You have linked a YouTube video that shows the [rubber sheet analogy for the curvature of spacetime][1]. 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 is described by the Schwarzschild metric:

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

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 is given by the Schwarzschild interior metric:

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

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. 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:

$$ ds^2 = -dt^2 + dr^2 + r^2 d\Omega^2 \tag{3} $$

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

  [1]: https://www.youtube.com/watch?v=0HnaLnUdYvs&feature=youtu.be&t=275