Accceleration due to gravity of a Black Hole

[Zyxer22]

Homework Statement

The radius R_h and mass M of a black hole are related by R_h = \frac{2GM}{c²}, where c is the speed of light. Assume that the gravitational acceleration ag of an object at a distance r_o = (1 + ε)R_h from the center of a black hole is given by a_g = \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 a_g at r_o for a mass M black hole, to first order in ε? (b) If an astronaut with a height of H is at r_o with her feet toward this black hole, what is the difference in gravitational acceleration between her head and her feet? Assume H << R_h. Express your answers in terms of M, G, H, c, and ε.

Homework Equations

a_g = \frac{GM}{r^{2}_{g}}

R_h = \frac{2GM}{c^{2}}

r_{0} = (1+\epsilon)R_{h}

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)

a_g = \frac{GM}{r^{2}_{g}}

differentiating gives da_g = \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 :)

 

[Zyxer22]

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

 

[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.