Why doesn't the Earth's rotation throw us off the surface?

The Earth is spinning at about a thousand miles per hour. Shouldn't we be thrown from its surface?

Yes I know someone would comment "gravity" but physically we should be thrown from Earths surface. Scientists say that gravity is holding us here but that seems to contradict physics. Maybe someone can better explain it to me.

• Weren't we just asked this? Nov 12 '17 at 3:15
• Yeah, we got nearly this exact same question a few days ago. Nov 12 '17 at 3:27
• The other question was closed. This one is asked a bit better. It could be a sincere question. astronomy.stackexchange.com/questions/23440/gravitational-field Nov 12 '17 at 4:27
• -1 for the comment "scientists say that...", Since the only answer you are going to get is based on science. Possibly this is an idiotic flat-earth question. Nov 13 '17 at 7:55
• -1 for "but that seems to contradict physics". Show us your physics then. And I mean the calculations. Otherwise there is a perfectly good answer by StephenG. Apr 30 '19 at 20:27

OK, the reason we don't get flung off the surface of the Earth is that the rotational forces are not large enough to do it.

Keep in mind that Earth formed because material was pulled together by it's own gravity. If Earth rotated so quickly, that material would be thrown off and Earth, as we know it, would not have formed. That's a bit simplistic, but it's reasonable.

The force of gravity we experience is :

$$F_g = g_e M_o = \frac {GM_eM_o}{R_e^2}$$

Where the "e" subscript is for Earth and the "o" for the person.

This is the basic Newtonian equation for gravitational force.

The gravitational force is always directly towards the center of the Earth.

Now the equation for the force we experience due to rotation is a bit more complicated looking because we rotate about an axis (north-south) and the force from rotation varies with latitude.

$$F_c = M_o \frac {v^2} r$$

And :

$$v = \frac {2\pi r} T$$

$$r = R_e \cos\theta$$

Where $$T$$ is the period of rotation and $$\theta$$ is the latitude.

That works out as :

$$F_c = M_o \frac {4\pi^2r}{T^2}=M_o\frac{4\pi^2R_e\cos\theta}{T^2}$$

And the direction of that force is perpendicular to the axis of rotation. So we need to find out how much force we experience from it radially - away from the center and that is :

$$F'_c = M_o \frac{4\pi^2R_e \cos^2\theta}{T^2}$$

For us to be flung off the Earth we'd need $$F'_c>F_g$$. So is it ?

The largest value for $$F'_c$$ is at the equator ($$\theta = 0$$) and we get (in SI units) :

\begin{aligned} F'_c &= M_o \times 0.034 \;{\rm m/s^2} \\ F_g &= M_o \times 9.8\phantom{00} \;{\rm m/s^2} \end{aligned}

So the rotational force is simply too small to overcome gravity. We can't get thrown off the Earth.

So how fast would Earth have to rotate that we would get thrown off the Earth ?

That would need :

$$\frac{4\pi^2R_e}{T^2} > g_e$$

So :

$$T^2 < \frac{4\pi^2R_e}{g_e}$$

$$T < 2\pi\sqrt{\frac{R_e}{g_e}} = 5066 \text{ seconds}$$

So we'd need a day shorter that one and a half hours to be flung off the planet. For the brief period we could cling on desperately, that would be dizzying.

If Earth rotated that fast it would not last long anyway - first the surface would be separated and then the inner layers would be flung off. Earth would not have formed in the first place (at least not in anything like it's current form) because we need gravity to be strong enough to keep things together.

Also note that it is because of the rotational forces that Earth is not quite a sphere - it's a bit swollen at the equator than the poles (because the net force inward at the poles is larger while at the equator there is this small but appreciable force from the rotation). This slight oblateness does change the numbers a tiny bit, but not much and not worth the effort of going into detail.

• For those who get dizzy when seeing too much math, here a tl;dr: Gravity is about 30 times stronger than the centrifugal force on the equator and even stronger elsewhere. Nov 12 '17 at 15:30
• TYPO (cite:) "a bit flatter at the equator" is incorrect and contradictory.. You rather mean: - a bit more swollen at the equator and a bit flatter at the poles. :) You're ~21km more distant from Earth's centre on the equator. Apr 30 '19 at 18:51

It doesn't contradict physics, though it can be confusing. I've read that Ptolemy rejected the rotating Earth hypothesis because he felt it should generate very high winds.

The "flying off" force, more correctly, the centrifugal force can be calculated and it's much smaller than gravity. (I'm not good at formulas, so I'll skip the math for now).

One way to think about this is to consider Newton's first law, and an object in motion stays in motion at the same speed and in the same direction. The Earth is large enough that standing on it's rapidly rotating surface feels like you're standing still because you're mostly moving in a straight line. It takes 6 hours to turn 90 degrees. Imagine being in a car and you feel the force to the side when the car turns. Now, imagine that car is moving very fast, but it takes 6 hours to turn left. The turn is slow enough that the "lifting" force in the case of the Earth or the sideways force in the theoretical car taking the 6 hour turn, is barely noticed.

The 1000 mph velocity at the equator isn't noticed at all because it's primarily in the linear direction with only a very slow turn, and being largely straight and constant at a constant velocity. The change in direction is too small to be noticable.

You need to take Earth's rotation into account for long distances. It affects Earth's prevailing winds and it needs to be taken into account with space-flight, missile launches and snipers, for very long distance shots, need to take the curvature and rotation of the Earth into account when shooting at far away targets.

But it's far to week to throw people off the Earth. Earth would need to rotate some 17 times faster for that to happen.