# Weight In Different Parts of the World

Is an Inuit who weighs 100 kg when he is camped on the North Pole during winter solstice significantly larger than a descendant of the Incas who weighs 100 kgs while living at 15,000 feet in Equador during the Equinox?. I see several factors that can affect the question, such as distance from the centre of mass which involves both elevation and increased diameter of the earth at the equator, relative proximity to the sun due to seasonal distance of the orbit and inclination of the axis as well as the anti-gravitational effect of the spinning earth. Or, we could just ask what a mass that weighs 100kg at the north pole, weighs in Equador under the conditions previously described.

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Or, we could change the name of this thread slightly from the very generic “Scientific Debate” and move it where it belongs, in General Questions.

David B, SDMB Great Debates Moderator

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Long answer: The altitude is nearly insignificant. Compare the radius of the earth, approx 8000 miles, to the distance to the center in the mountains, approx 8003 miles.

I don’t know about the seasonal changes, because I’m not sure of the position of the earth in it’s orbit, but my Gut instinct and WAG is that there is no significant difference.

The pole Vs. the equator is really minor, in terms of centrifugal force. There is a less than 0.1% difference in weight. I worked it out once.

grienspace, just to clear up the terminology, one doesn’t “weigh” in units of kilograms, one has mass in units of kilograms. One has weight in units of Newtons. That said, gravitational acceleration does vary over the surface of the Earth. One must consider factors such as obliquity, density and altitude. Despite these considerations, on the surface of the Earth one will only find a value of acceleration equal to approximately 9.80 m/s[sup]2[/sup] +/- 0.02 m/2[sup]2[/sup]. Given a mass of 100 Kg, one would only find a maximum difference of about 4 N.

kg is the unit for mass

Newtons (N) is the unit for weight.

Earth:
Polar R = 6.357 X 10^6 m
Eq. R = 6.378 X 10^6 m
Eq R + 5,000m (~15,000ft) = 6.383 X 10^6 m
Mass = 5.98 X 10^24 kg

G (universal constant) = 6.67 X 10^-11 (N)(m^2)/(kg^2)

Acceleration due to gravity (g) = G(M of Earth)/(R^2)

The weight of an object with mass = m?

``````w = (m?)g = G(M of Earth)(m?)/(R^2)
``````

So; a man with a mass of 100kg weighs:

at pole: 987 N
at equator + 5,000m: 979 N

At noon the sun would pull the man at the equator away from the Earth, while at midnight it would add to the Earths pull. The moon would have the same effect, but on an almost monthly basis. In all cases, I suspect that the moon and sun would have a negligible effect on something as small as a man.

As for the effects of spin: I have no idea, but I’m interested.

Oops! My previous reply would suggest that the moon orbits the Earth once a month while the Earth doesn’t spin. I do know better. Sorry!

I’m all excited that there are people like me who are interested in this sort of thing. I took high school physics as well 35 years ago, in Canada, and it was the only time I have ever come across the term for force as newtons. Around here, people generally weigh themselves in kilograms, not Newtons. Not scientific perhaps, but everyone knows what we are talking about. That’s why a person of 100kg weight at the pole may have a significantly different weight at the equator. The mass remains constant. For our American friends I could be talking in pounds, or slugs, and I have always found that difficult.

the angle of the dangle equals the heat of the meat and the mass of the ass remains constant.

F = GMm/r^2

F = weight
G = 6.67259 * 10[sup]-11[/sup] m[sup]3[/sup]/(kg*s[sup]2[/sup])
M = 5.975 x 10[sup]24[/sup] kg
m = 100kg
r[pole] = 6356775 m
r[equator] + alt = 6382732 m

F[pole] = 986.64 N
F[equator] = 978.63N

Now we find the effect the sun has on the weight of the two people
M = 1.99*10^30 kg
m = 100kg
r = 1.4959787 * 10^11 m
tilt of the earth at equinox = 0
tilt of the earth at winter solstice = 23.5

The angles of the two tilts are extremelly insignificant when finding the difference in force the sun adds depending on the time of day, it varies about 0.00005N.

As for centrifugal force. There is no tangential speed at the north pole, therefor no centrifugal force. Someone standing on the equator at a height of 15000ft relative to sea level is “moving” along at 464.165m/s, therefor F[c] = 3.3755N.

I imagine the moon, similar to the sun, would have a extremelly small effect on the outcome of the end result, and as I dont know the location of the moon relative to the earth, I will leave it out.

Therefor, taking into account the sun, centrifugal force, and gravitational force (all the factors you listed), the difference in the two is 11.38N. However, there are also other factors to consider such as the density of the material they are standing upon.

Also, the radius of the earth is 8000 miles? I assume thats just a typo…

And yes grienspace, people don’t weigh themselves in newtons (i’m from and live in Canada also), but that doesn’t mean its correct. Either way, the difference is small, if you wish to refer to “weight” as kilograms then if the Inuit was on a scale he would read 1.16 kilograms more than the person standing on the equator at 15, 000ft above sea level.

Oops, the scale would read 2.55 pounds more, if they were using a balance scale it would be exactly the same.

Thankyou Ian. I thought the centrifugal force would be more significant. However we could say that an Inuit weighs out on a scale 2 pounds more than an Inca of the same size. I’ll stick that in my memory bank.

aw, I know I read this somewhere! I forget if it was Unca Cecil or astonomer Sten Odenwald

But IIRC, the difference in your weight would be about 1 pound from the North Pole as compared to at the equator (lighter at the equator).