weightless condition at the surface.
No more dieting. I suspect, no more day and night, just twilight.
But, seriously, physics wonks, how fast would it have to be for centrifugal force to offset gravity, at the surface (sea level?)
Centripetal acceleration is V[sup]2[/sup]/R, where V is the velocity at the object’s location, and R is the radius measured from the object to the axis of rotation.
In this case, you want a centripetal acceleration of 1 g, or 32.2 feet per second squared. R at the equator is about 4000 miles, or 21,120,000 feet. So:
a = V[sup]2[/sup]/R
32.2 = V[sup]2[/sup]/21,120,000
V = 26,078 feet per second
or
V = 17,780 MPH.
Not coincidentally, this is roughly the speed required to achieve low earth orbit (just a couple hundred miles above the surface of the earth).
Or, if you’d prefer a rotational rate rather than a linear speed, it’d be about one rotation every 90 minutes. Interestingly, (given that the Earth is spherical) this figure depends only on the average density of the Earth, not on its size.
But won’t the surface layers float off, reducing the mass of the Earth, requiring a slower rotation?
They would only float away if V[sup]2[/sup]/R is greater than 32.2 feet per second squared. And even then, it will only float up to a higher altitude for a finite period of time, and eventually come back together again. Travelling greater than about 17K MPH at a tangent to the current surface of the earth means you are at the near the periapsis of an elliptical orbit; you’ll eventually swing out to the apoapsis, and then return to your original altitude and speed.
This of course is not true if your speed at the surface if the earth is in excess of about 25,000 MPH, the escape velocity. If you exceed this speed at the surface of the earth, you will indeed float off into outer space and never return.
If the earth were rotating at this critical angular speed, the “weightless” condition would only hold at the equator.
The effect of the rotation on local apparent gravity would decrease with latitude, dropping to zero at the poles.
But rotating that fast the Earth would be a lot flatter, so g at the equator would be lower. If you don’t take that into consideration before starting up the Earth-speeder-upper you will rip the planet apart.
Sounds complicated. Have you tried dieting?
Don’t diet, try spinning.
Given that the interior of the Earth is hot & pressurized I’m not sure of that. Drastically lower the weight of the crust by rotation, and you’d have plenty of earthquakes & volcanic activity to push or outright blast matter outward.
Yes, that matter would be blasted upwards. But not ejected into space unless it’s traveling at escape velocity, instead you’d reshape the earth into a sort of disk. Except now the raised area around the equator would be below the “weightless” criteria, and you’d have to speed the whole thing up again to continue the ride.
I think at the forces we’re talking about we can treat the Earth as essentially liquid. Sure there will be a bit of variation at the surface, but spin the Earth harder and it’s going to bulge at the equator more. My intuition would be that eventually the equilibrium state would lead to equal gravity being felt all over the now bulging Earth. Another way to put that is that the equator will rise and the poles will sink until the forces pushing the equator out are balanced by the forces pushing the equator in, and that means you’d have uniform gravity as you walked around your new diskworld.
Amazing answers, people. Thank you so very much! I wonder if any ‘heavenly bodies’ that we have data for meet this sort of scenario, weightlessness at the surface because of centrifugal force. ??? I guess pulsars rotate so rapidly that they are certainly tearing themselves apart, unless they are very massive with gravity 1000’s of times that of earth?
I guess, then, we do actually weigh just a little less than our mass would seem to indicate, as a result of the 4% of rotational spin necessary to produce weightlessness?? And, we weigh less at the equator than at the poles?
Not equal gravitational force everywhere; equal gravitational potential (which is also what we have now). Assuming in both cases that you consider centrifugal force to be one component of gravity, but why wouldn’t you?
A great many heavenly bodies meet this description, including all galaxies and star clusters. Neutron stars (including pulsars) do not, however: They are rotating extremely rapidly, but they do have extremely strong gravity to hold them together anyway. They’re fairly close to tearing themselves apart, but they don’t quite do so.
This is correct.
Also, because the earth is an oblate spheroid, there is less mass at the poles, so wouldn’t we less because of the gravity?
I think that effect would be cancelled out by the fact that at the poles you are closer to the center, increasing gravity.
I calculated the gravity on a rapidly rotating planet once; I had to find the shape of the planet first (I forget how for the moment, but I can probably find the equation again if necessary), then calculate the centrifugal force at the equator and take that away from the gravity at the equator. The roataion period was 2hrs 39 mins, gravity at the poles was 1.25 g and the gravity on the equator was 0.6 g.
If a planet rotates much faster than that it starts getting unstable, possibly elongating into a prolate speroid like Haumea
? Didn’t get this. They’re at the tippy-top, no?
Gravity does increase at the poles of a prolate speroid, since you are closer to the centre of mass. These is a slight decrease because some of the mass is not directly under your feet, but that only starts to be significant if the planet is much more oblate than it could possibly be in reality.
The center of the Earth is at the intersection of the axis of rotation and the plane of the equator. For the slightly flattened Earth the distance to this center is shorter from the poles than from a point at the equator. Since gravitational force depends on the distance between the centers of mass of objects it is larger at the pole than at the equator.