Calculating the Maximum Mass of a Black Hole at a Given Distance

[i_hate_math]

Homework Statement

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?

Homework Equations

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

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

 

[haruspex]

a=GM/(R+∂R)^2

Note the first word below:

difference in gravitational acceleration

 

[i_hate_math]

Note the first word below:

Oh I see, so it should be a=GM/(∂R)^2 instead of the whole thing

 

[haruspex]

Oh I see, so it should be a=GM/(∂R)^2 instead of the whole thing

No, it's the difference between two accelerations. Find the expressions for the accelerations and THEN take the difference.

 

[i_hate_math]

No, it's the difference between two accelerations. Find the expressions for the accelerations and THEN take the difference.

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?

 

[haruspex]

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?

Yes, except there's a good chance you have a sign error.

 

[i_hate_math]

Yes, except there's a good chance you have a sign error.

Thanks a lot. I just found another way to do this(should've read the textbook 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

 

[haruspex]

Thanks a lot. I just found another way to do this(should've read the textbook 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

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.

 

[i_hate_math]

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.

I see, i would go with the other method to avoid calculating errors, but thank you very much.