How Far Can Gravity Reach?

1. Jul 16, 2007

Liger20

I have always been told that the gravitational force of a body, like the sun, extends out all over the universe, but gets smaller with greater distances. I'm pretty sure that this is true, but according to general relativity, gravity is a warping of space(time). It's like the bowling ball on the trampoline analogy. Like the sun warps spacetime, the bowling ball warps the fabric of the trampoline. Here's what I'm getting at: Why wouldn't these creases in spacetime thin out and eventually reach a value of zero?

2. Jul 16, 2007

Schrodinger's Dog

OK let's start with your bowling ball on an elastic sheet analogy, is it impossible to comprehend a very large attraction close to the ball and an increasingly small but never zero curvature or attraction in terms of the Universes expanse? Is the universe actually infinite in size?

Mathematically using Newtons equation:

$$F_g=G_k\frac{m1m2}{r^2}$$

Fg=Force of gravity
Gk=the gravitational constant
m1,m2 are our masses.
r2=the distance between their centres of gravity

If we assume that the Universe is not infinite in size and we have no reason to believe it is. Then does the value of m1 and m2 reach 0 as r^2 approaches the Universes size or the limit of the equation? Let's put in the gravitational constant 6.7428x10-11, any value of m1,m2 and for convenience a size of about 156 billion light years wide, a sort of scientific estimate on the size of the universe.

As you will see no the value is not 0, gravity's extent is across the whole Universe, because it never reaches zero.

We use the infinity of the Universe in a very particular way.

Now we do not know this is the case, we cannot prove it but we can be sure that if the law is true and the Universes expanse is finite, given the equation gravitation has no value=0.

Last edited: Jul 16, 2007
3. Jul 17, 2007

Liger20

Hey, that helped a lot Schrodinger, but when I'm plugging in the masses and the distance, what units should I use? Or does it even matter? Sorry, I never was that great at math.......

4. Jul 17, 2007

Staff: Mentor

That's the metric/SI version, so that's kg and meters.

I think it is just important to remember that when you go far out on a hyperbolic curve (or a trampoline), you can't see with your eyes that it is non-zero, but it always is. But you can see it by plugging numbers into the equation.

Sometimes, though, seeing how small the effect is means you can ignore it. For example, when examining the interaction of our galaxy and the Andromeda galaxy, our solar system's mass is included, but it would not be meaningful to try to calculate our solar system's effect on the Andromeda galaxy its own.

Last edited: Jul 17, 2007
5. Jul 17, 2007

Liger20

Okay, thanks russ, that helped a lot.

6. Jul 17, 2007

Schrodinger's Dog

$$F_g=6.7428\times 10^{-11}\times \frac{1kg\times 1kg}{(1.474^{27}m)^2}\ \simeq\ 5.37^{-20}kgm^{-2}$$

5.37-20kgm-2

I've used mass1=1Kg,mass2=1kg, as you can see this is a very small figure, but it is not zero and in fact as long as r^2 is finite it will never be zero, just increasingly small.

Last edited: Jul 17, 2007