My Friend Has A Science Question

[Lorizael]

I would like to know how much effect the centripetal force caused by the rotation of the Earth has on our apparent weight. That is to say a 444.8 newton man would get a 100 lb reading when he stood on a bathroom scale at the equator. If the same man stood on the same scale on a non-rotating mass equivalent to the Earth, what would the scale say he weighs?

My friend has spoken.

[Ari Rahikkala]

((444.8 N) - ((100 pounds) * 6 378 137 meters * ((2 * pi) / day) * ((2 * pi) / day))) / gravity on earth = 99.6510599 pounds

e: which is, uh, obviously the wrong calculation (the result should be heavier, not lighter), but my internet connection is wonky and someone will have pointed and laughed at me about it before I get this edit in

[Lancer]

Whatever the answer, this is really good news.

[Lorizael]

Originally posted by Lancer
Whatever the answer, this is really good news.

What, that I have a friend?

Originally posted by Ari Rahikkala
e: which is, uh, obviously the wrong calculation (the result should be heavier, not lighter), but my internet connection is wonky and someone will have pointed and laughed at me about it before I get this edit in

[snoopy369]

According to physicsforums it is around 0.3% (1/290). The equation that determines it is here:




Ari, I don't know why you say it should be heavier. Centripetal acceleration increases weight (or rather increases the force a mass applies to the ground), not decreases.

[Kuciwalker]

I don't think that's technically an equation.

[Lancer]

Originally posted by Lorizael


What, that I have a friend?

Nah, because I'm heading to the tropics and could afford to lose a little weight.

[Ari Rahikkala]

Well, let's just get the damn symbols in (although I've forgotten what the appropriate symbols are so I'll just use words) and solve it, because I want to answer the original question: If the scale says you weigh 100 lbs, and you're on the equator, what would it say if the Earth suddenly stopped rotating?

Assuming scale manufacturers don't take rotation into account and just assume that the force exerted on the scale times mean gravity on sea level is the actual mass... if they did consider rotation somehow we'd have reportedMass=actualMass

reportedMass = (gravity * actualMass - angularRate^2 * radius * actualMass) * gravity
actualMass = reportedMass / (gravity - angularRate ^ 2 * radius) * gravity

actualMass = 100 lbs / (gravity on earth - (2 * pi / day) * (2 * pi / day) * 6378137 m) * gravity on earth = 100.345145 pounds

Snoopy: Doesn't this sound about right?

[snoopy369]

No, because if the earth stopped rotating you'd lose reported weight, not gain it, shouldn't you?

edit: Or are you saying that actual weight is what you weigh with the earth rotating?? Also, weight, not mass...

[Ari Rahikkala]

Mass is reported in kilograms (or pounds), weight in newtons, right? I think I'm consistent about that in my second post.

Snoopy, imagine accelerating the rotation of the earth. It seems to me when I visualise it that people would fly off at the equator if it's fast enough (). This motivates me to think that faster rotation means a lower weight and thus lower reported mass, and slower rotation a higher reported mass. If the scale manufacturers assume no rotation, then reported mass should approach actual mass as rotation approaches zero.

[snoopy369]

Mass = grams
weight = pounds or newtons

Pounds are used by the english system to measure mass, only because they did not differentiate between the two when it was invented.

In any event, mass does not change with gravity, or acceleration, or anything else ... mass is mass. Weight, or "the force applied by a body at rest on the earth or a proxy thereof", does change.

On the other hand, you're right, I just was thinking backwards

[SpencerH]

Why would centripetal force effect weight? IIRC from my physics classes some 25 years ago, the effects of centripetal force are perpendicular to the effects of gravity.

[snoopy369]

IANAP, but the way I understand it is that your (perceived) weight is affected slightly because some small portion of the force of gravity is redirected to keeping you on the earth (as explained above by Ari) instead of flying off; so instead of directly down, the line of the force of gravity goes just slightly diagonal - enough to lose 0.3% or so of its downforce.

[Kuciwalker]

Guys, this should be 100% physically intuitive

[Lorizael]

Originally posted by Kuciwalker
Guys, this should be 100% physically intuitive

Then why don't you answer the question yourself, ya big know-it-all?