earth rotation

[shadow2121]

If the earths rotation sped up gradually, then would the centripital force eventually overcome gravity and lead to zero-gravity on earth?

[ecoli]

but centripetal force points inward...

[Kyrisch]

I'm pretty sure he's referring to the pseudo-force, often called centrifugal. The force needed to KEEP an object on the surface of a spinning Earth would be equal to [math]F_{centripetal} = M\frac{v^2}{R}[/math]. If we set [math]R[/math] to the radius of the Earth and then set the equation equal to gravity you get [math]F_c = M\frac{v^2}{6 378 100\, m} = M \,9.8 m/sec^2[/math]. The [math]M[/math] cancels and you get [math]v= 7 906 \,m/sec[/math]. The Earth's approximate circumference (since it is a spheroid) is [math]40 000 000 \,m[/math]. This means the angular velocity would have to be [math]\frac{7 906\, m/sec}{40 000 000 \,m} = 0.000198 \,rad/sec[/math]. Doing some dimensional analysis...

 

[math]\frac{0.000198\, rad}{sec} \, \frac{3600\, sec}{hours} \, \frac{1 \,rotation}{2\pi \,radians} \, \frac{24 \,hours}{day} = 2.72 \,rotations/day[/math].

 

So yes, if the speed of the Earth's rotation increased to 271% of what it is now, you would experience "zero gravity" because all of the Earth's pull would be dedicated to keeping you in a circular path, and not pulling you down.

[honestdude14]

Ah.. centrifugal is probably what he meant.Yes, it COULD be possible and it would matter where you stood on the earth at the time. Remember the earth, just as anything that rotates, rotatates about an axis. If you were to stand at a pole you would feel nearly no difference. However, if you were to stand at, say, the equator, you should feel it at maximum. This would take more than a gradual increase though. Just imagine what would happen if we saw night and day that quick. People would be having epilepsies.

[swansont]

I'm pretty sure he's referring to the pseudo-force, often called centrifugal. The force needed to KEEP an object on the surface of a spinning Earth would be equal to [math]F_{centripetal} = M\frac{v^2}{R}[/math]. If we set [math]R[/math] to the radius of the Earth and then set the equation equal to gravity you get [math]F_c = M\frac{v^2}{6 378 100\, m} = M \,9.8 m/sec^2[/math]. The [math]M[/math] cancels and you get [math]v= 7 906 \,m/sec[/math]. The Earth's approximate circumference (since it is a spheroid) is [math]40 000 000 \,m[/math]. This means the angular velocity would have to be [math]\frac{7 906\, m/sec}{40 000 000 \,m} = 0.000198 \,rad/sec[/math]. Doing some dimensional analysis...

 

[math]\frac{0.000198\, rad}{sec} \, \frac{3600\, sec}{hours} \, \frac{1 \,rotation}{2\pi \,radians} \, \frac{24 \,hours}{day} = 2.72 \,rotations/day[/math].

 

So yes, if the speed of the Earth's rotation increased to 271% of what it is now, you would experience "zero gravity" because all of the Earth's pull would be dedicated to keeping you in a circular path, and not pulling you down.

 

This doesn't feel right. The current effect is less than 1% of g, so tripling the speed isn't going to put me in orbit — that will only increase the effect by a factor of nine.

 

The error here is in this step:

 

This means the angular velocity would have to be [math]\frac{7 906\, m/sec}{40 000 000 \,m} = 0.000198 \,rad/sec[/math].

 

This gives you rotations/sec, not radians/sec, so you're off by a factor of [math]2\pi[/math]. The speed would have to be increased by a factor of ~17 (7906 m/s vs 463 m/s) for the centripetal acceleration to be equal to the gravitational acceleration.

 

(and this all assumes the the earth doesn't deform and fly apart, which it would)

[Sisyphus]

(and this all assumes the the earth doesn't deform and fly apart, which it would)

 

Interesting to imagine how that would happen, though. If it's just enough spin to overcome gravity at the equator, then only a narrow equatorial band of material would actually fly off, right? The rest would just deform, flatten out into a more oblate spheroid. I guess if it happened suddenly there's no way anyone would survive, but after everything cools off and settles (in a few million years or so), you'd be left with a very peculiar planet, no? The image is so strange that I feel like I must be missing something important.

[swansont]

Interesting to imagine how that would happen, though. If it's just enough spin to overcome gravity at the equator, then only a narrow equatorial band of material would actually fly off, right? The rest would just deform, flatten out into a more oblate spheroid. I guess if it happened suddenly there's no way anyone would survive, but after everything cools off and settles (in a few million years or so), you'd be left with a very peculiar planet, no? The image is so strange that I feel like I must be missing something important.

 

You have to realize that the deformation will lead to a redistribution of the mass — we're assuming a rigidity that isn't in fact present. This is apparent in tidal deformation (the earth has tides, too, not just water). So you'd have to account for that as well to get an accurate representation of what would happen. We're also ignoring any tensile strength of the earth. But under those assumptions, if one were close to such a speed, then any slight perturbation (earthquake? meteor?) would be enough to send chunks of material off into an orbit, which may then intersect again with the earth, repeating the process. And what of the tides I had mentioned earlier? They would be causing much more dramatic changes in the planet, and greatly affect the Roche limit

[Kyrisch]

This gives you rotations/sec, not radians/sec, so you're off by a factor of [math]2\pi[/math]. The speed would have to be increased by a factor of ~17 (7906 m/s vs 463 m/s) for the centripetal acceleration to be equal to the gravitational acceleration.

 

(and this all assumes the the earth doesn't deform and fly apart, which it would)

 

Ah, yeah. The number didn't seem quite right to me either but it was late and I figured if I did anything wrong someone would correct me. I needed to use the radius there, not the circumference.

[honestdude14]

Is the math above shown for an object at the equator? I believe this problem goes beyond newtonian mechanics([math]

F_{centripetal} = M\frac{v^2}{R}

[/math]). As I mentioned before, the centripital/centrifugal forces should change as you approach the poles.Am I wrong? If so, please explain.

[Kyrisch]

Yeah, this is for the equator, I used equatorial radius and circumference. There's an error though so the end result is a factor of [math]2\pi[/math] too small.

[honestdude14]

Ah.. centrifugal is probably what he meant.Yes, it COULD be possible and it would matter where you stood on the earth at the time. Remember the earth, just as anything that rotates, rotatates about an axis. If you were to stand at a pole you would feel nearly no difference. However, if you were to stand at, say, the equator, you should feel it at maximum. This would take more than a gradual increase though. Just imagine what would happen if we saw night and day that quick. People would be having epilepsies.

 

I understand now.The circumference [math]

c=2\pi*r[/math] needs to be divided by [math]\cos\Theta[/math] in order to account for the latitude. On the equator that value ends up being 1 anyway. Say you head 5 degrees up in latitude.. the change in velocity becomes less effective just as it should. At the poles, latitude is equal to 90 degrees giving you 0 difference along that axis.

Edited July 12, 2008 by honestdude14
extra symbol

[Klaynos]

I understand now.The circumference [math]

c=2\pi*r[/math] needs to be divided by [math]\cos\Theta[/math] in order to account for the latitude. On the equator that value ends up being 1 anyway. Say you head 5 degrees up in latitude.. the change in velocity becomes less effective just as it should. At the poles, latitude is equal to 90 degrees giving you 0 difference along that axis.

 

Yep

 

A nice extra point you raised and solved