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More Density = More Gravity?

  1. Jan 11, 2009 #1
    Hi,

    I am a bit confused with how a neutron star or black hole has such a high force of gravity.

    As I understand the process of star death of a massive star (I'll use a neutron star for example),

    1) the star fuses heavier and heavier elements
    2) the core begins to contract
    3) this eventually causes a supernova in which the outer envelopes are blown off (star looses mass)
    4) the remaining contracted and dense core is supported from further collapse by neutron degenerate pressure
    5) this is what is known as a neutron star

    Yes, I know this is the simplified version but the thing I don't get is, if the original star looses mass in the supernova explosion (and in theory even if it didn't) why then does the remaining core (neutron star) have a higher gravitational force than the original star did? I thought that gravitational force only was dependant on mass not density. Two objects with the same mass curve spacetime equal amounts regardless of density or...? I don't understand that when a certain mass (in this case a star core) is shrunk into a object with higher density (a neutron star) but still has the same mass (or in this case even less mass)that it's gravitational field becomes stronger just because it is more dense.

    Obviously, I have misunderstood something about gravity or star death and if someone could help explain it would be great!!

    Thanks a lot!
     
  2. jcsd
  3. Jan 11, 2009 #2

    Janus

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    They don't have stronger gravitational fields. For examle, if the Sn were to suddenly collapse into a black hole, the Earth wouldn't notice any difference in gravitational effect.

    What does happen is that the radius of the body shrinks to a point where the surface gravity can become very strong, since surface gravity also depends on distance from the center of the body.

    Escape velocity also depends on distance from the center of the body. So when a black hole forms, what happens is that the body shrinks to the point that its physical radius is smaller than the distance from its center where the escape velocity is greater than the speed of light for that body's mass. Any light passing that point can't leave again, and any light originating inside these radius could not get out.
     
  4. Jan 11, 2009 #3
    Ok great, that clears things up then.

    One more thing, is there a equation that describes the relationship between an objects "surface gravity" and it's mass?

    Thanks again for the help!!
     
  5. Jan 11, 2009 #4

    Janus

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    [tex]A_g = \frac{GM}{r^2}[/tex]

    where Ag is the acceleration due to gravity
    G is the universal gravitational constant
    M is the mass of the body
    r is the radius of the body

    For comparison, Ag = 9.8 m/s^2 for the surface of the Earth. Otherwise known as 1g.
     
  6. Jan 11, 2009 #5
    Perfect!!

    That helped a lot!!

    Thanks!!
     
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