Why should you feel lighter at equator if normal force is weaker?

[fooly]

Ok forgetting everything about how Earth is not a perfect sphere,

the greatest centripetal acceleration is at the equator. So if you say N - Mg = -Mac, N should equal M(g-ac) which gives a weaker force than the gravitational force.

If the net force is a force towards the center of earth, why should you feel lighter at the equator?

 

[A.T.]

Ok forgetting everything about how Earth is not a perfect sphere,

the greatest centripetal acceleration is at the equator. So if you say N - Mg = -Mac, N should equal M(g-ac) which gives a weaker force than the gravitational force.

If the net force is a force towards the center of earth, why should you feel lighter at the equator?

Calculate N at the pole and compare to the equator.

 

[michojek]

has it got anything to do with traveling faster at the equator?

 

[haruspex]

Ok forgetting everything about how Earth is not a perfect sphere,

the greatest centripetal acceleration is at the equator. So if you say N - Mg = -Mac, N should equal M(g-ac) which gives a weaker force than the gravitational force.

If the net force is a force towards the center of earth, why should you feel lighter at the equator?

Your puzzlement puzzles me. You accept that the normal force on you from the ground is less, so your force on the ground is less also, so wouldn't you feel lighter? What do you think determines how heavy you feel? How heavy would you feel in orbit?

 

[fooly]

Your puzzlement puzzles me. You accept that the normal force on you from the ground is less, so your force on the ground is less also, so wouldn't you feel lighter? What do you think determines how heavy you feel? How heavy would you feel in orbit?

I accept that the normal force would be less, but that is the force that points away from the ground. So your net force would be stronger than if you were at the poles. That's why I'm confused, if the force that is holding you up is weaker, wouldn't you feel heavier, not lighter?

 

[fooly]

Calculate N at the pole and compare to the equator.

N - Mg = Mac = 0

N = Mg

So at the poles you would feel your true weight. I'm just confused, if N = M(g-ac) at the equator, what does that mean? Doesn't that mean there is a weaker force balancing you with gravity? (so you would feel heavier, not lighter?)

 

[Muphrid]

If the force that is holding you up is weaker, wouldn't you feel heavier, not lighter?

People in orbit are not being held up by any force at all, yet they do not feel heavier. They feel weightless.

Perceived weight is exactly the normal force being applied.

 

[fooly]

People in orbit are not being held up by any force at all, yet they do not feel heavier. They feel weightless.

Perceived weight is exactly the normal force being applied.

I understand that Normal Force is meant to equal the weight we feel, I just don't understand why. Mg is pushing you down and N is holding you up, right?

 

[HallsofIvy]

"Weight" is the force holding you to the ground (or floor). If that force is weaker, then you feel lighter- your weight is reduced.

 

[Muphrid]

Do you accept that people in orbit experience gravity but no normal force yet feel weightless?

Have you ever been on an elevator? You feel heavier when the elevator suddenly accelerates upward and lighter when it accelerates downward.

Part of this is explained by general relativity. People don't ever feel gravitational forces. They move under the influence of gravity, but as far as anyone is concerned, if they are moving solely under the influence of gravity, they're moving under a free trajectory. Gravity is "fictitious" like the "force" that pushes you against the side of your car in a turn is fictitious. Just as the force that frame of the car applies against you to keep you in the cabin and in the turn is the only real force in the problem, so is the normal force the only real force, and one only ever feels the normal, which is why the normal is exactly equal to one's weight.

 

[HallsofIvy]

You don't need relativity to "explain" that- it is perfectly well explained by Newtonian mechanics.

 

[Muphrid]

You don't need relativity to "explain" that- it is perfectly well explained by Newtonian mechanics.

Is it? I don't think Newtonian mechanics gives any good reason why weight should be only the normal force, not including gravity. It's not logical to say X force counts while Y force does not for the purposes of some quantity. Whereas in GR noting that the four-force of a particle under the influence of only gravity is zero makes it reasonable to say that gravity is fundamentally different from any true force, and as such, weight cannot include gravity.

My impression of the history is that this result from GR forced a rethinking of how to define weight in Newtonian mechanics to be consistent with this notion. For this reason, I don't think it makes sense to say Newtonian mechanics fully explains it. The idea has been put in more or less by hand to be consistent with GR.

 

[Drakkith]

I understand that Normal Force is meant to equal the weight we feel, I just don't understand why. Mg is pushing you down and N is holding you up, right?

N - Mg = Mac = 0

N = Mg

So at the poles you would feel your true weight. I'm just confused, if N = M(g-ac) at the equator, what does that mean? Doesn't that mean there is a weaker force balancing you with gravity? (so you would feel heavier, not lighter?)

Hold on. Do you understand that the centripetal force you are calculating is what is needed to keep an object moving at that velocity at that radius?

Per wiki: Centripetal force is a force that makes a body follow a curved path: its direction is always orthogonal to the velocity of the body, toward the fixed point of the instantaneous center of curvature of the path. Centripetal force is generally the cause of circular motion.

At the equator the Earth rotates at 465 m/s. A 1 kg mass at a distance of 6378 km from the center of the Earth only requires a force of 0.034 Newtons to provide enough acceleration to keep it at that distance as it moves along in it's circular path. HOWEVER, we are talking about something on the ground, so we have to include that this object is actually feeling FAR more than the required force. 30 times more in fact. This extra force is due to gravity, and the difference between the two is what you would feel as weight. If the ground was not there you would spiral into the center of the Earth since you have too much force pulling you in and not enough velocity to counteract that force.

Edit: The centripetal force is provided by gravity, it is not something separate. Satellites in orbit are balanced between their velocity and the force of gravity to stay in orbit. Because the two are balanced, they feel no weight. Because we are not balanced, and because we have something to resist our falling motion,(the ground) we experience a Normal Force.

 

[D H]

Is it? I don't think Newtonian mechanics gives any good reason why weight should be only the normal force, not including gravity.

Sure it does. Astronauts don't feel gravity because it affects every part of their bodies equally (ignoring tidal gravity). Gravity itself doesn't induce any stresses or strains. What we feel are those stresses and strains due to different parts of our bodies accelerating differently.

 

[Muphrid]

Sure it does. Astronauts don't feel gravity because it affects every part of their bodies equally (ignoring tidal gravity). Gravity itself doesn't induce any stresses or strains. What we feel are those stresses and strains due to different parts of our bodies accelerating differently.

So what we perceive as weight is really what, the pressure exerted by the normal force on our bodies, and the overall difference is that normal forces are contact forces while gravity acts more or less equally on all parts of the body? Is that a fair reading of what you're saying?

 

[Drakkith]

So what we perceive as weight is really what, the pressure exerted by the normal force on our bodies, and the overall difference is that normal forces are contact forces while gravity acts more or less equally on all parts of the body? Is that a fair reading of what you're saying?

Pretty much.

 

[fooly]

Hold on. Do you understand that the centripetal force you are calculating is what is needed to keep an object moving at that velocity at that radius?

Ah, so what you are saying is that the Centripetal Force does not reflect the force humans have towards Earth?

If that is so, what causes the weight of a person to be lighter at the equator? Gravity at the equator is lower than elsewhere on Earth because of the Centripetal Acceleration, correct?

 

[Drakkith]

Ah, so what you are saying is that the Centripetal Force does not reflect the force humans have towards Earth?

Well, break it down a bit. What is causing this centripetal force? Gravity. If gravity is the source of both the centripetal force, and our weight, and part of the gravity is JUST to counteract our motion around the center of the Earth at 465 m/s, then the amount of force "left over" is what we have forcing us into the ground.

Sorry for the non-mathematical explanation.

 

[D H]

If that is so, what causes the weight of a person to be lighter at the equator?

Short answer to your question: It's the Earth's rotation. There's a direct effect and an indirect one.

First, we need a working definition of weight. In some physics texts you'll see weight defined as the product of gravitational acceleration and mass. In a few, you'll see weight defined as the sum of all real forces except gravity. Some call this latter definition "apparent weight". A few call it "scale weight" because this is what a spring scale measures. It's this scale weight that you are asking about. If you are standing still on the ground, your scale weight is the normal force the ground exerts on you (plus a tiny bit from buoyancy; I'll ignore that).

Why not use weight as mass times gravitational acceleration? The answer is that don't feel gravitational acceleration. From the perspective of general relativity, gravitation is a fictitious force; we don't feel fictitious forces. They aren't "real". From the perspective of Newtonian mechanics, gravitation is a real force, but we don't feel it because it affects every cell equally. Gravitation alone doesn't induce any stresses or strains in our bodies. Or in an accelerometer, for that matter. An accelerometer at rest on the surface of the Earth will register an acceleration of about 1g upward because accelerometers don't sense gravitational acceleration.

Back to the problem at hand, I'll look at things from the perspective of an Earth-fixed frame. This is a frame rotating with the Earth. In this frame, a person standing still on the surface of the Earth has zero velocity. The person isn't moving, so the normal force, gravitational force, and fictitious centrifugal force must add to zero. As noted above, it's the normal force that we feel as weight. Even if the gravitational force was the same at the equator and the poles, this outward centrifugal force will mean that the normal force at the equator is smaller than it is at the poles.

Now look at things from the perspective of an inertial frame. Here our person standing still with respect to the (rotating) surface Earth is no longer standing still. The person instead is undergoing uniform circular motion, one revolution per sidereal day about the Earth's rotation axis. The net force acting on the person is exactly that needed to result in this circular motion. In this frame, there are only two forces acting on the person, the gravitational force exerted the Earth as a whole and the normal force exerted by the Earth's surface. Do the math and the normal force from this perspective will be exactly the same as the normal force calculated from the perspective of the rotating frame. Real forces are the same in all reference frames. It's only the fictitious forces that are frame dependent. Once again, even if gravitational force is the same at the poles and the equator, the normal force will be smaller at the equator compared to at the poles.

Bottom line: Regardless of perspective, the Earth's rotation is directly responsible for a reduction in the normal force.

The Earth's rotation is also indirectly responsible for an additional decrease in the normal force. This direct cause assumes that gravitational acceleration is uniform over the surface of the Earth. This is not the case. The Earth's rotation makes the Earth take on the shape of an oblate spheroid. There's a bulge at the equator. The poles are closer to the center of the Earth than is the equator. This in turn means that gravitational acceleration is greater at the poles than at the equator.