How much energy is needed to disassemble the galaxy and scatter the stars?

Question

There are N=10e10 identical stars in a galaxy, mass of each star is M=10e30kg and rms velocity of stars is V=100 km/s.

How much energy is needed to disassemble the galaxy and scatter the stars (but not disasseble the stars themselves)?

>Escape velocity = sqr 2 * Oribital Velocity

Only for a small body in external 1/R potential in rms sense.

In galaxy individual potentials of stars are different, and in additon there is no external potential. Escape velocities vary too. Simply put mechanical energies of individual stars are not conserved.

Answer 1

The same answer as Remo Aviron's, only for a different reason.

If we neglect all the fancy stuff like dark matter, it will be the virial theorem. If the pairwise interaction potential U depends on the distance r as r^n, then the average kinetic energy is equal to (n/2) times average potential energy . For the gravitational interaction n=-1. Then = -2 .

The average mechanical energy of a star is + = - . The mechanical energy of all the stars in the galaxy is -N.

The energy is zero in the "disassembled" state. Hence, we need
N* = N*M*/2 = 10E10*10E30*(10E5)^2/ 2[kg*m^2/sec^2] = 5*10E49 J.

Addendum: The classical derivation of the virial theorem refers to time averaging. I use ensemble averaging. Time and ensemble averaging are the same for systems with large number of particles (ergodic hypothesis).

Answer 2

If a star already has (1/2)10^40 J kinetic energy, it would need twice that to escape the galaxy. In other words, the extra energy needed is (1/2)10^40 J. However, after one star escapes completely, the mass of the galaxy has gone down by a factor of (10^10-1)/10^10, so that the next star wouldn't need quite as much energy to escape. Doing a sum (10^10-n)/10^10 for all 10^10 stars gets you 2.5 x 10^49 J, not 5 x 10^49 J that I see in some answers.

No, this isn't right either. I'll get back to this later.

Addendum: This analysis needs to be adjusted. If the galaxy lost half of its mass, then all the remaining stars have sufficient velocities to escape from each other. Therefore, we only need to consider the extra energy required by half of the stars to escape, and that comes out to 1.25 x 10^49 J, or 1/4 of the total kinetic energy of the stars in the galaxy.

Answer 3

Escape velocity = sqr 2 * Oribital Velocity

E star=1/2 m*V.final^2 -1/2 mv^2
=1/2 m [(100/s *sqr(2))^2 - 100k/s^2)]
=1/2 m *(100k/s)^2
=5x10e39 J

Total Energy = E*N
=5*10e49 J

>>>>>>>>>>>>>
Edit 12/16/07

Alexander, you are smart and tricky.

Zo Maar correctly pointed out that Potential Energy at infinity is twice KE of orbit. But everytime you remove a star, the PE goes down by the pro rata amount, e.g., remove 1% of the stars and the PE goes down by 1%. (This also follows from Newton's law of Universal Gravitation). Since PE at infinity is 2 * KE of orbit, the energy necessary to disassemble goes down by 2% for every 1 % of stars sent packing.

And Scyth pointed out that after 50% of the stars are gone, voila, escape velocity = current orbital velocity. (This also follows right out of Newton's law and the Zo Maar derivation).

What this means is that the above answer was off by a factor of 4, or,

E=1.25*10e49 J

As an aside, I found it easy to visualize the problem as simply a thin ring of stars all travelling at v=100km/s. And as you removed stars, the ring would expand and rotate slower, so instead of disassembling at 50% star removal, it would just be a bigger, less massive and slower rotating ring. But the ultimate answer as to the amount of energy necessary to dissassemble remains the same. This analogy can be applied to an ordinary galaxy.

>>>>>>>>>>>>>>>

Scythia, below, beat me to the answer. But I didn't realize it until after I made the above derivation. Give Scythia credit.

Answer 4

Love your questions, Alex, keep'em coming.

I'd go the conservation of energy route here. So we need to audit the total energy (TE) of your galaxy and then identify those terms in TE that we'd need to overcome to split it apart, but keep the stars intact.

Let KE = total kinetic energy in the system of N = 10^10 stars, each with mass M = 10^30 kg, and rms (average) velocity V = 100 kps = 100,000 mps. KE = 1/2 NM v^2. We assume the KE came from stellar gravity forces acting over average distance of motion. Thus, KE comes from the motion of each star as well as from the spin of the galaxy if any.

These are the external energies we'd need to overcome when scattering the stars. [NB: The before and after splitting KE from the BB would cancel out; so it was not considered. Thus, all the KE comes from forces internal to the galaxy...e.g., gravity and, perhaps, thermal.]

Let IE = NMc^2, which is the potential or internal energy of the galaxy due to total mass NM. [Dark matter was discounted, but as this is the majority of mass in a galaxy, it should not have been. If you want Dark Matter including then NM/(DM + NM) ~ .05; so that .05(DM + NM) = NM and DM + NM = NM/.05 = TM, the total matter of your galaxy, and use TM instead of NM.] IE is an internal energy and would not change upon rendering the galaxy and keeping the stars intact.

However, it is unclear what happens to DM when a galaxy splits. As it represents .95 of the internal energy, if DM does not follow the splitting stars, then the galaxy will lose 95% of its internal energy probably to heat entropy. And that represents work that needs to be done WE(DM) = (TM - DM) c^2 to lower the internal energy to NM c^2. If the DM follows the NM, then no such work would be required.

Let WE = the work energy needed to render the galaxy, but keep the stars intact. This is the answer you are looking for.

Then TE = TE(X) + TE(I) with TE(X) = KE is the total external energy if we assume the KE came from the interactions of the forces of gravity upon the stars, and TE(I) = IE the total internal energy. It's the total external energy, the KE converted from gravity, that we'd need to overcome.

Once the galaxy has been ripped apart, we assume the rms v = 0; so that TE(R) = TE(I) + WE; where the total energy of the split off stars is just their internal energies plus the work that went into splitting them off. As the TE's are equal before and after, we have TE = KE + IE = IE + WE = TE(R) where R is the average distance R from what used to be the center of the galaxy.

Bottom line, KE = WE = 1/2 NM v^2 is the energy you'd need to overcome. Or, if DM is considered, WE = 1/2 TM v^2; where TM = NM + DM.

[NB: This is full of approximations and simplifying assumptions, but when seeking out orders of magnitude, this should do the trick.]

Answer 5

Hmmm... wouldn't you really have to tell me the average escape velocity? See... if I am in low Earth orbit, my velocity relative to the center is 8.8km/s or so. But to get out of Earth orbit, I only need a little less than 3km/s more... so without telling me what the escape velocity from the central body is, I am kind of at a loss.

Or am I overlooking something?

Answer 6

Ask some logical question, fool.Don't try to be oversmart
help saving the world"geeks are coming".dont believe me ,look above and below me.

Answer 7

Graham's number in gigawatts would be more than sufficient.

Answer 8

A lot

Answer 9

More than you and I can imagine...

Answer 10

a butt load?