How is gravity proportional to space-time curvature in the rubber-sheet analogy?

You have linked a YouTube video that shows the rubber sheet analogy for the curvature of spacetime. The trouble is that while the rubber sheet analogy is not a bad way for beginners to get a rough idea what is going on, it can be very ,misleading if you push it too far. In this case I suspect it has lead you to imagine the sheet wrapping round the surface of the Earth, and this would mean that the size of the Earth would affect the curvature. However this is not the case.

Outside the Earth the curvature at a distance $r$ from the centre of the Earth is described by the Schwarzschild metric:

$$ ds^2 = -left(1-frac{2M}{r}right)dt^2 + frac{dr^2}{left(1-frac{2M}{r}right)} + r^2 dOmega^2 tag{1} $$

Note that this only depends on the mass of the Earth $M$. The radius of the Earth, $R$, does not appear in the equation.

If we assume the Earth has uniform density and zero pressure then the curvature inside the Earth at a distance $r$ from the centre is given by the Schwarzschild interior metric:

$$ ds^2 = -left[frac{3}{2}sqrt{1-frac{2M}{R}} - frac{1}{2}sqrt{1-frac{2Mr^2}{R^3}}right]^2dt^2 + frac{dr^2}{left(1-frac{2Mr^2}{R^3}right)} + r^2 dOmega^2 tag{2} $$

I'll leave it as an exercise for the reader to show that both equations (1) and (2) give the same result at the surface of the sphere i.e. at $r = R$.

The curvature inside the Earth does depend on the radius of the Earth $R$, but also on the mass, $M$. In your question you imagine reducing the mass while keeping the size of the Earth fixed. To calculate the gravitational force we would need to calculate the four acceleration, and while this isn't that complicated it's a tedious calculation so I'm not going to do it here$^1$. Instead let's reduce the mass all the way to zero, in which case equation (2) becomes:

$$ ds^2 = -dt^2 + dr^2 + r^2 dOmega^2 tag{3} $$

and equation (3) is the the flat space (Minkowski) metric in polar coordinates.

So if we reduce the mass of the Earth to zero while keeping the radius the same the gravitational field disappears. This shouldn't be a surprise as without any mass there won't be any gravity.

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$^1$ If we have any volunteers to calculate the Christoffel symbols for the interior metric I'd be very interested to see the results

Using a rubber sheet to visualize gravity may be confusing in some cases, as the deformation of the sheet is affected by the size of an object, while gravity is not. The rubber sheet analogy is only a visual representation of gravitation outside a massive body. As gravity is proportional to mass, not size, the analogy becomes awkward if applied to a large hollow object. A hollow object affects the objects outside it as though all its mass were concentrated at a point in its center. Objects inside it can not so easily be accommodated by the rubber sheet analogy.

It may help to understand why the rubber sheet analogy is inappropriate in some cases if one considers what happens gravitationally inside a hollow massive object. Take a look at this website:

http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/Mathematical_Thinking/grvtysp.htm

Also, see the wikipedia article on the Shell Theorem:

http://en.wikipedia.org/wiki/Shell_theorem

Isaac Newton proved the Shell Theorem, which says that (1) if a massive object is a hollow sphere, it's gravitational effect on external objects is as though all its mass were concentrated in the center of the sphere, and (2) no net gravitational force is exerted by the sphere on any object INSIDE, regardless of the object's location within the shell.

An interesting corollary to this is that if there were a gravitating body which could somehow move unobstructed in a hole through the center of the solid Earth, it would spend its time oscillating from outer surface to outer surface through the center. At least, this is what I got from a Leonard Susskind lecture on gravity. See this website for a "hole through the earth" example:

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html

Don't forget that where gravity is concerned, size makes no difference!

Assuming that the curvature is due to size is the problem here because the curvature is due to mass.

So if an object were taken off of the earth, the earth's curvature would decrease and there would be some curvature away from the wart, due to the mass of the object.

In General Relativity, Einstein established that gravity is due to the curvature produced by objects in space.

That's not quite right I'm afraid. The YouTube video shows the rubber-sheet analogy, which isn't ideal, but it will do. Because the force of gravity depends on the slope, not the curvature. The steeper the slope the stronger the force of gravity. You need that curvature to have a slope, but the curvature itself relates to the tidal force rather than the force of gravity.

We all know that gravity is proportional to mass. The picture Einstein painted looks like this: The bigger the object, the bigger the curvature and hence the stronger gravity.

Again, that's not quite right. You need a very massive but very small object for the force of gravity to be strong. Think of a small dense 1kg lead weight in the rubber sheet analogy, as opposed to a 1kg beachball.

Let's say that the material inside the Earth is taken out and Earth is made less dense. Now the gravitational pull will be decreased according to Newton's formula due to reduction in mass.

Yep.

But as per Einstein's picture the dimensions of curvature will be same, as the size of Earth is the same still and the gravity of Earth should not be altered.

Nope. Imagine a beachball full of sand, and a beachball full of air. They don't deform the rubber sheet the same.