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Finding gravity on a star/planet

  1. Jan 29, 2009 #1
    1. The problem statement, all variables and given/known data

    "The mass of a certain neutron star is 3.0x10^30kg (1.5 solar masses) and its radius is 8000 m (8km). What is the acceleration of gravity at the surface of this condensed burned-out star?"



    I have no idea where to start
     
  2. jcsd
  3. Jan 29, 2009 #2

    LowlyPion

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  4. Jan 29, 2009 #3
    Nothing!

    So from that sight, I gather this:

    F=Gx(m1m2/d^2) .....right?

    So:
    G=F/(m1m2/d^2)

    meaning:
    F/((3.0 x 10^30) x m2/8000^2) ..... Whats m2 and F?
     
  5. Jan 29, 2009 #4

    DaveC426913

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    The m1m2 will get you the total attraction between both bodies. You only need one, so lose the m2 (just set it to 1).

    You don't need to flip the equation around. You're solving for F. That's your answer.

    G is the gravitational constant, which you are given on that page (6.67x10^-11).
     
    Last edited: Jan 29, 2009
  6. Jan 29, 2009 #5
    Ok, so...

    F = (6.67x10^-11) x [ (3.0x10^30)/(8000^2) ]

    so F = 3.1265625 x 10^12


    Isn't that a bit strong considering earth is ^-11 and this star has only an 8km radius? Or am I looking too much into it?
     
  7. Jan 29, 2009 #6

    DaveC426913

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    But you should try to understand what you're doing with this equation, don't just plug the numbers in.

    The force an object feels under gravity of a massive object is directly proportional to the mass of the massive object.
    i.e.: F is proportional to m : as m gets bigger the force gets bigger

    and inversely proportional to the square of its distance
    i.e. F is proportional to 1/(d^2) : as d gets bigger the force get smaller - and quickly

    Put those together and you've got F proportional to m / (d^2).

    This is only proportional though, which means you have no units. The gravtitational constant allows you to convert it to units.
    So:
    F = G * m / (d^2)
     
  8. Jan 29, 2009 #7

    DaveC426913

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    I haven't checked the numbers but this is a neutron star, and you are on its surface. Millions or billions of g's is par for the course.

    What you need to do is check every one of your numbers to make sure you've got the units right. Is everything in kg or metres, not grams or kilometres? That's a really easy mistake to make.
     
  9. Jan 29, 2009 #8
    I'm sure its all in the correct units.
    Thank you both for your help
    =]
     
  10. Jan 29, 2009 #9

    DaveC426913

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  11. Jan 29, 2009 #10
    Well, isn't earth (6.67x10^-11)?
    So this star being (3.12x10^12) seems a bit extreme
     
  12. Jan 29, 2009 #11

    DaveC426913

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    No. Earth is 9.8x10^0.

    The 6.67x10^-11 is the universal gravitational constant.

    Think of this:

    Your car's engine revs are proportional to your car's speed (let's pretend) but to apply units to that you'd have to have a constant that says "X revs equals Y miles per hour". Say, maybe your constant is q=.05
    i.e.: velocity = q * revs

    revs = 1000

    So, v = .05 * 1000
    v = 50mph

    the consant there is q
     
    Last edited: Jan 29, 2009
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