Due to centrifugal force, is there a difference in weight between an object at the equator and at a pole?

Answer 1

yes, centrifugal force is a fictictious force which arises in a rotating frame of reference. For simplicity, at the equator F_c = m*w^2*r where w is the rate of rotation of the reference frame in radians per second and r is the distance from the axis of rotation....m is the mass of the object being accelerated {technically it's a vector quantity F_c = -m w x (w x r) where the x is the cross product}

Note: there is another reason things weigh less at the equator. The distance between the center of gravity and the surface of the earth is greater at the equator because the earth is an oblate sphereoid (flattened sphere) and the force of gravity is inversely proportional to the square of the distance.

Answer 2

If there were only gravitational force, and there were no spin about its axis, the earth will be a sphere.

In that case, there will not be any difference in the weights of an object at different places on the surface of earth.

But the spin of the earth had made the net force pressing the earth surface act at angle and not along the vertical when the earth was formed.

That is the CENTRIPETAL force acted so as to flatten the Earth; i.e. shift the matter toward the equator.

Such flattening has actually taken place, and the Earth’s shape is close to an ellipsoid of revolution.

As a result of this the equatorial radius is 1/300 greater than the polar radius.

When the process was completed the flattening ceased to be effective.

Now, the force exerted on the “globe” is directed normal to the surface.

(Remember once again that if the mass is not redistributed like this the force on the surface will be such as to flatten the earth toward equator due to the spin of the Earth)

At poles, the weight of the body is equal to the gravitational force. Let it be mg.

The weight at any place will be equal to mg - mR (omega) ^2 cos^2 (fi).

(fi) is the latitude of the place.

Or g’ = g - R (omega) ^2 cos^2 (fi).

At equator g’ is 1/300 less than g.

If we use an appropriate value for the acceleration of a freely falling body at each of the various latitudes, we do not have to calculate the effect of Earth’s rotation on the weight of the body.

So far we have seen only the weights of stationary objects with respect to earth. But when objects move relative to earth, they do have additional or less speed with respect to earth.

Due to this relative speed , the weight ( the pressure they exert on the surface)will varry.

If a body moves approximately with a speed 8km/s then its weight will be zero. I t will move round the surface of earth without pressing the surface.This is called orbital speed.

Answer 3

Yes. And because of the centrifugal force, the Earth bulges slightly at the equator. This puts the surface at the equator a little farther from the center of the Earth than elsewhere, further lessening the force of gravity. So at the equator, you weigh about 0.6% less than at the poles.

Answer 4

Yes, but indirectly. The earth isn't a perfect sphere; it actually bulges at the equator somewhat. Therefore, an object at a pole is closer to the Earth's core than an object at the equator, meaning that the object at the pole will weigh slightly more.

Answer 5

Yes but very little.. but the main difference is due to the variations in gravitation forces and the fact that the earth is flat at the poles and rounder near to the equator.

Answer 6

placed purely, centrifugal is the incorrect theory that a mass compelled to go back and forth in a circle (weight on a string etc) ought to if released, go back and forth immediately faraway from the centre of the circle regardless of the indisputable fact that that's not authentic because the mass ought to go back and forth in a immediately line contained in the route it became shifting, from the point it became released (a tangent to the sting of the circle) and that is centripetal A drawing can make it a lot a lot less complicated!!

Answer 7

There is, but it is so slight as to be washed out by the difference in gravity caused by varying densities of matter around the globe.

Answer 8

slightly because the gravitational constants are slightly different at 2 places

Answer 9

yes there is a difference.it is because of distance and velocity(speed)