Weight At North Pole And Equator

[MedPR]

Why do you appear to weigh more at the north pole than at the equator? I'm not really understanding TBR's explanation.


Edit: "way more" What the hell.

Sorry.

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[milski]

- the Earth is not an ideal sphere - its radius at the equator is larger, so you'll be further away from its center there and gravity will be lower.

- there is centripetal force on the equator which decreases your weight, there is not one on the pole.

I'm not which one of the two has a magnitude worth talking about.

 

[gettheleadout]

My very first thought was milski's second reason. Didn't even consider the non-ideal sphere, but I'm sad he/she beat me in responding so quickly.

 

[Danlee07]

Centripetal force is stronger at the equator, but you're still closer to the center of Earth's mass at the pole than at the equator. Since the equator is farther out, gravity would be stronger at the pole and then weight would be higher.

 

[milski]

And by centripetal force I actually mean what's usually referred to as centrifugal force - the opposite of centripetal. I should consider getting some real sleep one of these days.

 

[gettheleadout]

Centripetal force is stronger at the equator, but you're still closer to the center of Earth's mass at the pole than at the equator. Since the equator is farther out, gravity would be stronger at the pole and then weight would be higher.

This "but" is confusing. The centrifugal force lowers the body's weight at the equator, while the distance from the center of mass at the pole increases weight there. The different conditions of the two locations act to complement each other in exaggerating the difference in perceived weight.

 

[milski]

This "but" is confusing. The centrifugal force lowers the body's weight at the equator, while the distance from the center of mass at the pole increases weight there. The different conditions of the two locations act to complement each other in exaggerating the difference in perceived weight.

I typed 'centripetal' in my first post. Serves me right for typing too fast without thinking.

 

[MedPR]

See I actually thought the radius would be less at the equator, thus increasing the acceleration due to gravity..

F=GMm/r^2=mg

g=GM/r^2. Decrease radius = increase gravity?

 

[gettheleadout]

I typed 'centripetal' in my first post. Serves me right for typing too fast without thinking.

See I saw your post but didn't realize that Danlee was responding to it taking what you said literally instead of acting like you said centrifugal. No worries. The "but" makes more sense if he actually means centripetal like you "said."

See I actually thought the radius would be less at the equator, thus increasing the acceleration due to gravity..

F=GMm/r^2=mg

g=GM/r^2. Decrease radius = increase gravity?

The rotational motion of the earth causes centrifugal force on itself, causing it to expand around the equator, resulting in a larger radius. Increase the radius and the denominator in "g" gets bigger, resulting in a lower F.

 

[Danlee07]

I'm confused by what you guys are referring to... so I should've said "and" not "but"? Or the centrifugal instead of centripetal?

 

[milski]

I'm confused by what you guys are referring to... so I should've said "and" not "but"? Or the centrifugal instead of centripetal?

centrifugal & and

The centripetal is actually the gravity force in that case.

 

[Danlee07]

Sweet, fried brain

 

[milski]

Centripetal is the force that makes you go around in circle. In the case of an object on the equator, it's gravity that makes it go in that circle and stay on the surface of Earth.

Let's assume that the Earth is an ideal sphere.

An object on the equator is in a uniform circular motion. That means that the net force on it is not 0. The only way for that to happen is for the normal from the surface on which the object is to be slightly less than the gravity force on it. Since the apparent weight is actually the normal that the object exerts on the ground, on the equator it will be slightly less than the gravity force.

On the pole you can consider the object to be stationary and the normal and the gravity have equal magnitude. It will have slightly higher apparent weight.

The non-ideal shape of Earth only exaggerates that effect by further decreasing the gravity on the equator.