

Subject:
Weight compared between equator and pole
Category: Science > Physics Asked by: eiermaaga List Price: $4.50 
Posted:
14 Mar 2006 03:35 PST
Expires: 13 Apr 2006 04:35 PDT Question ID: 707092 
If a 200 pound person stands on a scale on the equator does the scale show the same weight as if the person would stand on one of the poles? Stefan, Switzerland 

Subject:
Re: Weight compared between equator and pole
Answered By: pafalafaga on 14 Mar 2006 04:44 PST 
Stefan, Nice question. Gravity varies around the world, so that a person's weight varies as well. At the poles, the force of gravity is about onehalf of one percent greater than at the equator, so that a person weighing in at 200 pounds at the equator would add about a pound, and would weigh in at about 201 pounds at the poles. Of course (and as noted below, in the comments), it's unlikely that such a change would be noted, since a person's weight varies throughout the day anyway. But for a fixed weight of 200 pounds, the variation would be noticeable if the scales were carefully calibrated. Here's a site discussing the variation in gravity at the equator and poles: http://www.seed.slb.com/qa2/FAQView.cfm?ID=991 How much does gravity vary around the Earth and why? Hope that's what you needed...let me know if there's anything else I can do for you. pafalafaga search strategy  Google search on: [ gravity variation pole equator ] 

Subject:
Re: Weight compared between equator and pole
From: probonopublicoga on 14 Mar 2006 04:19 PST 
A person's weight varies anyway over the course of a day so it would be impossible to prove empirically. Moreover, clothing is a factor and a person at the pole might be well wrapped up whereas he/she might opt for nudity at the equator. 
Subject:
Re: Weight compared between equator and pole
From: liviofloresga on 14 Mar 2006 06:41 PST 
Depending on the type of scale used, your are comparing masses, and the mass does not vary, so an scale like the above ones will show the same weight, I guess: http://healthjournal.upmc.com/0105/Images/PhotoWeightLossSurg.jpg http://www.ancasterjewellers.com/diamonds/carat%20weight/scale.jpg 
Subject:
Re: Weight compared between equator and pole
From: qed100ga on 14 Mar 2006 07:44 PST 
"Depending on the type of scale used, your are comparing masses, and the mass does not vary, so an scale like the above ones will show the same weight, I guess: http://healthjournal.upmc.com/0105/Images/PhotoWeightLossSurg.jpg http://www.ancasterjewellers.com/diamonds/carat%20weight/scale.jpg" No, not really. If this were true then scales would indicate the same weight even if they were, for example, situated on the surface of the Moon. In fact, were this true, bathroom & hospital scales would show the same weight even under conditions of freefall, i.e., "weightlessness". Indeed, on the equator or on a pole, the mass (inertia) of an object is the same either place. But the average radius of the equator (from Earth's mass center) is greater than at the poles. Newton's law of gravitation is F = GMm/r^2 r is the radius from the mass center. And if that's the only variable to change between the two locations, then F will be less on the equator than on a pole. And yet, by all accounts the inertial and gravitational masses of an object are the same amount. They are the same thing. So if a hospital scale measures my mass, how does it show two different weights for different places? The answer is that such devices don't measure mass. They measure the value of F in the equation above, and F is determined not just by r, but also by the product of two gravitating masses: Mm. If M is Earth, and m is either me or my daughter, then indeed the number shown on the scale will be different (given a certain location) from that of my preteen child, who is smaller and has less body mass. The product Mm for me is greater than the product Mm for my kid. This is even though there is a law of physics that, under otherwise identical conditions, two different masses fall due to gravity with equal acceleration. So how do you measure the actual mass of a body? How do astronauts falling freely in a space station determine their body masses? (They need to know this for purposes of collecting medical data.) They measure not their weight, but their inertia. Newton's basic force law is this: F = ma / Force equals the product of mass & acceleration This says that if I push on an object with a given mass and it changes speed with a certain acceleration, then multiplying the mass and the acceleration tells the force. But this can be rearranged to be F/a = m So if I give an astronaut a shove with a known force and measure how much its speed changes over a given time, then the acceleration can be derived. Doing the division F/a then tells the astronaut's body mass. 
Subject:
Re: Weight compared between equator and pole
From: inigomontoyaga on 14 Mar 2006 14:52 PST 
Don't forget to take into account the centrifigal force of a rotating planet. If you're wondering about the reading on the scale (and not just the differing gravitational pull), this has some effect, however small it may be. Given a equatorial circumference of 24902 miles, a person standing on the equator is traveling around the earth's axis at roughly 1038 miles per hour. The guy standing on the pole isn't moving any more than someone standing on the centerpoint of a merrygoround. Somebody else will have to do the math. My high school physics and calculus are haven't gotten much use these past twenty years. Cheers. 
Subject:
Re: Weight compared between equator and pole
From: manukaga on 15 Mar 2006 18:45 PST 
qed100, you have completely missed livioflores' point! Did you even look at the pictures linked? They show two types of balance scales, which work by comparing two mass against each other. These scales will indeed give the same reading on the Moon as they would on the Earth, because the 200pound person (who now weighs 33 pounds) will still be, for example, 10 times heavier than the 20pound reference weight (which now weighs 3.3 pounds). They won't work in freefall, however, contrary to your opinion; the gravitational pull has to be strong enough to overcome frictional forces in the pivot and allow the arms to swing. 
Subject:
Re: Weight compared between equator and pole
From: qed100ga on 16 Mar 2006 04:39 PST 
Hi manuka, Yes, I did look at the illustrations in the links. My intuition told me that there must be a continuous change in the behavior of the scales with the strength of gravity. After thinking about it I see that you & livioflores are correct that the balance scales, upon reaching equilibrium, will tend to show the same value both on Earth's surface and the Moon's. But, my intuition was correct; the scales will still behave differently, and it'll still be possible to measure the differences in weight for the two different surface gravities. Here's why: The two masses on a balance scale oppose each other with their respective torques. The masses on the ends of the moment arms stay equal. The lengths of the arms stay equal. The accelerations due to gravity change. Thus, the torques on the masses change. And although the ratio of the two masses' torques remains the same in both situations for all horizontal projections of the moment arms, the actual torques scale with gravity, and the time required for the scale to reach equilibrium will also change. It is this length of time which will tend to approach zero as the value of g approaches zero, and by measuring the time, the mass in question can be determined as a multiple of the reference mass. Thanks. I appreciate seeing the problem more clearly now. 
Subject:
Re: Weight compared between equator and pole
From: bipolarmomentga on 24 Mar 2006 17:01 PST 
"Don't forget to take into account the centrifigal force of a rotating planet. If you're wondering about the reading on the scale (and not just the differing gravitational pull), this has some effect, however small it may be." Great point, It's not insignificant. Equatorial radius is roughly 6378135 meters and the velocity is about 468.5 m/s. Angular acceleration is vē/r which offers an opposing acceleration of 0.0344 m/sē. That's about a third of a percent of the calculated acceleration due to gravity at the equator/sealevel of 9.822, and agrees closely to the measured value of g at the equator of 9.789 m/sē. Though upon reference to the link the answer, the value of g used in the response was from a measurement and not a calculation and thus was included in the result. Oh well, fun anyway. 
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