A question about gravity.

General Questions
#1

A sudden thought occured to me. Gravity is theoretically infinite. Thus, there is no such thing as escape velocity. Anything leaving a gravitational field will always have some net force pointing back to its origin. Ergo, what goes up, must come down.

Now, big bang theorists wonder about the nature of the “bang.” Will the universe collapse back into itself or will it diverge?

My question is, isn’t it inevitable, because gravity is infinite, that the universe will eventually collapse back into itself?

¾È ³ç, ÁÖ µ¿ ÀÏ

#2

Jeez Beeruser… you think too darn much! First that dimensional thing now this? To be honest, there is no answer to your questions… only theory and speculation. One argument can be as good as any other. (I happen to think there is escape velocity in the fact that there is a limit to a gravitational field, so if you pass the limit you are not affected by gravity, hence the term “escape”) All I keep reading is that the universe is ever expanding… so collapsing back in on itself sounds more like a cool plot for the next Star Trek movie. Hmmm… maybe to work that in they could have the timeline shift in reverse?

#3

Gravitatonal force is theoretically infinite, but fades out indefinitly. There is always an escape velocity. If an object has enough speed and mass it will escape the gravitational well. The gravity pull will always have effect how infinitly small that is. The only object in space where no things escape are black holes. Past a certain point called, event horizon, nothing escapes, not even light.
It all comes down to the mass of the Universe. If the Universe has enough mass, the expansion will eventually recede, and the Universe will collapse unto itself. If not it will only slow the expansion indefinetly, but never reach the point of collapse. Right now it looks as if there isn’t enough mass to stop the expansion, but if recent theories of “shadow matter” (I don’t know the exact wod in english) are correct, it may be enough to reverse the expansion. But thats way far in the future.

As for the reversal in time: Stephen Hawking wrote in his book, “A Breef History of Time”, that it’s not a far out idea.

#4

You might find this on quantum gravity helpful:
http://www.damtp.cam.ac.uk/user/gr/public/qg_qc.html
Current cosmological thinking tends toward an infinite, ever-expanding universe.

#5

If a ship reaches escape velocity, gravitational force will dimish faster than his speed, so the ship can get as far as is desired from the gravity well.

As for the big bang, its the exact same reason. As the universe gets bigger, the gravitational force between bodies gets weaker. If the universe lacks the required mass (as is widely believed today), than the expansion of the universe will continue to be slowed by gravitation, but the gravitational pull between bodies will reduce faster than the rate the bodies are slowed. Thus the expansion of the universe will asymptotically reduce towards a constant rate (which would be the point at which gravitation across the universe as a whole becomes negligible).

Indeed, the boundary between infinite expansion and an eventual contraction is much like an escape velocity for the entire universe.

#6

What it all comes down to is the balance between energy and gravity. Escape velocity is defined as the speed at which one has to go so that one never falls back into the gravitational well. This velocity decreases as you get farther away from the gravitational source. On the surface of the earth, this number is on the order of 10 m/s. Several light-years away, this number is probably something like a few m/year. To get this velocity, one has to put energy into the system. If you attempt to jump off the earth, you will inevitably come back down as the force of gravity is stronger than the energy put into counteracting it. If you are already several light-years away, all you really need to do is blow in the correct direction and you will never fall back into the gravity well because the amount of energy you put in is far greater than the force of gravity at that distance.

So the question for the universe becomes: Is there enough energy causing the expansion to counteract the combined force of gravity. The answer: we don’t know, but probably. There just doesn’t seem to be enough matter in the universe to counteract our current rate of expansion.

TheDude

#7

I think you’re missing my point. There are two things here: velocity vs. acceleration. You can have escape velocity of .99c leaving the earth, but you will still have a net force of some small fraction of a Newton pointing back to the earth. We will always have a constant deceleration of that object, albeit infinitesimally small. Perhaps over eons (I’m pulling figures out of the air, here) the object will inevitably fall back to zero velocity, and since gravity is theoretically infinite, it will still have a net force (and net acceleration) pointing back to the earth, no matter how many googles of light years it is away.

Never say never when concerning gravity.

¾È ³ç, ÁÖ µ¿ ÀÏ

#8

I’m afraid your definition of “escape velocity” is off a bit, Beeruser, though this is quite understandable since a number of web sites also get it wrong. For example, “Sea and Sky’s” Glossary of Astronomy Terms mistakenly (or at least misleadingly) defines it as: “The speed required for an object to escape the gravitational pull of a planet or other body.”

But escape velocity is not the speed at which the planet’s gravitational attraction is zero. It’s the velocity at which your kinetic energy is equal to your gravitational potential energy. It’s the speed you must reach in order to counter the body’s gravitational effects, not eliminate them.

You can see from the real definition that escape velocity exists as a finite quantity and is an entirely valid concept.

<BLOCKQUOTE><font size=“1” face=“Verdana, Arial”> quote:</font><HR> since gravity is theoretically infinite, it will still have a net force (and net acceleration) pointing back to the earth, no matter how many googles of light years it is away.
[/quote]

Though this is more or less correct in the context you used it, remember that gravity is not really infinite, since it propagates at the speed of light (though you’re absolutely right that no escaping object can outrace gravity). Outside the light cone of the originating astronomical body (i.e., googles of light years away), its gravity is zero.

Much more to the point, your statement is really only valid if you and the Earth were the only two bodies in the universe. I think it would be more accurate and far more useful to say that the vector pointing back to your origin is effectively zero as soon as you enter another gravitational gradient that counter-balances it. This is the way vectors normally work in physics.

#9

Hey, Beeruser. You’re a man of math. Lemme put it this way.

When you leave the planet, if your velocity is greater than escape velocity, the acceleration towards the planet will decrease as you go in such a way that your velocity will approach a constant (positive) value at infinite distance from the planet.

If your velocity is precisely escape velocity, that constant value at infinite distance will be zero.

So ya see the point? The acceleration of gravity will always be insufficient to turn you around, if you have made escape velocity.

#10

Ok, guys, tell me if I have this straight: If I leave a body at greater than escape velocity, then the inertia which carries me forward is stronger than the gravity which slows me down. Both forces are indeed always working, but the gravity weakens faster than the inertia. Ooops, no, the inertia never weakens. What I meant was that the deceleration due to gravity is always less than the current forward velocity.

Therefore, although the object will continue to lose velocity, the velocity will never actually reach zero. If the velocity would reach zero, then gravity would cause it to start falling again. But because the effects of gravity are weaker and weaker the further you get, the object never really stops going forward.

#11

That’s the way I understand it! :slight_smile:

#12

It’s a limit! I see now. But I’d like to see the equations. I’m sure someone here can show me.

By the way, ambushed, I was using an isolated system for this scenario. There are of course other bodies of mass out there and having such a tiny net force pointing towards the earth would be unrealistic.

¾È ³ç, ÁÖ µ¿ ÀÏ

#13

Beeruser:
I’m sorry if I misunderstood that you intended to limit your scenario to an isolated system; I guess I missed where you made that explicit. But I wasn’t trying to be a jerk or anything – just trying to help out in my own small way…

As for the equations you asked for, I “borrowed” the following from an online physics page at NYU.

(I hope the message board html formatting doesn’t mess this up too much)
<!-- MATH: \begin{equation}
\mbox{K.E} + \mbox{P.E.} = \mbox{Constant}
\end{equation} -->
Gravitational Potential Energy and Escape Velocity: The
gravitational potential energy of two point masses (m[SUB]1[/SUB] and m[SUB]2[/SUB])
separated by a distance r is
<!-- MATH: P.E. = - G m_1 m_2/r -->
P.E. = - G m[SUB]1[/SUB] m[SUB]2[/SUB]/r. Here, the
zero of potential energy is defined to be at
<!-- MATH: r = \infty -->
<IMG
WIDTH=“54” HEIGHT=“15” ALIGN=“BOTTOM” BORDER=“0”
SRC=“http://www.physics.nyu.edu/course_pages/20th_century_concepts/img45.gif”
ALT="r = \infty">.
If you
throw an object (mass m) up (with velocity v) from the surface
of the earth it has total energy:
<!-- MATH: E = m v^2/2 - G m_E m /R_E -->
E = m v[SUP]2[/SUP]/2 - G m[SUB]E[/SUB] m /R[SUB]E[/SUB]. If
v is too small, E<0, and the object will eventually fall back
down to earth. If v is sufficiently large, E>0, and the object
will never return. The critical velocity, or escape velocity
v[SUB]e[/SUB] is given by
<!-- MATH: m v_e^2/2 - G m_E m /R_E = 0 -->
m v[SUB]e[/SUB][SUP]2[/SUP]/2 - G m[SUB]E[/SUB] m /R[SUB]E[/SUB] = 0.

#14

As for the specific formulas for velocity and such, I’m too lazy to do the integration. If someone else wants to do it, I’ll give the basis for it:

a = dv/dt
v = dx/dt
x = distance from the Earth

a = Gm[sub]E[/sub]/x[sup]2[/sup]

dv = Gm[sub]E[/sub]dt/x[sup]2[/sup]

Since x is the only independent variable, I would think that this shouldn’t be too hard to format into a fairly straightforward integration.

Obviously what we are really looking for is a definite integral with initial conditions that will represent a case for escape velocity at t=0 or faster.

#15

I guess I was looking at the problem wrong. I was thinking velocity vs. acceleration, or static vs. dynamic. Perhaps this is true if the acceleration was constant.

The exchange of kinetic and potential energy does make sense, and I can see it clearly now.

Well, sorry to bother everyone again. I think it’s time I go back and hit the books.

¾È ³ç, ÁÖ µ¿ ÀÏ