Kharakov

Does anyone know if the point in a body (Such as the earth or the sun) where the gravitational force pulling towards the center of the body is balanced by the mass pulling towards the surface of the body is close to the circumference of the event horizon of a black hole of the same mass as the whole object?

For the celestial body you would need to take into account all gravitational force pulling away from the object's center to find the place of equilibrium.

(ie the supposed black hole at the center of the milky way would be pulled outwards by all the mass of the whole milky way galaxy- and by the mass of the galaxies around the milky way galaxy- ad infinitum)

One more question:

Did astrophysicists account for the the gravitational acceleration of light towards a galaxy (or any mass which has a large / strong gravitational field for light to escape from) when they measured the apparent acceleration (using Doppler shift) of galaxies moving away from us?

tronvillain

Kharakov:
Perhaps you should try and make your question clearer. To me, it appears to be gibberish - I don't think that such a point exists at all, making relating it to black holes somewhat difficult.

DB_Hunter

Yeah, I don't understand the first point, either... I'm not even sure what the idea really is...

This is just wrong - light doesn't change it's (scalar) speed (not velocity, which is a vector) unless it goes through a change in medium. There's no acceleration involved in red-shifted or blue-shifted light - doppler effect is caused by the motion of the object emitting the light or other EM radiation relative to the observer, but the speed of the radiation is a constant.

DB

Asha'man

Hu? You lost me here.

Remember that gravity is an inverse square law. The effect of the outside universe is trivial compared to the local mass.

Also, remember that the gravitational force at the center of a solid sphere is effectively nil, since the attraction from one side of the sphere is balanced by the opposite side. Similarly, an object at the center of the galaxy would experience no net acceleration from the rest of the galaxy, assuming that the galaxy is symmetric. Since the universe as a whole appears pretty homogeneous, cumulative effects from the rest of the universe are pretty much canceled as well.

Well, yes and no. Astronomers look for (and have found) cases of gravitational lensing, where one heavy object is clearly bending the light from a more distant object. They are very aware of these effects.

However, the effect of gravitational lensing on a Doppler redshift is nil. While the light is shifted in one direction as it enters the gravity well, the exact same shift is applied in reverse as it leaves the gravity well. Think of a frictionless roller coaster: you would gain speed as you go down the slope, but lose it coming back up again.

Asha'man

However, there is a time dilatation effect within a gravity well. This is why light cannot escape a black hole: while the speed of light (in a vacuum) is a constant, time is distorted by gravity, making the effective speed (to an outside observer) seem slower. I’m pretty sure that there is a measurable Doppler shift generated from entering or leaving a gravity well. However, the effect is small for everything except superdense objects.

JB01

It can be proven that for any spherically symmetric mass distribution (i.e., one whose density is a function only of radius), the net gravitational force at any point in the interior is due only to that portion of the mass lying within the radius of the point. Thus, in reality, there is effectively no outward gravitational pull at any point within a large, spherical body such as a star or planet (at least not due to the mass of the body itself).

Friar Bellows

You're asking what the equilibrium point is between the gravitational force due to all other matter in the universe and the gravitational force due to the celestial body in question? And whether this is the same or close to the event horizon of a black hole the same mass of the body?

Just consider the Earth-Sun system. At what point between the Earth and the Sun is the gravitational force due to the Sun the same magnitude as the gravitational force due to the Earth? Well, I calculate it to be about 260,000km from the Earth. As for the Schwarzschild radius which defines the event horizon, for a black hole the mass of the Earth you get about 1 centimetre.

You could do the same calculation for a black hole of a million solar masses at the centre of our Galaxy. The Schwarzschild radius in this case is 3 million kilometres. I don't know how to accuractely calculate the net force due to all other matter in the Galaxy. But I assure you that its equilibrium point (where it equals the gravitational force of the black hole) is nowhere near the event horizon of the black hole. In fact, there is probably no such equilibrium point, given what we know about the gravitational field inside a spherically symmetric distribution of matter. Of course, the Galaxy is not a spherically symmetric distribution of matter!

You need to read up some more about black holes. An event horizon does not define the physical extent of a black hole. It defines the point at which not even light can escape the gravitational field of the black hole. It's not a physical singularity, and not even the tidal forces are infinite there. Read the following book:

<a href="http://www.amazon.com/exec/obidos/ASIN/0393312763/104-6501227-9031115" target="_blank">Black Holes and Time Warps: Einstein's Outrageous Legacy</a> by Kip S. Thorne.

Yes. It's contribution to the redshift is exceedingly small. Light entering our Galaxy and striking the Earth's surface receives a gravitational redshift due to our Galaxy of about -0.001. That is ignorable when you're talking about redshifts of .4 or 2.3. In most cases, you can ignore gravitational redshift. In a few cases, you can't, and then astronomers can use general relativity and observational data to calculate its contribution.

No, not Doppler shift! The redshift due to the cosmological expansion of space is not a Doppler effect. Peculiar velocities of objects relative to the "Hubble flow" produce Doppler effects, but not the expansion of space. There really are three types of redshift:

1) Doppler, due to peculiar velocities of observer and object.
2) Gravitational, due to the gravitational fields of the bodies which the light passes near.
3) Cosmological, due to the expansion of space.

These are 3 different effects, usually present together. Read the following book for more information:

<a href="http://www.amazon.com/exec/obidos/ASIN/052166148X/104-6501227-9031115" target="_blank">Cosmology: The Science of the Universe</a> by Edward R. Harrison.

[ March 27, 2002: Message edited by: Friar Bellows ]</p>

Undercurrent

In laymans terms, Kharakov, this means that the point you speak of, for spherically symmetric objects, like the Earth and Sun, the point you speak of is always at the centre of the body. Anywhere else in the body, the field will be pulling towards the centre of the body, and there is no "balancing point" partway down where the mass above you pulls you up with the same force that the mass below you pulls you down.

m.

Coragyps

Light leaving the Sun has a gravitational redshift of 2.12 x 10^-6. Detectable, but not just real big. The equation is

w/w(0)= 1- GM/rc^2

where w and w(0) are shifted and rest wavelengths, G is 6.67 x 10^-8 (the gravitational constant) , M is the mass of the object emitting the light, r its radius, and c the speed of light.
Simplified a little,

w/w(0)= 1- 7.41 x 10^-29 * M/r

with mass in grams and r in centimeters.
(The redshift is just GM/rc^2 )

[ March 28, 2002: Message edited by: Coragyps ]</p>

Kharakov

The first question is completely wrong- I realized that about 4 hours after leaving my computer lab. There will always be mass on the other side of the object which is having gravitational influence upon any point on the other side of the object.

In reference to the second question, doesn't the 'wavelength' of light change when leaving a gravity well?

Instead of altering the wavelength/frequency of light in your equation (while maintaining a constant velocity), alter the wavelength and velocity/ or frequency and velocity. Does it still work out mathematically when you do this?

C= wavelength * frequency

So if we hold the frequency of light constant, while we switch the wavelength and the velocity we end up with a whole new way of looking at things, which is sometimes useful.


Thanks for clarifying the red shift problem, although now I have to read about cosmological expansion effects.

[ March 29, 2002: Message edited by: Kharakov ]</p>