# Why don't I weigh more at the poles

1. Jan 11, 2005

### notsureanymore

Hi all

Another thought I have been pondering is why don't I weigh significantly more at the poles than I do at the equator

The earth rotates at n kilometers per hour, I think around ~1300 kmph

Would this not create a centrigual effect countering earths gravity. thus making me lighter at the equator than the poles where no real rotation is evident.

I know the earth has an equatorial bulge. but to me. it seems to be a very small bulge, does this bulge increase the earths mass at the equator enough to counter the centrifugal forces.

I read elsewhere pole to pole the earth is 22 km's less in diameter than at the equator. this seems marginal.

And.. will the earth eventually flatten ?

Chris H.

2. Jan 11, 2005

### vincentchan

good question... the earth rotation speed is about
$$2 \pi R / T$$
6 378100 m (radius of earth) * 2 pi / 1 day ( period of the earths rotation)
=2*3.14*6378100m/(60*60*24)sec
=464m/s at the equator
the centrifugal force is $$mv^2/R$$
for a 70kg man, the force will be
70*464^2/6378100
=2.36N...
F=mg
m=F/g=2.36/9.8=0.2kg
so, at the equater.. your weight will be about 0.2 kg less than the weight at the pole

3. Jan 12, 2005

### rachmaninoff

There were several similar discussions recently, the major point of which was that the gravitational potential at sea level is the same everywhere.

4. Jan 12, 2005

### notsureanymore

Ok the cetrifugal force is less than I anticipated. so the smallish bulge is enough to balance the gravitational forces.

This of course occurs at all points on the surface.

The upshot is that if the earth was spherical and I wanted to lose weight I would move to the equator.

but as the earth is suffering middle age spread I will have to go back to the gym.

Damn shame that .

Chris H.

5. Jan 12, 2005

### errorist

Actually, the force of gravity is a little different at the pole. The pole is about 14 miles below sea level when compared to the Equator do to rotational velocity forcing the mass at the equator to bulge out. The difference is due to the altitude difference from the center of the Earth. Gravity changes with altitude.
This means there is more downward force at the poles.So, this must be added to the effect of rotational velocity.

6. Jan 12, 2005

### DaveC426913

"And.. will the earth eventually flatten ?"

No. The two forces ('centrifugal' and gravitational) reach equilibrium at some point and the flattening does not continue.

I expect that happened during the formation of the planet. I don't know of any evidence to suggest that the bulging is changing in present day.

7. Jan 12, 2005

### errorist

The gravity is different due to the altitude difference. You can't get away from that. More downward force is there due to the elevation.

8. Jan 12, 2005

### BobG

The force of gravity is more at the poles both due to being closer to the center of the Earth and due to no centrifugal force. The use of 9.8 m/sec^2 is used as an approximation regardless of your latitude, or whether you're standing on Mt Everest or at sea level, because the difference is so small.

Btw, you wouldn't 'weigh' .2kg less at the poles because kg is a measure of mass, not weight. Regardless of how a particular scale uses or misuses units, you won't notice any difference on a balance scale since that's really measuring mass - you might notice a difference on a spring scale, if it's sensitive enough.

9. Jan 12, 2005

### errorist

The use of 9.8 m/sec^2 is used as an approximation regardless of your latitude, or whether you're standing on Mt Everest or at sea level, because the difference is so small.

Correct. But, there is still a gravity difference because you are closer to the center of the Earth at the poles. No way around that.

10. Jan 12, 2005

### motswac

hi, there,
i was just reading your message. are you suggesting that earth's gravity has something to do with the earth's rotational speed. if so, since we would not be attached by anything to earth when it is stationary, wouldn't we be pushed away from earth when it start's rotating? Newton's law of motion."we will continue moving on a straight line unless an external force acts upon us". If you suggest that this gravitaional force is created by our earth's rotation, what force (external) is changing our course (from straight line to circular, along with the earth's rotation?)

Cyp Mot

11. Jan 12, 2005

### JesseM

He wasn't saying gravity is caused by rotation, he was just saying that at the equator the centrifugal force would oppose the gravitational force, decreasing your weight slightly (assuming we define 'weight' in terms of the total force you feel, not solely in terms of the force due to gravity).

12. Jan 12, 2005

### notsureanymore

Ok this is interesting you are saying gravity is not a function of mass so much as a function of the centre of mass.

The following totally ignores other forces...
Wouldnt gravity be less at the poles due to less mass between you and the centre of the earth?

Wouldn't this also mean gravity is less at sea level over a deep portion of the ocean. than at the same level over land due to the fact that sea water is less dense than rock and magma?

What if you climbed a high mountain. Wouldnt gravity be more for the mountain climber than for the photographer hovering nearby in a helicopter. due to the large mass of the mountain below the mountaineer? (assuming the helicopter is not over the mountain)

OR

Is gravity the same at all points of equal radius from the centre of the earth regardless of intervening mass?

13. Jan 12, 2005

### Gokul43201

Staff Emeritus
The gravitational equipotentials are spherical shells centered on the center of the earth. The field increases outwards as a linear function of the radial distance from the center.

$$E = \frac {GM}{r^2} = \frac{G\rho}{r^2}~\cdot~ \frac {4}{3} \pi r^3 = Kr$$

This expression holds from r=0 to r=R(poles). For R(poles) < r < R(local surface), you would expect the increase in the field to be less than linear, because the spherical shells are no longer completely filled with earth. Nevertheless, there must be an increase in the field.

But due to the centrifugal effect, the net field sees a reduction as described in an earlier post here.

The two effects should nearly cancel each other, making the surface of the Earth an equipotential.

EDIT : Googling, I've found that I'm wrong...but I can't see how.

Last edited: Jan 12, 2005
14. Jan 13, 2005

### errorist

Is gravity the same at all points of equal radius from the centre of the earth regardless of intervening mass?

In theory yes. But, in actuality no.

http://www.csr.utexas.edu/grace/gallery/animations/ggm01/

At the poles there is a fourteen mile difference in elevation than at the Equator. The amount of missing land mass at the pole does have an effect of the gravity. It is not enough to change or equal out the total difference.

Last edited: Jan 13, 2005
15. Jan 13, 2005

### Andre

OK Gokul was very close, I think.The most simple expression of Newtons law only accounts for point masses and for homogeneous spheres. Earth is not a homogeneous sphere by far. The most remarkable abaration is the equatorial bulge. This means that if you are at the equator, three things must be accounted for gravity correction.
First you're farther away of the centre oif the Earth this tends to decrease the gravitional force w/Newtons universal law. Next you're spinning, creating an apparant centrifugal force that is opposing gravity as well. But the excess mass is directly beneath you and this tends to increase gravity. The nett result is a very slight decrease.

At the poles the reverse is true. Closer to the centre, less R more gravity. No centrifugal force opposing gravity, but the excess mass is off to the sides instead of beneath you, that would decrease the gravitational force, compared to the sphere.

Therefore the gravity at sea level is not equal everywhere but there is a relation between gravity, and total pressure on the surface area, balancing the Earth.

But gravity is not the biggest anywhere on the surface of the Earth. So where would that be instead?

16. Jan 13, 2005

### BobG

You would think so, but the only time the spherical shells model truly works for mass 'outside' your radius is if your mass really is a perfect sphere.

The perfect sphere model makes the real problem simpler. There is a certain amount of mass in front of you (towards the center of Earth) and a certain amount of mass behind you. Instead of calculating the amount of mass in front of you and the amount behind you, you can figure out how much mass way out on the opposite side is cancelled by the mass behind you.

For non-spherical Earth, even though you're closer to the center of mass at the poles, all of the mass is still in front of you .... And you're closer to the mass on the opposite side of the sphere (and all the rest of the mass), as well.

Edit: Actually, not all of the mass is closer. Some of the mass near the equator is slightly further away. Your average distance from every point in the oblate sphere is closer and, more importantly, all of the mass is still 'in front' of you.

Last edited: Jan 13, 2005
17. Jan 13, 2005

### errorist

Seems to me if the Earth stopped rotating it would all equal out??

18. Jan 13, 2005

### BobG

Eventually.

The Earth's rotation rate is slowing (ever so slightly), but the redistribution of mass to match the rotation rate is even slower. A lot of the motion of tectonic plates, etc is the Earth trying to redistribute its mass to match the current rotation rate.

So, yes, if the Earth stopped rotating, eventually the Earth would form a virtually perfect sphere.

19. Jan 13, 2005

### pervect

Staff Emeritus
The Earth isn't spherical, so integrating out the spherical shells won't give the right answer.

The Newtonian potential of the non-spherical Earth is used in the GPS calculations, for instance.

http://xxx.lanl.gov/abs/gr-qc/9508043 has an expression for it, which is their eq 2)

$$V = -(\frac{GM}{R})(1- \frac{J2}{2}(R/r)^2\,(3 cos^2(\theta)-1)))$$

Theta here is the lattitude.

J2 is a commonly used measure of the earth's departure from sphericity. It's definitinon is given in the NASA planetary data sheet

http://nssdc.gsfc.nasa.gov/planetary/factsheet/fact_notes.html

The origin of this expression may appear mysterious but it's not really, though it's a bit involved.

It's basically the result of a series expansion. There is an unfortunately not very legible derivation online

http://www.apl.ucl.ac.uk/lectures/3c37/3c37-6.html [Broken]

Some keywords if you want to go Google-fishing "Gravitatioanl Potential Theory, Legendre polynomials"

Goldstein also derives the gravitatioanl potential for an approximately spherical body in "Classical mechanics", pg 226.

Last edited by a moderator: May 1, 2017
20. Jan 13, 2005

### notsureanymore

Slightly OT

I was wondering about the changing shape of the earth. as its rotational speed changes it needs to readjust its shape to maintain equilibruim.

Hence the plate shift that recently killed so many.

When a thought entered my cranium. the oceans act like a balancer. Now I am sure its only slight but without it wouldnt there be a risk of the earth developing a wobble like an out of balance washing machine.

This could increase to the point of throwing us out of orbit (unlikely to ever get that severe of course), but without the oceans rebalancing the earth everytime a major event occurs we would most likely be in a major p**.

now my thoughts wander to other planets without oceans. how do they cope? perhaps the atmosphere is dense enough to do the job????

Any thoughts?
(perhaps should this go to another thread.)