Compressing the Earth to a 20-meter-wide sphere would only subtract one hour of time? (General relativity) The Internet has exploded with news that the Applied Physics Institute of New York published a paper claiming that superluminal speeds can be achieved using a type of Alcubierre Drive, but without the (probably nonexistent) negative mass or negative energy Alcubierre called for...
Instead we just need a superdense spaceship that uses relativity to bend Space-Time...
From New Scientist:  (and elsewhere):
If we take the mass of the whole planet Earth and compress it to a shell with a size of 10 metres, then the correction to the rate of time inside it is still very small, just about an extra hour in the year,” says Bobrick.
Read more: https://www.newscientist.com/article/2269544-a-warp-drive-that-doesnt-break-the-laws-of-physics-is-possible/#ixzz6oZQO2Yo3
Is that true?
If Earth's mass is condensed to a diameter of 10 or 20 meters (not sure if Bobrick meant radius or diameter) it would have a width only 1/1,274,200th or 1/637,100th of the real Earth, and a volume a few quadrillionths or even quintillionths of Earth's....
An increase in density by a factor of a quintillion or more only changes the time differential by an hour or so?
 A: The spacetime surrounding the tiny Earth is the Schwarzschild metric:
$$
ds^2=\left(1-\frac{r_s}{r}\right)dt^2-\left(1-\frac{r_s}{r}\right)^{-1}dr^2-r^2d\Omega^2,
$$
From this, we can see that passage of time is $r$-dependent. The difference in the passage of time between the two different $r$'s (related to the surfaces of normal Earth and tiny Earth) is equal (to high precision; see the other answer) to the passage of time at infinity and the passage of time for small $r$ of, say, $20(m)$  If we compare both passages you get:
Time passage at $r=\infty$: 1
Time passage at $r=20$: $\sqrt{\left(1-\frac{r_s}{r}\right)}$
Filling in the values for $r_s$, the Schwarzschild radius of the Earth [about $9(mm)$] and the value of $r=20$ gives the time passage at $r=20$ as compared to one second at infinity:
Time passage at $r=20$: $\sqrt{\left(1-\frac{r_s}{r}\right)}=\sqrt{\left(1-\frac{0.009}{20}\right)}\approx0.9998$
So if one second has passed at infinity, about $0.9998$ second has passed on the surface of the small Earth. A year contains $32536000$ seconds, which means that $0.9998x32536000=32529493$ seconds have passed on tiny Earth. The difference is $ 6507$ seconds. Which is about $2$ hours.
Following the same procedure, it follows that the time passed on the surface of normal Earth, if one year [$32536000(sec)$] has elapsed at infinity, will be $32535999.756(sec)$, about $0.02(sec)$ less.
A: An hour per year is actually a much higher time dilation factor than is currently experienced on the surface of the Earth relative to a distant observer (which is on the order of 0.02 seconds per year, if my calculations are correct). Note that if you compressed the Earth as described, it would not change its mass at all, so its gravity would remain the same (gravity doesn't depend on density). To a good approximation the only change in time dilation would come due to being able to get closer to the center of mass.
