How fast would the Earth have to spin to cancel gravity at the equator?

[Raider5678]

I was reading a wikipedia article (https://en.wikipedia.org/wiki/Gravity_of_Earth) that said that gravity at the equator is weaker then gravity at the poles for a few reasons. First off the poles are flatter than the rest of the planet, meaning you're closer to the center of the earth, hence more gravity. The reason that they are flatter is because the earth is spinning, and the centrifugal force (if your into puzzle boxes: http://www.yot.com/ it involves centrifugal force) is making the equator bulge out. This bulge leads to our next reason the gravity is weaker. SInce you are farther from the center of gravity the gravity is weaker there, and since the planet is spinning the centrifugal force also decreases the gravitational pull on you. Now my question is how fast would the earth have to be spinning to cancel out gravity at the equator, or would it simple turn into a flat disk before the gravity canceled out. And if so, are there any known cases of a planet spinning so fast it's a flat disk without it flying apart?

[dimreepr]

Unless the spin is under constant acceleration then you seem to have got speed and acceleration all mixed up; the acceleration due to gravity should be thought of as a potential acceleration whilst a constant speed has no acceleration even in potential.

 

 

Edit to add/ The poles are only 43 km closer to Earths centre than the equator which equates to 0.03% off a perfect sphere, so any difference with regards to gravity is almost unmeasurable.

 

Another addition, the speed of the Earth at the equator is 1040 mph whilst at the pole the speed is effectively zero with almost no difference to gravity other than being 43 km closer to its centre.

Edited March 11, 2016 by dimreepr

[StringJunky]

Accounting for the oblate shape and gravitational acceleration at the equator and poles, the difference is is 0.5%.

 

 

Taking into account both of the above effects, the gravitational acceleration is 9.78 m/s² at the equator and 9.83 m/s² at the poles, so you weigh about 0.5% more at the poles than at the equator.

 

http://curious.astro.cornell.edu/about-us/42-our-solar-system/the-earth/gravity/94-does-your-weight-change-between-the-poles-and-the-equator-intermediate

[J.C.MacSwell]

Unless the spin is under constant acceleration then you seem to have got speed and acceleration all mixed up; the acceleration due to gravity should be thought of as a potential acceleration whilst a constant speed has no acceleration even in potential.

 

 

Edit to add/ The poles are only 43 km closer to Earths centre than the equator which equates to 0.03% off a perfect sphere, so any difference with regards to gravity is almost unmeasurable.

 

Another addition, the speed of the Earth at the equator is 1040 mph whilst at the pole the speed is effectively zero with almost no difference to gravity other than being 43 km closer to its centre.

The spin is not accelerating, but all points off the rotational axis are. There is a net centripetal force as the resultant of gravity and other forces. There are tidal effects as well, so the effective gravity is not constantly the same at any point.

[EdEarl]

 

Universe Today

Space agencies take advantage of the higher velocities at the Earth’s equator to launch their rockets into space. By launching their rockets from the equator, they can use less fuel, or launch more payload with the same amount of fuel. As it launches, the rocket is already going 1,675 km/hour. That makes it easier to reach the 28,000 km/hour orbital velocity; or even faster to reach geosynchronous orbit.

The Earth would have to spin about 28,000 km/hr at the equator to launch things into space, which is almost 17 times faster than it now spins.

[swansont]

You have to assume the earth is a rigid body and these deformations we see aren't happening. Because the earth would fly apart.

 

But it's not just that gravity is weaker, even though that's the way it's presented (only the changes in radius affect the gravity). Newtonian gravity is unaffected by spin. What's happening is that if you are moving in a circle, a force is required to keep you in that path, known as a centripetal force. Gravity provides this. But what we feel and call gravity is actually the earth pushing up on is, (the normal force, since it's normal, or perpendicular, to the surface)

 

Ignoring rotation, for the moment, you have W + N = 0. The normal force and the weight are the same magnitude and opposite in direction. The net force is zero. If we're spinning, though, the net force is no longer zero, it's mv^2/r, which reduces the normal force. We feel lighter, but gravity is unchanged. The condition we're looking for is when N=0

 

mv^2/r = GMm/r^2

 

mv^2/r = GMm/r^2

 

v^2 = GM/r

 

A little less than 8,000 m/s at the equator, or about 28,500 km/hr

[Raider5678]

The Earth would have to spin about 28,000 km/hr at the equator to launch things into space, which is almost 17 times faster than it now spins.

Thank you!

 

You have to assume the earth is a rigid body and these deformations we see aren't happening. Because the earth would fly apart.

 

But it's not just that gravity is weaker, even though that's the way it's presented (only the changes in radius affect the gravity). Newtonian gravity is unaffected by spin. What's happening is that if you are moving in a circle, a force is required to keep you in that path, known as a centripetal force. Gravity provides this. But what we feel and call gravity is actually the earth pushing up on is, (the normal force, since it's normal, or perpendicular, to the surface)

 

Ignoring rotation, for the moment, you have W + N = 0. The normal force and the weight are the same magnitude and opposite in direction. The net force is zero. If we're spinning, though, the net force is no longer zero, it's mv^2/r, which reduces the normal force. We feel lighter, but gravity is unchanged. The condition we're looking for is when N=0

 

mv^2/r = GMm/r^2

 

mv^2/r = GMm/r^2

 

v^2 = GM/r

 

A little less than 8,000 m/s at the equator, or about 28,500 km/hr

Thank You! This helps a lot!