1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Blackhole and Gravity

  1. Apr 22, 2016 #1
    1. The problem statement, all variables and given/known data
    The radius Rh of a black hole is the radius of a mathematical sphere, called the event horizon, that is centered on the black hole. Information from events inside the event horizon cannot reach the outside world. According to Einstein's general theory of relativity, Rh = 2GM/c2, where M is the mass of the black hole and c is the speed of light.

    Suppose that you wish to study a black hole near it, at a radial distance of 43Rh. However, you do not want the difference in gravitational acceleration between your feet and your head to exceed 10 m/s2 when you are feet down (or head down) toward the black hole. (a) Take your height to be 2.0 m. What is the limit to the mass of the black hole you can tolerate at the given radial distance? Give the ratio of this mass to the mass MS of our Sun. (b) Is the ratio an upper limit estimate or a lower limit estimate?

    2. Relevant equations
    Rh = 2GM/c^2
    Fg=GmM/R^2
    Fg=m*a

    3. The attempt at a solution
    Using the two expressions of the gravitational attraction: a=GM/(R+∂R)^2=10
    and we have R=43*Rh=86GM/c^2
    and my height is ∂R=2 metres
    now sub the expression for R in, and do the massive calculation I got M=1.0125*10^32 kg
    the sun has mass Msun=1.99*10^30 kg
    thus the ratios is M/Msun=50.879...
    This is not however the correct answer
    Please let me know where I might have gone wrong
     
  2. jcsd
  3. Apr 22, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Note the first word below:
     
  4. Apr 22, 2016 #3
    Oh I see, so it should be a=GM/(∂R)^2 instead of the whole thing
     
  5. Apr 22, 2016 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    No, it's the difference between two accelerations. Find the expressions for the accelerations and THEN take the difference.
     
  6. Apr 22, 2016 #5
    a(head)=GM/(R+∂R)^2
    a(feet)=a=GM/(R)^2
    and a(difference)=10=GM/(R+∂R)^2-GM/(R)^2
    does that look right?
     
  7. Apr 22, 2016 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Yes, except there's a good chance you have a sign error.
     
  8. Apr 22, 2016 #7
    Thanks a lot. I just found another way to do this(should've read the text book thru thoroughly), they used the differential eq, a=GM/(R)^2 becomes da=-GM/(R)^3 dR where dR=h=2 and subbing in everything this is much easier
     
  9. Apr 22, 2016 #8

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The next step from post #5 (after fixing the sign error) would have been to use the binomial expansion of (R+∂R)-2 and make the approximation for small ∂R/R. That comes to the same as taking the differential.
     
  10. Apr 23, 2016 #9
    I see, i would go with the other method to avoid calculating errors, but thank you very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Blackhole and Gravity
  1. Temp of a Blackhole (Replies: 3)

Loading...