# Fg and planetary motion

1. Dec 6, 2007

### juliusqueezer

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/c^2, where M is the mass of the black hole and c is the speed of light. Suppose that a person who is 1.7 m tall wishes to study black holes near them, at a radial distance of 50Rh. However, the person doesn't want the difference in gravitational acceleration between their feet and head to exceed dag = 10 m/s^2 when they are feet down (or head down) toward the black hole.

(a) As a multiple of our sun's mass, what is the limit to the mass of the black hole the person can tolerate at the given radial distance?

2. Relevant equations
Rh=(2GM)/c^2
Fg=G((Mm)/r^2)

3. The attempt at a solution

I can solve for Rh and eventually get a mass, but that in no way accounts for the acceleration difference of the person. My question is, how must I account for this?

2. Dec 6, 2007

### dotman

Hello,

Well, the problem states that the person is 1.7m tall. So the person is going to experience some acceleration due to gravity at their feet:

$$Fg_{feet} = \frac{GMm}{r^2}$$,

and some different acceleration due to gravity at their head, which is further away, if they are feet first:

$$Fg_{head} = \frac{GMm}{(r+1.7)^2}$$

The problem has said they wish to study at a distance of $50 R_h$. You need to find the maximum mass of the black hole, that will not create more than a 10 m/s^2 difference in accelerations, above.

Hope this helps.

3. Dec 7, 2007

### Shooting Star

In the above formulae, the accn due to gravity is just GM/r^2. The 'm' is not necessary.

4. Dec 7, 2007

### juliusqueezer

thanks for all the help. Eventually it came down to Afeet-Ahead=10. It ended in a massive equation with alot of arithmetic.

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