Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gravitational Potential Enery Problem

  1. Mar 2, 2007 #1
    Gravitational Potential Energy Problem

    I have this problem and it has been bugging me
    The problem (if i recall correctly) went like this:
    Ive got an earth like object ( m = 6E24 kg) that will fall 1 A.U = 1.5E11 m
    towards a neutron star with a radius of 1E4 m. At this point it will presumably break up and form an accretion disc around the neutron star.
    The goal is to calculate roughly the resulting temperature of the accretion disc.

    so i started by assuming the virial theorem applies
    Kinetic energy = 1/2 * Potential energy
    3/2 kT = G M m / 2 r (1)

    I am trying to solve for Temperature but I do not know the mass of the neutron star M. I know m = 6E24 and i know r = 1.5E11 m
    how do I solve for M?

    well heres what i did
    M = density * volume
    I took a typical density of a neutron star, say, 1E18 kg/m^3
    I know volume = 4/3 Pi r^3 and r = 1E4 m
    I get mass of neutron star = 4E30 kg

    plug this into (1)

    I find the temperature to be 10^*(56) kelvin???

    thats ginormous! where did i go wrong? or perhaps that was the point of the equation?

    Last edited: Mar 3, 2007
  2. jcsd
  3. Mar 3, 2007 #2
    Hi. Perhaps this post belongs in the Astrophysics or General Astronomy category.

    Could the moderator please move it if doesn't belong here to increase the chances of reply?

    thank you
  4. Mar 3, 2007 #3


    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    in your equation (1)
    some factor is missing in the LHS because that is just the kinetic energy of a single particle

    but on the RHS you have a bulk energy distributed among many particles

    there are other things one could say, like have you calculated the energy derived from the planet falling approx one AU
    and thought about where that energy went and if some of the mass of the planet was blown away

    but I would not worry about the other issues first, first thing is to make sure your equation (1) is right
    and I think it is missing a term on the LHS corresponding to N, number of particles
    Last edited: Mar 3, 2007
  5. Mar 3, 2007 #4
    thank you for the reply. well this question was asked in an introductory astronomy class so im not sure how complicated he wanted us to do things. he doesnt teach that well and claims we have learned what we need to know to solve this problem.

    i took your suggestion and used

    3/2 n K T = G M m / 2 r

    and i found n to be about 1 *10^(26) particles
    using this i still calculate the temperature of the accretion disc to be

    10^(30) Kelvin.

    which is still enormous!

    i should mention something that i forgot to mention earlier that the teacher made a specific point in saying that the the earth sized object falling towards the neutron star was made of iron.

    thats how i calculated n in the first place
    n = M total / Miron where Mtotal = mass of earth and i looked up the mass of one iron atom in kilograms.

    but the point is that the temperature before I didnt include n was 10^(56) K.
    In order to reduce that to a "reasonable" number you'd the number of particles to be like 10^40 or higher. so i think there might be something more fundamental wrong here?

    thank you
    Last edited: Mar 3, 2007
  6. Mar 3, 2007 #5


    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    In this equation what does m stand for? I thought you were using it for the mass of the earth. In that case n should be more like 10^50
    and you have made it 10^26
    so that is one reason you get way too high a temperature.

    But maybe you want m to be the mass of an iron atom? If you are just solving the equation for ONE SINGLE IRON ATOM in orbit around the neutron star then the n (equals the number of particles) really should be ONE the way you had it at first!

    I don't understand your approach, so maybe someone else will step in and give you some coaching. In the meantime, in any case it would be a good idea to learn to use the google calculator. it is a great laborsaving device. You want the mass of earth divided by the mass of an iron atom (which is 56 atomic mass units) then there is NOTHING TO LOOK UP you simply type this into the google box:

    mass of earth/(56 atomic mass unit)=

    You type that into the window and press return and it will give back
    mass of Earth / (56 atomic mass unit) = 6.4245496 × 10^49

    TRY IT, it's great! However it doesnt round off, and seven places is unreasonable precision, so round it off to 6.4 or simply to 6
    (dont give your teacher something that says 6.42454... when he just wants an approximate answer)

    So something with one earth mass made of iron atoms would contain 6 x 10^49 atoms.

    Last edited: Mar 3, 2007
  7. Mar 3, 2007 #6
    wow thank you yeah if it is like 10^50 particles then the answer is VERY reasonable only 10^6 Kelvin
    Last edited: Mar 3, 2007
  8. Mar 5, 2007 #7


    User Avatar
    Science Advisor
    Gold Member

    I would be very suspicious of any answer that exceeds the Planck temperature. It's a clear indicator of a faulty assumption.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook