Human weight changes on your location

[Crepitus]

Yes the further from the center of mass ..you receive less gravity force

Yes, but that is a different discussion. If the earth were a perfect sphere, you would still weigh less at the equator than at the poles, per a scale sitting on the surface.

Because the rotational speed is faster at the equator; hence , the planet is trying to “ fling you off more “

Technically gravity is warping you momentum into a circle. If gravity were to suddenly stop you go flying off at a tangent to the surface.

 

[Quasar44]

In a free fall from orbit to surface : you would be weightless and have zero weight

*assuming the scale's frame of reference is the same as your own

Given recent confusion, i think it bears repeating.

Yes the scale is also falling with you but on your feet

 

[ReinyDays]

In a free fall from orbit to surface : you would be weightless and have zero weight

If I was falling ... then the force of gravity is pulling me down ... thus I would have weight ... the confusion comes when I stand on a scale, which is also in free fall, and the scale reads "0" ... all that means is both the scale and I are experiencing the same acceleration ... and the forces applied is proportional to our respective masses ...

Perhaps it would help to understand that orbits are free falls ... an object in LEO must maintain a speed of about 17,000 mph ... as this object falls to Earth, the Earth curves away from the object at the same rate ... thus keeping the object at the same elevation above the Earth ... this is caused by gravity, thus the object has weight, without weight or gravity, the object would fling out into space in a straight line ...

 

[ReinyDays]

Very true, in that the scale is also under acceleration. I admit, in my comments, i ignored this. But i ignored it for a reason. Scales are calibrated to register force. As it is true in our scenario that the scale also has the same inertia as the person on itt, it is not it is not doing extra work on the person at the equator. They are in the same frame, moving identically. Both will experience less.force in the downward component. Calibrated scales register force.

*assuming the scale's frame of reference is the same as your own
Given recent confusion, i think it bears repeating.

What you're not taking into consideration is that the observer is also in this same frame-of-reference ... and is also under acceleration ... if the observer doesn't take this into consideration, and assumes they are not accelerating, then everything the observer sees with show the centrifugal force ... of the same magnitude, but opposite direction of the force the observer to experiencing ... thus the observer sees the person being weighed as weighing less ... when in fact, the same mass with the same acceleration due to gravity will give the same weight ... always ...

Inertia isn't changing, but momentum is ... the momentum vector has the same magnitude, but it is constantly changing direction ... our observer's momentum is also constantly changing direction exactly the same as the scale and person being weighed, and so to the observer it appears momentum is constant, and they conclude no work is being perform to change momentum, all the work is used to depress the scale ... and appears to be doing less work on the scale and the weighed ... giving a lower reading ... but this lower reading is strictly an artifact of the observer's non-inertial frame-of-reference ...

ETA: The reason this all sounds confusing is because it is confusing ... it was the end of the 17th Century before this was discovered ... and why we call this the Newtonian Revolution ... these were truly revolutionary ideas at the time ... perhaps could only have happened in England after she broke away from Rome ... profoundly non-intuitive ...

 

[Quasar44]

In a free fall from orbit to surface : you would be weightless and have zero weight

If I was falling ... then the force of gravity is pulling me down ... thus I would have weight ... the confusion comes when I stand on a scale, which is also in free fall, and the scale reads "0" ... all that means is both the scale and I are experiencing the same acceleration ... and the forces applied is proportional to our respective masses ...

Perhaps it would help to understand that orbits are free falls ... an object in LEO must maintain a speed of about 17,000 mph ... as this object falls to Earth, the Earth curves away from the object at the same rate ... thus keeping the object at the same elevation above the Earth ... this is caused by gravity, thus the object has weight, without weight or gravity, the object would fling out into space in a straight line ...

You just nailed it !! You clearly know your stuff . If you were going 16,000 mph then you would just crash to the earth and die ??
7 miles per sec would break orbit

 

[ReinyDays]

You just nailed it !! You clearly know your stuff . If you were going 16,000 mph then you would just crash to the earth and die ??
7 miles per sec would break orbit

Pretty much, it would take longer, but hitting the atmosphere at 16,000 mph would burn up about anything ...

Of course, the further the orbit is from the surface, the slower orbital speed is, Kepler's Third Law ... for a 23,000 mile orbit along the equatorial plane, one need only go 6,800 mph; and this would keep us over the same point along the equator throughout the orbit ... this is where weather satellites are, and why they photo the same spot on Earth every shot ... the so-called "geostationary orbit" ...

 

[Fort Fun Indiana]

when in fact, the same mass with the same acceleration due to gravity will give the same weight ... always ...

But the scale is not measuring this. It is measuring total force perpendicular to the surface. While the force imparted by gravity is constant, the amount of net force perpendicular to the surface is not.

I think where you and i diverge is that you are defining weight by the force imparted on the person by gravity, while i am defining it as the downward force measured by a scale. We are both correct, in these different contexts.

 

[ReinyDays]

when in fact, the same mass with the same acceleration due to gravity will give the same weight ... always ...

But the scale is not measuring this. It is measuring total force perpendicular to the surface. While the force imparted by gravity is constant, the amount of net force perpendicular to the surface is not.

I think where you and i diverge is that you are defining weight by the force imparted on the person by gravity, while i am defining it as the downward force measured by a scale. We are both correct, in these different contexts.

Ah ... I see ... my nose has been buried in my dog-eared physics textbook ... which defines weight as equal to mass times gravitational acceleration ...

So you'd say an object in orbit is "weightless", as well as someone floating in water ... and indeed we generally speak of a ship's displacement, rather than it's weight ...

Yours is certainly the intuitive, or "common sense", approach ... and there's nothing wrong with that ... I step on a scale, I see 161 lbs, I'm not going to bother multiplying some centrifugal factor by cosine 43.5º to get my weight due to gravity ... and in any other forum, I would be considered as "splitting the hare" and making a bloody mess of a simple question ... however, with so much unknown in science, and all the controversy this allows, I do think in this forum we should stick to the "proper" definition of these words, or maybe I should say the deductive definition ...

The OP is correct, even using gravitation weight ... due to the equator being further from the center of gravity than the poles ... as well as the fact that the Earth's density is irregular ... one would weight more standing over a denser spot and weigh less over a less dense spot ...

(Pseudo-forces is something of a pet peeve of mine ... but more so the "other" one ... I'd rather a red hot poker be stuck up my ass than hear anyone say Coriolis force causes cyclonic motion ... even when the world's foremost expert in cyclogenesis, Mark Landsea of the NHC, says it ... and by "pet peeve" I mean I let it up on my bed and night to cuddle, provide it a healthy diet and take it to the vet's office regularly) ...

 

[Fort Fun Indiana]

when in fact, the same mass with the same acceleration due to gravity will give the same weight ... always ...

But the scale is not measuring this. It is measuring total force perpendicular to the surface. While the force imparted by gravity is constant, the amount of net force perpendicular to the surface is not.

I think where you and i diverge is that you are defining weight by the force imparted on the person by gravity, while i am defining it as the downward force measured by a scale. We are both correct, in these different contexts.

Ah ... I see ... my nose has been buried in my dog-eared physics textbook ... which defines weight as equal to mass times gravitational acceleration ...

So you'd say an object in orbit is "weightless", as well as someone floating in water ... and indeed we generally speak of a ship's displacement, rather than it's weight ...

Yours is certainly the intuitive, or "common sense", approach ... and there's nothing wrong with that ... I step on a scale, I see 161 lbs, I'm not going to bother multiplying some centrifugal factor by cosine 43.5º to get my weight due to gravity ... and in any other forum, I would be considered as "splitting the hare" and making a bloody mess of a simple question ... however, with so much unknown in science, and all the controversy this allows, I do think in this forum we should stick to the "proper" definition of these words, or maybe I should say the deductive definition ...

The OP is correct, even using gravitation weight ... due to the equator being further from the center of gravity than the poles ... as well as the fact that the Earth's density is irregular ... one would weight more standing over a denser spot and weigh less over a less dense spot ...

(Pseudo-forces is something of a pet peeve of mine ... but more so the "other" one ... I'd rather a red hot poker be stuck up my ass than hear anyone say Coriolis force causes cyclonic motion ... even when the world's foremost expert in cyclogenesis, Mark Landsea of the NHC, says it ... and by "pet peeve" I mean I let it up on my bed and night to cuddle, provide it a healthy diet and take it to the vet's office regularly) ...

Hah...i laughed about the "coriolis force" (just say "effect", people)

 

[ReinyDays]

when in fact, the same mass with the same acceleration due to gravity will give the same weight ... always ...

But the scale is not measuring this. It is measuring total force perpendicular to the surface. While the force imparted by gravity is constant, the amount of net force perpendicular to the surface is not.

I think where you and i diverge is that you are defining weight by the force imparted on the person by gravity, while i am defining it as the downward force measured by a scale. We are both correct, in these different contexts.

Ah ... I see ... my nose has been buried in my dog-eared physics textbook ... which defines weight as equal to mass times gravitational acceleration ...

So you'd say an object in orbit is "weightless", as well as someone floating in water ... and indeed we generally speak of a ship's displacement, rather than it's weight ...

Yours is certainly the intuitive, or "common sense", approach ... and there's nothing wrong with that ... I step on a scale, I see 161 lbs, I'm not going to bother multiplying some centrifugal factor by cosine 43.5º to get my weight due to gravity ... and in any other forum, I would be considered as "splitting the hare" and making a bloody mess of a simple question ... however, with so much unknown in science, and all the controversy this allows, I do think in this forum we should stick to the "proper" definition of these words, or maybe I should say the deductive definition ...

The OP is correct, even using gravitation weight ... due to the equator being further from the center of gravity than the poles ... as well as the fact that the Earth's density is irregular ... one would weight more standing over a denser spot and weigh less over a less dense spot ...

(Pseudo-forces is something of a pet peeve of mine ... but more so the "other" one ... I'd rather a red hot poker be stuck up my ass than hear anyone say Coriolis force causes cyclonic motion ... even when the world's foremost expert in cyclogenesis, Mark Landsea of the NHC, says it ... and by "pet peeve" I mean I let it up on my bed and night to cuddle, provide it a healthy diet and take it to the vet's office regularly) ...

Hah...i laughed about the "coriolis force" (just say "effect", people)

Hah...i laughed about the "coriolis force" (just say "effect", people)

The two terms are used interchangeably in the literature ... and it's pointed the wrong direction to have a roll in cyclonic motion ... just too stupid ... we have the convective force along our north-south axis and the pressure force along our east-west axis ... these two forces together create the torque we need to initiate a cyclone, as torque always requires two non-parallel forces ... and without torque, we have no rotation ... Physics 102 ...