# Homework Help: Gravitation - Galaxies

1. Nov 10, 2009

### Wellesley

1. The problem statement, all variables and given/known data
The Andromeda galaxy is at a distance of 2.1 X 1022 m from our Galaxy. The mass of Andromeda is 6 X 1041 kg and the mass of our Galaxy is 4 X 1041 kg.

(a) Gravity accelerates the galaxies toward each other. As reckoned in an inertial reference frame, what is the acceleration of Andromeda? What is the acceleration of our Galaxy? Treat both galaxies as point particles.

(b) The speed of Andromeda relative to our Galaxy is 266 km/s. What is the speed of Andromeda and what is the speed of our Galaxy relative to the center of mass of the two galaxies?

(c)What is the kinetic energy of each galaxy relative to the center of mass? What is the total energy (kinetic and potential) of the system of the two galaxies? Will the two galaxies eventually escape from each other?

2. Relevant equations
G=6.67X10-11
T2= 4* PI2/GMGalaxy

3. The attempt at a solution

I've been stumped on this question for a while now. Usually, I try and come back to it the next day (in this case the past few days) and try anew, but it didn't work.\

a.) I got the answers for this part pretty easily. I used F= GMm/r2 and solved for F, and then acceleration.
I got 9.07X10-14 m/s2 for the Milky Way, and 6.05X10-14 m/s2 for Andromeda.

b.) Here is where I the trouble starts..... I first came up with the center of mass of 1.26X1022 meters and 8.4X1021 meters, with the larger one being the distance between the Milky Way and the center of mass (since Andromeda has more mass, the center of mass is closer to it).

From there, I've tried using v2=GMCOM/r and KE/PE for the system, but I haven't come up with the book answers of 1.1X105 m/s (speed of Andromeda) and 1.6X105 m/s (speed of the Milky Way).

c.) I've tried to do this part, using the answers from part b. I managed to sum up the KE somewhat correctly, but I failed to sum the total energy correctly. Is there only one gravitational potential energy calculation, or two (one for each radius from the center of mass)?
KEAndromeda= 1/2mv2 = 1/2 * 6X1041 * (1.1X105)2 = 3.63 X 10 51 J ---> Answer in back= 3.4X1051 J

KEMilky Way= 1/2mv2 = 1/2 * 4X1041 * (1.6X105)2 = 5.1X1051 ---> Answer in back = 5.1X1051 J

For the summation I have:

3.6X1051 + 5.1X1051 - (6.67X10-11*6X1041*4X1041) / 1.26X1022
- (6.67X10-11*6X1041*4X1041) / 8.4X1021

Total Energy = 8.75X1051 - 1.27 X 1051 - 1.9057X1051 = 5.574X1051 ---> Answer in back = 7.7X1051

The book I have is Ohanian Physics 2nd edition which doesn't have many examples that deal with this type of problem.

Any help would be greatly appreciated. Thanks.

2. Nov 10, 2009

### ideasrule

Have you tried the conservation of momentum? There are no external forces, so you can express the velocity of one galaxy in terms of another using the conservation of momentum equation.
Gravitational potential energy is -GMm/r, where r is the distance between the two masses. You can use the center of mass as one of the "masses", but then you'd need to use an equivalent mass for the mass of the center. It's easier to just use the two galaxies for M, m, and r.

3. Nov 10, 2009

### Wellesley

Thanks. I did get the total energy, but the KE for Andromeda is still off for some reason. Any ideas?

6X1041* 266,000 m/s = 1X1042 * v2
Velocity = 159600 m/s ---> the answer for the speed of the Milky Way.

Yet, this will not work for the other answer: 110,000 m/s....

So close!!!

Last edited: Nov 10, 2009
4. Nov 10, 2009

### ideasrule

It's just a rounding error. Using the unrounded figure for the speed, v=1.064*10^5 m/s, gives you the book's answer.

I don't quite get this. If Andromeda has a velocity of 266 km/s, that means you're using the reference frame of the Milky Way. But if that's the case, and 1*10^42 is the total mass of the system, that means v2 is the collective speed of the system. It's less confusing to consider the situation from the ref. frame of the center of mass. In that ref. frame, the total momentum of the system is 0, so m1v1-m2v2=0. We already know that v1+v2=266 km/s, so solving the two equations gives you the answer.

5. Nov 10, 2009

### Wellesley

Sorry, my post was vague. Your explanation makes more sense to me now, after looking at the two equations v1+v2=266000 and m1v1-m2v2=0, instead of just the one I originally thought I needed (m1v1-m2v2=0).

I knew the right method was something much simpler than what I was originally trying to do, but I never thought to use momentum. What really confused me, was where the book put this question....it was placed under the section where kinetic energy and gravitational potential energy was introduced. I though I had to use something along those lines to solve part b.

Thanks again for all the help, the problem made a lot more sense after your posts!

Last edited: Nov 10, 2009