The Earth is moving with a speed of about $1670$ $km/h$ around its axis. This speed is more than the sound speed. So the Earth is always breaking the speed of the sound. How much at least should the Earth be smaller so that we notice this $1670$ $km/h$ speed as it is? Also, how much at least should the Earth be smaller so that we notice it is faster than the sound? (Breaking the speed of the sound is when you hear a loud and harsh sound; like when you are in a plane and it moves faster than the sound's speed).
The Earth is moving with a speed of about $30$ $km/s$ around the sun. How much at least should the Earth be smaller so that we notice this extremely fast speed?
How much at least should the Earth be smaller so that the planes reach their destinations just with floating in the air and not flying?
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4$\begingroup$ I think you are misunderstanding the effects of Earth's size on these phenomena - or that you are assuming something in your word choice that makes the rest of us very confused. $\endgroup$– Anders SandbergCommented Mar 9, 2020 at 22:33
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$\begingroup$ Stand at the equator, in a place like Quito, Ecuador. You and the earth beneath your feet are traveling eastward at about 1670 kmh. The land, mountains, ocean, are all doing that. So is the air you are breathing. Of course, it's not exact. There are currents in the ocean, and winds in the air. But basically it is all turning together. And the whole world is orbiting the sun at high speed. The sun is traveling even faster. Try to understand this. I hope you're not falling for the kind of nonsense that flat-earthers are spreading around as much as they can. $\endgroup$– Mike DunlaveyCommented Mar 10, 2020 at 17:07
2 Answers
The only thing affected by the size of the earth is the speed at the equator (1670 km/h). That is simply the speed of a point at the equator of a sphere with a radius of km, which rotates once every 24 hours. If the radius were halved, and the earth still rotated in 24 h, the speed would be halved as well, i.e. 835 km/h.
A sonic boom is caused when an object moves through the air at a speed greater than the speed of sound in the air. As all the air, and all stationary objects, all move with the earth (ignoring winds), there cannot be a sonic boom. Even if the earth rotated much faster, everything would move with it, and there would not be a sonic boom.
All of the earth (the rock, the air, and us) all move around the sun at 30 km/s. Hence we cannot experience that movement. Also, the sun (and earth) move around the galaxy at something like 230 km/s and the whole galaxy moves at even greater speeds. However, you need to remember that speed is relative: it's always measured against a reference point, such as the earth vs the sun, or you vs the earth. When measuring the speed of the galaxy, it varies, depending on which reference point (another galaxy?) you use.
Finally, unless the earth has 0 mass, it will always attract other objects. Hence, a plane "floating" above a tiny earth will fall towards the earth. It will not float sideways to another point. For that to happen, you need to give the plane some sideways momentum. If the earth has very small mass, then a tiny momentum will cause the plane to fly away completely, but it will not cause it to drift around the earth.
Two problems with hypothetical questions that either tweak the laws of physics or make sufficient changes to familiar objects (Earth, e.g.) is this: what remains 'normal' and how far can we go before nothing makes sense anymore?
In the context your last inquiry, "what does is take to make a plane float?", one can do the following:
Since you mention the speed at the equator ($v=464\,$m/s), we'll keep that constant. With that, we can compute the centripetal acceleration using the only two parameters that I can remember off-line: the angular speed of the Earth is:
$$ \omega = \frac{2\pi}{\rm day} = \frac{2\pi} T $$
where $T=$86,400s.
The centripetal acceleration is:
$$\alpha = \omega^2 R$$
With the circumference of the Earth given by $2\pi R = Tv $, that becomes:
$$\alpha = \big[\frac{2\pi} T\big]^2 \frac{Tv}{2\pi}=\frac{2\pi v} T=\frac g {290}$$
where I have used the other parameter I know by heart, the gravitational acceleration at the Earth's surface:
$$ g = 9.8\,{\rm m/s}^2 = \frac {GM} {R^2}$$
(Aside: any coincidence that the inverse flattening of the Earth's ellipsoid is 277...which is a 3rd parameter I remembered from SRTM days).
At this point we run into one of the problems mentioned at the outset: what to do about the Earth?
I will be keeping the Earth's mass, $M$, fixed and change the radius.
The goal is to shrink the radius until the centripetal acceleration ($\alpha \propto 1/r$) equals the surface gravity ($g\propto 1/r^2$), and we immediately see we have a problem. The surface gravity at fixed mass grows with inverse radius faster than the centripetal acceleration at fixed velocity. We need to increase the radius.
Since:
$$ \frac{\alpha} g \propto \frac{1/r}{1/r^2}=r $$
we need to increase the radius by a factor of 290 for the plane to float at the equator. Quito to Singapore would be an easy, if not very long, flight.
Had we instead kept the length of the day fixed, along with fixed mass, I suspect the ratio of forces would scale as $r^3$, since $$\sqrt[3]{290}=6.6...$$ and geosynchronous orbits are at 6.6 Earth radii...which appears to be a 4th Earth fact...
Another choice is fixed density, $\rho$, so that surface gravity grows as:
$$ \frac{GM}{R^2}\propto \frac{\rho R^3}{R^2}\propto R $$
so that if we fix the length of day, then $\alpha/g$ is constant, and we're stuck.
In summary: truly hypothetical questions rarely have one correct answer.
