Back of the envelope
So, lunch conversation at work being typically eclectic, the subject came up of how fast we're moving (relative to the axis of the Earth) due to the rotation of the Earth.
Well. Here in sunny Cleveland, the lattitude is about 40 degrees. The radius of the Earth is about 6378 kilometers, and the rate of rotation of the Earth is 1 revolution per 24 hours, or 2pi rad/24 hr, which is 0.262 rad/hr.
The linear velocity of everything here in sunny Cleveland is then r*omega, where r is the distance to the rotational axis (not the center) of the Earth, and omega is the rate of rotation. So, linear velocity v = 6378 km*cos(40 deg)*0.262 rad/hr = 1279 km/hr, or about 795 miles/hr.
That inevitably raises the question, if the Earth abruptly stopped rotating, and everything on the surface (this means you) kept going at 1279 km/hr (795 mph) tangent to the surface of the Earth, how high above the surface of the Earth would you get before you face-planted going that speed? After all, the Earth is curved. Seems like you should get some altitude.
However, gravity is not your friend. Let's assume that a person has a center of mass 1 m above the ground. Let's also assume that the person's feet have been yanked out from under them by the stopping of the Earth's rotation, so they're in free-fall. We can break the person's motion up into two components, one normal (perpendicular) to the Earth's surface and one tangential (parallel) to the Earth's surface. Now, we can assume for the moment that the gravity ("normal") component of the resulting velocity is perpendicular to the linear velocity ("tangential") component calculated above, and do a calculation based on that. If the distance travelled by our hapless victim becomes significant compared to the radius of the Earth, that means our assumption that these components are always nearly perpendicular was bad, and we'll have to do a more complicated thing to figure distance.
So. How long would it take for our hapless victim's center of mass to fall 1 m, resulting in a face-plant? That's pretty easy: from highschool physics, d = (1/2)at^2, where d is distance, a is acceleration, in this case the acceleration of gravity at the surface of the Earth (9.8 m/s^2), and t is time. So t = sqrt(2d/a), or in this case, t = sqrt(2*(1 m)/(9.8 m/s^2)) = 0.45 s.
That's not quite the end of the story, since the curvature of the Earth could hypothetically save our victim from face-planting for a bit longer. So we need to know if we can safely approximate the Earth as flat over the distance the victim would travel.
Well, the linear velocity component v = 1279 km/hr = 355 m/s. The horizontal distance you'd travel before face-planting, approximating the Earth as flat, is d = vt. Using our time-of-fall calculation above for t, this is d = 355 m/s * 0.45 s = 160 m (about 160 yards, for you all still stuck in English units). That's 0.16 km, which is 4 orders of magnitude smaller than the radius of the Earth, 6378 km. We can therefore safely ignore the Earth's curvature as being insignificant.
In conclusion, what would happen if the Earth suddenly stopped rotating is that our hapless victim, resident of sunny Cleveland, would be thrown (at most) 160 yards and face-plant in (at most) 0.45 seconds going about 795 mph.
Well. Here in sunny Cleveland, the lattitude is about 40 degrees. The radius of the Earth is about 6378 kilometers, and the rate of rotation of the Earth is 1 revolution per 24 hours, or 2pi rad/24 hr, which is 0.262 rad/hr.
The linear velocity of everything here in sunny Cleveland is then r*omega, where r is the distance to the rotational axis (not the center) of the Earth, and omega is the rate of rotation. So, linear velocity v = 6378 km*cos(40 deg)*0.262 rad/hr = 1279 km/hr, or about 795 miles/hr.
That inevitably raises the question, if the Earth abruptly stopped rotating, and everything on the surface (this means you) kept going at 1279 km/hr (795 mph) tangent to the surface of the Earth, how high above the surface of the Earth would you get before you face-planted going that speed? After all, the Earth is curved. Seems like you should get some altitude.
However, gravity is not your friend. Let's assume that a person has a center of mass 1 m above the ground. Let's also assume that the person's feet have been yanked out from under them by the stopping of the Earth's rotation, so they're in free-fall. We can break the person's motion up into two components, one normal (perpendicular) to the Earth's surface and one tangential (parallel) to the Earth's surface. Now, we can assume for the moment that the gravity ("normal") component of the resulting velocity is perpendicular to the linear velocity ("tangential") component calculated above, and do a calculation based on that. If the distance travelled by our hapless victim becomes significant compared to the radius of the Earth, that means our assumption that these components are always nearly perpendicular was bad, and we'll have to do a more complicated thing to figure distance.
So. How long would it take for our hapless victim's center of mass to fall 1 m, resulting in a face-plant? That's pretty easy: from highschool physics, d = (1/2)at^2, where d is distance, a is acceleration, in this case the acceleration of gravity at the surface of the Earth (9.8 m/s^2), and t is time. So t = sqrt(2d/a), or in this case, t = sqrt(2*(1 m)/(9.8 m/s^2)) = 0.45 s.
That's not quite the end of the story, since the curvature of the Earth could hypothetically save our victim from face-planting for a bit longer. So we need to know if we can safely approximate the Earth as flat over the distance the victim would travel.
Well, the linear velocity component v = 1279 km/hr = 355 m/s. The horizontal distance you'd travel before face-planting, approximating the Earth as flat, is d = vt. Using our time-of-fall calculation above for t, this is d = 355 m/s * 0.45 s = 160 m (about 160 yards, for you all still stuck in English units). That's 0.16 km, which is 4 orders of magnitude smaller than the radius of the Earth, 6378 km. We can therefore safely ignore the Earth's curvature as being insignificant.
In conclusion, what would happen if the Earth suddenly stopped rotating is that our hapless victim, resident of sunny Cleveland, would be thrown (at most) 160 yards and face-plant in (at most) 0.45 seconds going about 795 mph.
posted by Craig at 7:29 AM
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