Homework Help: Accceleration due to gravity of a Black Hole

1. Nov 3, 2011

Zyxer22

1. The problem statement, all variables and given/known data

The radius Rh and mass M of a black hole are related by Rh = $\frac{2GM}{c2}$, where c is the speed of light. Assume that the gravitational acceleration ag of an object at a distance ro = (1 + ε)Rh from the center of a black hole is given by ag = $\frac{GM}{r2}$, where ε is a small positive number. (This formula is valid for certain values of ε. Exactly how large or small ε must be depends on the size of the black hole.) (a) What is ag at ro for a mass M black hole, to first order in ε? (b) If an astronaut with a height of H is at ro with her feet toward this black hole, what is the difference in gravitational acceleration between her head and her feet? Assume H << Rh. Express your answers in terms of M, G, H, c, and ε.

2. Relevant equations

ag = $\frac{GM}{r^{2}_{g}}$

Rh = $\frac{2GM}{c^{2}}$

r$_{0}$ = (1+$\epsilon$)R$_{h}$

3. The attempt at a solution

a$_{0}$ = $\frac{GM}{r^{2}_{0}}$

= $\frac{GM}{[(1+\epsilon)^{2}R_{h}]^{2}}$

= $\frac{GM}{(1+\epsilon)^{2}(\frac{2GM}{c^{2}}^{2}}$

=$\frac{c^{4}}{4(1+\epsilon)^{2}GM}$

part b)

ag = $\frac{GM}{r^{2}_{g}}$

differentiating gives dag = $\frac{-2GM}{r^{3}}$dr

= $\frac{-2GM}{r^{3}}$dr

=$\frac{-2GM}{[(1+\epsilon)R_{h}]^{3}}$

=$\frac{-2GMH}{(1+\epsilon)^{3}\frac{2GM}{c^{2}}^{3}}$

=$\frac{-Hc^{6}}{4(1+\epsilon)^{3}GM^{2}}$

I'm not sure why my answers are wrong, but my hint says

The gravitational acceleration depends on the mass of the black hole and the distance from its center. Here the distance is a not fixed value but a certain multiple of the radius Rh. Use Taylor series or the binomial expansion to determine the leading order dependence on ε.

Which doesn't help me. I'd appreciate any hints/explanations :)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 4, 2011

Zyxer22

If there is anyone who could at least give me a point in the right direction I'd be grateful.

3. Nov 5, 2011

Zyxer22

So, the solution to this problem was posted on my course website, but I'm still having issues. Basically, the answer is the same as mine until:

a$_{0}$ = $\frac{c^{4}}{4(1+\epsilon)^{2}GM}$

The (1+ε)2 is simplified as (1+2ε)... ok, if ε is really small, I suppose assuming ε2 = 0 is alright.

But, then they move the term to the numerator and switch the sign:

a$_{0}$ = $\frac{c^{4}(1-2\epsilon)}{4GM}$

I don't understand that at all...

For part b the answer is the same with the exception of the ε term again. Why is it ok to ignore this part of the equation? Even if it's small, it appears large enough to be a factor in the initial equation to be notable.

**Edit**
Ok... so I think I see what they did for part a. They multiplied by (1-2ε)/(1-2ε) which made (1+4ε2) on the bottom ε2 = 0
I think it's ridiculous that we're expected to just know we need to get the answer in that form... but eh.

I'm still curious why the
(1 + ε)3 term can be ignored for part b.

Last edited: Nov 5, 2011