Accceleration due to gravity of a Black Hole

In summary, the given equations and problem involve calculating the gravitational acceleration for a black hole of mass M and radius Rh, at a distance ro = (1 + ε)Rh from its center. For part (a), the first order approximation of ag is given by c^4 / 4(1+ε)^2GM, and for part (b), the difference in gravitational acceleration between an astronaut's head and feet is given by -Hc^6 / 4(1+ε)^3GM^2. The solution for part (a) involves simplifying the (1+ε)^2 term to (1+2ε), and for part (b), the (1+ε)^3 term can be ignored due
  • #1
Zyxer22
16
0

Homework Statement



The radius Rh and mass M of a black hole are related by Rh = [itex]\frac{2GM}{c2}[/itex], 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 = [itex]\frac{GM}{r2}[/itex], 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.

Homework Equations



ag = [itex]\frac{GM}{r^{2}_{g}}[/itex]

Rh = [itex]\frac{2GM}{c^{2}}[/itex]

r[itex]_{0}[/itex] = (1+[itex]\epsilon[/itex])R[itex]_{h}[/itex]

The Attempt at a Solution



a[itex]_{0}[/itex] = [itex]\frac{GM}{r^{2}_{0}}[/itex]

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

= [itex]\frac{GM}{(1+\epsilon)^{2}(\frac{2GM}{c^{2}}^{2}}[/itex]

=[itex]\frac{c^{4}}{4(1+\epsilon)^{2}GM}[/itex]




part b)

ag = [itex]\frac{GM}{r^{2}_{g}}[/itex]

differentiating gives dag = [itex]\frac{-2GM}{r^{3}}[/itex]dr

= [itex]\frac{-2GM}{r^{3}}[/itex]dr

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

=[itex]\frac{-2GMH}{(1+\epsilon)^{3}\frac{2GM}{c^{2}}^{3}}[/itex]

=[itex]\frac{-Hc^{6}}{4(1+\epsilon)^{3}GM^{2}}[/itex]

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 :)
 
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  • #2
If there is anyone who could at least give me a point in the right direction I'd be grateful.
 
  • #3
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[itex]_{0}[/itex] = [itex]\frac{c^{4}}{4(1+\epsilon)^{2}GM}[/itex]

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[itex]_{0}[/itex] = [itex]\frac{c^{4}(1-2\epsilon)}{4GM}[/itex]

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:

Related to Accceleration due to gravity of a Black Hole

What is the acceleration due to gravity of a Black Hole?

The acceleration due to gravity of a Black Hole is extremely strong, as it is directly related to the mass and density of the Black Hole. In general, the closer an object is to the center of a Black Hole, the stronger the acceleration due to gravity will be. This can result in extremely high speeds for objects falling into a Black Hole.

How does the acceleration due to gravity of a Black Hole compare to that of other celestial objects?

The acceleration due to gravity of a Black Hole is much stronger than that of other celestial objects such as stars or planets. This is because the mass and density of a Black Hole is concentrated into a much smaller space, making the gravitational force much more intense.

Can the acceleration due to gravity of a Black Hole be measured?

Yes, the acceleration due to gravity of a Black Hole can be measured by observing the movements of objects around it. By studying the orbits of stars and gas clouds near a Black Hole, scientists can determine the strength of the gravitational force and calculate the acceleration due to gravity.

Does the acceleration due to gravity of a Black Hole change over time?

Yes, the acceleration due to gravity of a Black Hole can change over time as the Black Hole's mass and density may change due to the absorption of matter. However, this change is often negligible and does not significantly affect the intense gravitational force of a Black Hole.

Is the acceleration due to gravity of a Black Hole the same at all points within the event horizon?

According to Einstein's theory of general relativity, the acceleration due to gravity of a Black Hole is infinite at the center, known as the singularity. However, at the event horizon (the point of no return where the escape velocity exceeds the speed of light), the acceleration due to gravity is constant and equal to the speed of light. Beyond the event horizon, the acceleration due to gravity continues to increase towards the singularity.

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