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Relating Newton's Force to Black Holes (Calculation?)

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Muscle can be torn apart by a force of 100,000 N applied across an area of 1 m2. A 10 cm2 muscle therefore will be torn by a force of 100 N.

    If a student of average size were being lowered into a black hole of 1 solar mass, at about what distance from the hole's center will he be torn apart?

    2. (What I thought was a) Relevant equation

    RSchwarzchild= (2MG)/c2

    Where M is the mass of the black hole; G is the gravity constant; c is the speed of light.
    (seen here: http://hyperphysics.phy-astr.gsu.edu/hbase/astro/blkhol.html#c2)

    3. The attempt at a solution

    I did some background research on black holes and I thought that if I calculated the Schwarzchild radius, I would be able to identify the event horizon (defined as the last distance at which light can escape the pull of a black hole) of the black hole, and thus, that would tell me the distance from which the student would be torn apart.

    RSchwarzchild= (2MG)/c2
    = [2(1 solar mass)(6.67 x 10-11 Nm2/kg2]/(2.998 x 108 m/s)2

    R = 1.48x10-27 m

    I think this would be the correct answer, but it doesn't use any of the numerical givens other than the mass of the black hole.

    When I consider the givens regarding surface area and force; I can tell that I am supposed to use an estimate of an 'average student's size' to attain my answer. If I estimate that the average person is about 1.75 m2 (estimate using numbers from BSA https://en.wikipedia.org/wiki/Body_surface_area), then I believe I can say that 100000 N is to 1 m2, as x N is to 1.75 m2 and cross multiply to find that it would take x=175000 N of force to tear apart the student's muscles. But, I am unsure of how to relate this force to black holes and distance.



    Any help or hints would be appreciated! Thank you.
     
  2. jcsd
  3. Oct 10, 2015 #2

    SteamKing

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    I don't see units of "solar mass" in the rest of this equation. How can the result for the Schwarzschild radius be in meters?

    Always check the units of your calculations. The resulting units should be the same on both sides of the equation.

    You looked up some other numbers related to this problem on the internet. Why didn't you do the same for the mass of the sun?
     
  4. Oct 10, 2015 #3
    Thank you for the input!
    I'm not sure why I would need the mass of the Sun to evaluate this problem?
     
  5. Oct 10, 2015 #4

    SteamKing

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    I thought I explained it very clearly. Units of "solar mass" in the formula for the Schwarzschild radius are not compatible with the units for the constant G.

    Your calculation does not result in units of meters for the Schwarzschild radius.

    10-27 m is to the size of a hydrogen atom as a hydrogen atom is to the size of a beach ball.
     
  6. Oct 10, 2015 #5
    I see, thank you for clarifying. Sorry to make you repeat yourself! I'm going on my third week ever in a Physics class, so thanks again, your input is invaluable.
     
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