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Mass of blackhole, volume.

  1. Mar 12, 2013 #1
    A clump of matter does not need to be extraordinarily dense in order to have an escape velocity greater than the speed of light, as long as its mass is large enough. You can use the formula for the Schwarzschild radius Rs to calculate the volume 4/3piRs^3 inside the event horizon of a black hole of mass M.

    What does the mass of a black hole need to be in order for its mass divided by its volume to be equal to the density of water (1 g/cm^3) ?


    So I have the density which = mass/volume = 1 g/cm^3

    I know I'm suppose to find the mass but then how do I know what the volume is?
    If the volume formula is 4/3pir^3, I'm having trouble with finding radius now, since I don't have the volume, I'm really frustrated about this question.Also using Schwarzchild radius formula Rs = 2Gm/c^2 seems useless because I don't have mass now?! Err so I need mass and volume but all I have is density?!
     
    Last edited: Mar 12, 2013
  2. jcsd
  3. Mar 12, 2013 #2
    SOLVED IT....

    Rs = (3km)x M/Msun convert 3km to 3x10^5 cm...

    density = mass/volume = M/ 4/3 piR^3

    density = 3M/ 4pi ((3x10^5 cm)M/Msun)^3

    ...

    M/Msun = root (3Msun/4pi(density)(2.7x10^16))

    = 1.33x10^8 Msun

    Took a while but finally got it, and this is how if anyone is interested.
     
  4. Mar 12, 2013 #3
    With that you could calculate the minimum possible mass of a black hole. Is there one? Could you just shove two neutrons together really hard and have them form an event horizon?
     
  5. Mar 12, 2013 #4
    Thank you Oldspice1212! I was looking for how to solve this question too!

    Could you tell me where you got Rs = 3km x M/Msun? and where the 3km came from?



    As well, for this section of what you wrote: "M/Msun = root (3Msun/4pi(density)(2.7x10^16))"

    I plugged this in as 3rdroot(3*(2*10^30) / 4pi(1)(2.7*10^16)) but I did not get 1.33*10^8 Msun, Did I do it correctly?
     
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