On which part of the earth does a person weigh the most?

Question

North pole, south pole or at any part of the equator of the earth

Answer 1

sigh, i can't let this one go unanswered... some of the answers are just too painful (ie "increases linearly along the radius..." blah blah blah. why BS so much if your answer is wrong?).



anyway, the answer... you would weigh the most at either of the poles. the earth is not round (NOR IS IT PEAR SHAPED!!!), but is rather flattened slightly into a disk. whoever said death valley is the lowest point on earth apparently doesnt understand the difference between height compared to sea level and height relative to the center of the earth. for this reason the poles are, on average, closer to the center of the earth than any point along the equator.



cutting any fancy pants BS out of the answer and playing it to you straight, gravity is a function which is INVERSE (or divided by) the distance between the two objects. that means that the closer the centers of mass of the two objects are, the stronger the gravitational force between them. in other words, you wouldn't want to be on the equator at all because you'd be farther out... and to say that those at the equator have more mass under them is ridiculous. for calculating gravitational effects, all objects can be treated as a point mass located right at the center of mass, so being at the equator doesnt help AT ALL!!!



second, if you've ever ridden a childs little spinning toy at the park (can never remember what those things are called) and you stand at the center, you spin around but there's not any great pull out to the sides. that's akin to standing on one of the poles of the earth. now, if you stood out to the side of this childs toy, we all know that the inertia of spinning (what we call centrifugal force) causes you to feel like you're being thrown out to the sides. this is like standing on the equator, where that feeling of being thrown out to the sides makes you feel lighter. thus being at the equator both increases your distance from the earths center of mass AND has the added negative caveat that centrifugal force is throwing you out from the planet as well.

Answer 2

The Earth is an oblate spheroid. That means very, very slightly pear-shaped.



You weigh more if you are closer to the center of mass, so I would say you weigh the most at the South Pole.



We are talking about a ridiculously tiny fraction of a gram, remember. You will never feel the difference.



I did not make up the pear-shaped thing. Quote:



"Even the oblate-spheroidal notion of the earth is wrong, strictly speaking. In 1958, when the satellite Vanguard I was put into orbit about the earth, it was able to measure the local gravitational pull of the earth--and therefore its shape--with unprecedented precision. It turned out that the equatorial bulge south of the equator was slightly bulgier than the bulge north of the equator, and that the South Pole sea level was slightly nearer the center of the earth than the North Pole sea level was.



"There seemed no other way of describing this than by saying the earth was pear-shaped."



Unquote.



Maybe the information is old, but that's what I found. Now that Promethius (below) has his degree in physics, perhaps he can move on to charm school.



As for David M.'s novel definition of weight, he seems to be talking about mass. Mass is constant. Weight changes depending on the strength of the gravitational field acting on a mass, which is why you'd be only 1/6 your normal weight if you went to the Moon. Don't tell David that, though.

Answer 3

I have to disagree with David. A pound is a unit of force. A person's weight is the force with which their body is attracted toward the center of the earth. Their mass is always compared to a standard. A 1 kilogram steel mass has a different weight when measured on the moon.



So, distance from the center of the earth (yes, you do weigh less on top of Mt. Everest) and a small amount of centrifugal force (equator vs. poles), can both have a measurable effect on your weight.



However, since we are measuring weight and not mass, your weight changes significantly if you are floating in water or in some other medium where your bouyancy would have an effect.



Rephrasing your question to be, where on the surface of the earth will apparent gravity me minimized, the answer is somewhere near the North Pole.

Answer 4

Kalel's answer is interesting, but Promethius is right on target. He knows what he's talking about. Some of the answers are in "outer space", especially those that say you weigh more at the equator because there's "more mass" beneath your feet.



Actually, there are three things going on here. To see all of it, imagine that the spinning earth is more flattened at the poles than it is, and more bulging at the equator.



First, because the force of gravity varies as the square of the distance to the center of mass, you'd weigh at the poles because you're closer to earth's center.



Second, the centripetal force tending to make you fly off the earth (due to the earth's rotation and your distance from the axis of rotation) will increase as your distance from the axis of rotation increases. This is zero at the poles, and reaches a maximum at the equator. Note that this force, which weight, is "horizontal", directed away from the axis of rotation, not away from the earth's center of mass.



Third, imagine that you're neither at a pole, nor at the equator, but somewhere in-between, such as at 45 degrees latitude. If you resolve all the acting forces into two components -- one directed toward the axis of rotation, and the other directed towards the equatorial plane (sort of like cylindrical coordinates) -- then the latter force will be slightly greater than the former, with the effect that the earth (if it were gaseous) would tend to flatten out like a spinning pizza crust. (It wouldn't flatten out completely, because the gravitational component directed toward the axis of rotation greatly exceeds the oppositely directed centripetal force; that's why Jupiter is not flat.)



This last is the reason that the solar system is flat.

Answer 5

I believe it would be at the equator. The Earth is not a perfect spheroid, but rather is a bit flattened out at the poles due to its own rotation. So the equatorial diameter is greater than the polar diameter. The Earth is sufficiently large and uniform that I believe it behaves locally like a sphere with radius equal to the local radius. Since the mass of a body with uniform density increases cubically with radius, and gravity weakens with the inverse of the square of the radius, the gravitational force due to a body's mass at the body's surface increases linearally with radius. So at the point of greatest radius, in this case the equator, a person's weight is probably the greatest. (In other words, you're further from the Earth's center of mass, but there's also more mass directly under you.) Since the equatorial radius is only .34% greater at the equator than at the poles, the differences are negligible on the scale of a typical individual's weight.

Answer 6

Congratulations!!!!!! You asked a question and got 100% wrong answers!!!



People are confusing the shape of the earth, gravity, density of the air etc and failing to examine the defined properties of weight.



Weight on earth is measured with a number of scales but lets use the pound. A solid piece of metal is made that at a fixed position and height on earth is held to be exactly one pound. That will be the only place on earth that it will weigh exactly that but the importance is that that bilt of metal is the defined pound.



Where ever you weigh a body on earth, whether it be at the poles, in heat, in cold, below sea level or at the top of Everest. The definition of your weight that is used is the comparison between you and that same reference bilt of metal at the exact same spot you are at, at the time. It therefore cancels any difference in gravity, air density or any other variable. Those are only apparent variations and are false.



The right answer therefore can only be that you weigh exactly the same wherever you are on earth.

Answer 7

Theoretically it should be at the highest point of the equator ,depending on the distance between the body and the center of the earth ,so I supposed that this distance will be greatest at the equator,but it could be anywhere else like the mount Everest ,as I said depending on the distance to the center of the earth.

The earth is oblate spheroid ,the radius to the equator is greater than its to the poles and the gravity comes from the mass so more mass means more gravity .

Answer 8

the Jennie Craig Clinic.



The answer is not so much determined by whether you are at the poles or the equator, but rather inconsistencies in the Earth's spherical shape or uniformity.



Centrifugal force also plays a role, I'm not sure how much, so maybe the poles.

Answer 9

It might be interesting to know the Heaviest Man instead...



The heaviest person in medical history was Jon Brower Minnoch (USA, 1941–83), who had suffered from obesity since childhood. He was 185 cm (6 ft 1 in) tall and weighed 178 kg (28 st) in 1963, 317 kg (50 st) in 1966 and 442 kg (69 st 9 lb) in September 1976.

In March 1978, Minnoch was admitted to University Hospital, Seattle, where consultant endocrinologist Dr. Robert Schwartz calculated that Minnoch must have weighed more than 635 kg (100 st), a great deal of which was water accumulation due to his congestive heart failure.



In order to get Minnoch to University Hospital, it took a dozen firemen and an improvized stretcher to move him from his home to a ferry-boat. When he arrived at the hospital, saturated with fluid and suffering from heart and respiratory failure, he was put in two beds lashed together. It took 13 people just to roll him over.



After nearly two years on a diet of 1,200 calories per day, he was discharged at 216 kg (34 st) – the greatest weight loss for a human being. In October 1981, though, he had to be readmitted – after putting on over 89 kg (34 st). When he died on September 10, 1983, he weighed more than 362 kg (57 st).

Answer 10

I hope you are referring to the weight as in the one we take on the scales...I believe its the same everywhere as the earth's gravitonal pull is almost the same all across the planet, however slight modification on the same could be found at different places on earth but usually the differences are not more than a few grams of "weight"

Answer 11

The US, why? because we are "the fat factory"! we eat fries, work 24/7/365, and are slaves to McDonalds, and just sit our fat butts all day and all night! its otrotious, and degrading to the rest of the world, no other nation has the health problems we do like diabeates, heart disease, and high cholesterol. In america, obesity is the second leading cause of death, next to smoking. at least drink some water, thats a start!

Answer 12

Short of the magnetic forces of the poles, I would assume standing in Death Valley California, because that is the lowest area of land that I know of. The lower you are, the stronger the force of gravity.

Answer 13

I live mear the equator and sometimes - like today -it like feels like i"m just going to float off and kiss the sky. So my answer is the north pole it's cool.

Answer 14

the person weighs the most along the equator as gravity is at max strength at the equator so it pulls u down harder makin u heavier than u r somewhere else on earth.

Answer 15

Any of the poles. Centrifugal force is weakest there.

Answer 16

Northern hemisphere up in the United States.

Answer 17

McDonalds

Answer 18

In the Marianas Trench possibly, but very few people have ever gone there.

Answer 19

. The spinning Earth is flinging us away from its surface a tiny bit, so that we weigh a little less than we would otherwise, simply because we are not being held down to the surface as tightly.



However, this is a very small effect. At the equator, your weight is reduced (compared to a non-spinning Earth) by about 0.3%; the effect weakens as you go north or south, and once you reach the north or south pole it disappears completely because the Earth is not spinning there. So if you can feel differences in your weight of 0.3% (about half a pound for a 150 pound person), and if you travel from the equator to the north or south pole, then technically speaking, you could feel the effects of the Earth spinning. On the other hand, people's weight naturally fluctuates more than 0.3%, so it's unlikely that you'd be able to disentangle this from other effects (like whether or not you had just eaten lunch).



http://curious.astro.cornell.edu/question.php?number=665

Also



The answer is not as obvious as we usually think, because the density of the earth is not constant throughout its volume. If we do take the density as constant then we may calculate that the acceleration of gravity will be maximum at the surface of the earth. In reality, because the density is not constant, the maximum acceleration occurs at a depth of about 2000 kilometers below the surface!!!!



I will start the calculations by referring to a law that is called the "Gauss Law of Gravitation" (see any freshman-level college physics textbook. My favorite book, because I was in undergraduate school in the late 1960's, is Halliday and Resnick's physics text). This law states that, for a spherically symmetric distribution of mass, the gravitational effect is equivalent to the total mass of the shell residing at the center of the sphere!! One of the effects is that if one goes below the surface of the earth, the spherical shell above has no effect on the net gravitational acceleration. Again, this assumes a spherically symmetric distribution.



So we will assume that the distribution of mass of the earth is spherically symmetric, which is a reasonable assumption. This is not precisely true because the earth is just not entirely smooth in its distribution of mass. But any differences are not important to this discussion. The main thing that is important is the radial distribution of the mass of the earth. That is, the earth is more dense as we go toward the center.



I consulted a book called "Smithsonian Physical Tables" (published by the Smithsonian Institution Press. My copy is 9th edition, 1969). I found a table of the density of the earth versus depth, which is based on calculations such as the average density, the precession of the earth, the flattening of the earth, and the known elastic behavior of the earth. Anyway, I put the table data into a spreadsheet, plotted it, and then fitted a sixth-order polynomial to the data (that's the highest-order polynomial that my version of Excel supports). Using that polynomial to represent the density versus depth, I did the following calculations, using the spreadsheet because it's easy.



Here is the table I started with: radius [cm] density [g/cm^3]

637000000 2.7

636000000 2.7

634000000 3.0

631000000 3.4

625000000 3.5

597000000 3.75

557000000 4.0

517000000 4.25

467000000 4.4

437000000 5.8

392000000 7.25

347000000 9.0

317000000 9.6

157000000 10.25

0 10.7





And here are the coefficients of the sixth-order fit: exponent of r coefficient

0 1.07000000E+01

1 1.58174000E-08

2 -5.30174000E-16

3 4.69666000E-24

4 -1.70152000E-32

5 2.63425000E-41

6 -1.46403000E-50





The law of gravitation (not Gauss's law) states that the gravitational force between two masses is

F = GmM/r2

where G is the universal gravitational constant, m is one of the masses, M is the other mass, and r is the distance separating the centers of mass of the objects. We can "remove" the test object (the one that we want to experience the acceleration, and usually the smaller object) by dividing its mass through in the equation, and we obtain an acceleration:

F/m = a = GM/r2

We get into the nuts and bolts of the details when we want to compute the mass, M, and for that we use the polynomial from above. We can represent a thin spherical shell of mass as dM (which means "a little itty bitty thin shell of M"), and, assuming that the earth is spherically symmetric then the density will be constant over a little itty bitty thickness of the earth, dr. So dM = r dV where r is the density and dV is the volume of the thin shell. So now we have

a = G/r2 S (r dV)

(I am using the symbol S to represent an integral)

where the integral is from zero to r.

Now, the area of the surface of a sphere of radius r is 4pr2 so for a thin shell of thickness dr the volume is

dV = 4pr2 dr

Putting this together we have

a = G/r2 S (r 4pr2 dr) = 4pG/r2 S (rr2 dr)

It is tempting to cancel out the r dimension in this equation!! But that is the wrong thing to do!! If we were to cancel out the r2 terms we would be off by a factor of 3 in the final calculation. This is because the r outside the integral is the final radius at which we are calculating the acceleration, but inside the integral the variable r is a function of r and needs the r2 term to be multiplied times the polynomial.



Let's first use the assumption that the earth's density is constant. If we do that then r can be taken outside the integral and we obtain

a = 4pGr/r2 S (r2 dr)

and when we do the integral we obtain

a = 4pGr/r2 (r3 / 3) = 4pGrr/3

Put all the numbers in, using the accepted average density of the earth as 5.522 gm per cubic centimeter, and we get a = 983 cm per second per second at the surface of the earth. As you probably know, the measured value for a is about 981 cm per second per second, so we aren't far off. Also, note that the function for a is a steadily increasing monotonic function in r, so the maximum value is at the surface of the earth. Once you leave the surface the acceleration declines by r-2 because there is no additional mass to contribute and the distance between the centers of the masses increases.



Now, to re-do this using the non-constant density of the earth, the variable r is replaced by the sixth-order polynomial, which is a function in r. Doing the integral causes each term of the polynomial to increase by one in its order of r and to be divided by that final order of r. (I know this is a bit technical, and you can ignore it if you want, but I wanted to provide this level of detail in case any other reader of this answer wants to work through the problem!)



Finally, using the coefficients of the polynomial and the proper order of the terms of r from the integrand, the terms of the equation are computed in the spreadsheet and added up, and multiplied by all the constants (such as 4 and G and p). Amazingly, the total mass of the earth comes out to be about 6.0x1027 g, which is only about 0.3% high, and the acceleration due to gravity at the surface comes out to be about 985 cm per second per second, which is only about 0.4% high!!! That's amazing since we started out with a table of the estimate of the variation of the density of the earth at different depths! A plot of the acceleration of gravity at various depths using the computed data shows that the maximum acceleration of about 1052 cm per second per second occurs at a depth of about 2000 kilometers. That acceleration is about 6.8% higher than at the surface.



Here's a table summarizing the acceleration information. (Notice that I have changed the units of the acceleration to m/sec2 instead of cm/sec2. I did this to have the same units as the table that follows in the next section below.)



depth [km] acceleration [m sec-2]

0 9.85

10 9.86

30 9.87

60 9.89

120 9.91

400 9.94

800 9.97

1200 10.14

1700 10.43

2000 10.52

2450 10.37

2900 9.78

3200 9.15

4800 4.50

6370 0.0







--------------------------------------------------------------------------------



After I finished the work above, I found in the Halliday & Resnick textbook a chapter problem that gives some apparently useful information. Problem 35 of Chapter 16 gives the following table of gravitational acceleration versus depth. It is apparently real data from the 1960's.



depth [km] acceleration [m sec-2]

0 9.82

33 9.85

100 9.89

200 9.92

300 9.95

413 9.98

600 10.01

800 9.99

1000 9.95

1200 9.91

1400 9.88

1600 9.86

1800 9.85

2000 9.86

2200 9.90

2400 9.98

2600 10.09

2800 10.26

2900 10.37

4000 8.00





The overall structure of the data that this table represent is very very close to what I calculated above. Any differences are likely due to the rounding of the corners of the 6th-order curve that was found for the original table of information. That is, I think the density of the earth changes more rapidly than what my 6th-order polynomial allows. According to the table from Halliday & Resnick the maximum acceleration is at a depth of about 2900 kilometers, which isn't far from the 2000 kilometer estimate that my original calculations found! Also, Eder Molina (who is a PhD geophysicist in Sao Paulo Brazil at the Dept. of Geophysics, Institute of Astronomy and Geophysics, and one of our answering scientists) reviewed my calculations (at my request!) and found in his library a graph of both the density and gravity versus depth, and his information is very close to what I found in Halliday & Resnick.

Answer 20

I think Kalel is right on

Answer 21

equator. Gravity is strongest there.

Answer 22

in the parts where ther is more cold

Answer 23

sea level or below where gravity is its strongest.

Answer 24

on the u s of a

then uk then australia