)$$
$$m(w-v)(w+v) = M(V-W)(V+W)$$

And
$$m(w+v) = M(V-W)$$

$$w-v = V+W$$

Note finally that W has changed only a tiny little bit due to the planet's very large mass, so we can say:
$$V+W = 2V$$

So finally:
$$w = v + 2V$$

I hope this is right :)

Last edited: Jun 22, 2008
3. Jun 22, 2008

### levi2613

Thank you for taking the time to answer....

I can actually follow along with the math pretty well. And I intuitively understand how the spaceship would pick up speed.... What I can't reconcile is that the total mechanical energy of the spaceship would be conserved. If it picks up twice the speed of the planet, its KE is multiplied by 4. Since the total mechanical energy = KE + PE, it would seem that the PE would have to drop dramatically. But the only potential energy I see is the gravitational enery, which would be the same at equidistant points before and after the collision. So if the potential energy is the same, but the KE has increased by a factor of 4, where's the conservation??

Thank you!!

4. Jun 23, 2008

### Nick89

I guess that in the question above they assumed the sun to be standing still, not moving (there is no mention of the comet picking up speed). I can't think of any other answer lol...