# Homework Help: Tidal Forces Near a Black Hole

1. Apr 15, 2006

the questions is that:

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5 times the mass of the sun and has a Schwarzschild radius of 15 km. The astronaut is positioned inside the spaceship such that one of here 0.03 kg ears is 6 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6 cm closer.
a) What is the tension between her ears? Would the astronaut find it difficult to keep from being torn apart by the gravitational forces? (Since her whole body orbits with the same angular velocity, one ear is moving too slowly for the radius of its orbit and the other is moving too fast. Hence her head must exert forces on her ears to keep them in their orbits.)
b) Is the center of gravity of her head at the same point as the center of mass?. Explain

my question is that can i use the formula F=GMm/r^2 directly? So i could find two force acting to the ears respectively. Then get the difference,
is it the way to work out part 1 of (a)?

Also , why one ear is moving too slowly and the other is moving too fast?

Finally, i totally do not how to answer part (b)........

thx for attention~

2. Apr 15, 2006

### Andrew Mason

Yes.

Work out the orbital period for each ear from: $F_g = F_c = m\omega^2r$

The definition of the centre of mass is the vector sum of all the mass of all elements times their displacement vector from the origin divided by the object's total mass. The centre of gravity is the the vector sum of the weight times their displacement vector from the origin divided by the object's total weight. So if the weights of the equal masses differ, the centre of mass and the centre of gravity will differ.

AM

3. Apr 15, 2006

I have found the period , but the Period is almist the same for both ears....
i have calculated it in this way, F=the gravitational force from black hole,
r = the orbital distance from black hole, m=the mass of one of the ears,
is it necessary to use the tension in part a?
also, the astronaut find it difficult to keep his ears from being torn apart by the gravitational forces, is it?

4. Apr 15, 2006

### Andrew Mason

The period of both ears is the same. The question merely points out that if the ears were not joined together they would orbit with different periods because their gravitational forces are not the same (the ear closer to the black hole experiences stronger gravity so its angular speed is greater).

What is the tension that you worked out? Compare that to the equivalent weight hanging on one's ear on earth. That will give you some idea of what the force would feel like.

AM

5. Apr 16, 2006

yes, if the ears are separated, the period and the angular speed should be the same, but why the question mentioned that the ears would have different speed, one is moving too fast while another is moving too slow?

6. Apr 16, 2006

### Andrew Mason

If the ears are joined they obviously must have the same angular speed - otherwise they would seperate. It is only if they were free to move independently that they would have different angular speeds - because they experience different forces of gravity.

AM

7. Apr 16, 2006

but if i assume the ears are separated, the period of ears are almost the same, there is no big difference..........
or i have done something wrong?

8. Apr 16, 2006

### Andrew Mason

What do you define as a big difference? What do you get for $\delta\omega$?

The difference in angular speed is determined by:

$$\omega^2 - (\omega-\delta\omega)^2 = GM(\frac{1}{R^3} - \frac{1}{(R+\delta R)^3})$$

Ignoring higher order terms of $\delta\omega$:

$$2\omega\delta\omega \approx GM(\frac{3R^2\delta R}{R^6})$$

So:

$$\delta\omega \approx GM(\frac{3\delta R}{2\omega R^4})$$

This also shows why tidal forces in a small body are not signficant unless you have an extremely large gravitational gradient (large M, small R).

AM

Last edited: Apr 16, 2006
9. Apr 16, 2006

i know what i should do, thank you

10. Apr 17, 2006

### DaveC426913

Considering the OP appears to have gotten what he needs, I'd like to hijack this thread.

Larry Niven wrote a Hugo Award winning short story called Neutron Star, wherein the hero of the story found himself in a hyperbolic path around a very massive object. There was such a huge gradient along the length of his ship that it pulverized every bit of equipment.

It seemed to take several seconds though. I rather suspect that it would more likely have happened so fast that the acceleration would be more like ramming a brick wall (Equivalence Principle: gravity is indistinguishable from (de)acceleration).

So, to my question:

Just how short is the orbital period of an object that is orbiting fast enough to experience a gravitational gradient? Seconds? Fractions of seconds?

11. Apr 18, 2006

### pervect

Staff Emeritus
Presumably this really means orbiting at a Schwarzschild r=coordinate of 120 km

Essentially, though it would be better to say that the tidal force is given approximately by 2GM/r^3 than to talk about the gravitational force. The result is the same.

GR actually gives an exact value for the tidal force of 2GM/r^3 but that's not for an orbiting body, it's for a stationary one. Probably the author of the quesiton didn't really expect you to consider this point, though, it's quite involved calculationally.

I presume we are supposed to think that one ear is always pointing towards the black hole. I calculate that the person is spinning around at 100 revolutions/second, though - probably something else the author didn't intend you to consider.

The extra force due to this spinning isn't particularly large compared to the tidal force from the black hole in any event.

Another advanced point - the period (proper time) for circular orbit is, unless I've made an error, Tau = (2*Pi*r/c)sqrt(2*r/r_s - 3)

where r_s is the Schwarzschild radius.

(This is an exact GR solution, found by taking the derivative of the effective potential and setting it to zero). r / r_s > 3 for stable cirular orbits, and r/r_s = 1.5 is the photon sphere where the orbital speed is the speed of light. This formula was what I used to get the .01 second orbital period).

12. Apr 18, 2006

### DaveC426913

Yeah, that's what I suspected.

The story had the event happening over an interval long enough for the pilot to actually experience it: "Hey, my navel and my spine are trying to pull themselves apart". (He had positioned himself flat so as it provide the smallest gradient across his body).

I realized that the actual experience should feel more like a car crash - perceptually instant deceleration. Since he was on a hyperbolic path (otherwise he'd've never escaped), he would have experienced an instantaneous hypervelocity 90 degree turn. It would have been indistinguishable from an impact and bounce - as if he'd hit mile think brick wall at a 45 degree angle.

13. Apr 18, 2006

### pervect

Staff Emeritus
This was a reply to the original post - I didn't work out the numbers for "Neutron star". I have read the story (and I may even have it lying aruond somewhere) - but I haven't looked up the details, and I'm not sure if Niven gives us enough information to "plug & chug" to get all the gory details.

Sticking with the original problem, though, we can ask what happens if the observer in the 120km orbit isn't rotating.

Rather than having 1000g forces trying to pull his ears apart, he'll have a rotating force that for 5 milliseconds is trying to tear them apart at 1000g, then for the next .005 seconds is trying to compress them at 500 g or so. (If you work out the details, there is a tidal compresive force of half the magnitude of the stretching force). At that point, the force will be tryin to rip his nose off at 1000g, though.

I expect that his head would fail quite spectatularly under such forces, but I don't know the details - I definitely don't want to find out firsthand :-).

Last edited: Apr 18, 2006