Why is gravity stronger on North pole??

Question

I just need some help on this.

Why is there are stronger effect of gravity on the North Pole than anywhere else in the world?

All help will be appreciated!

Answer 1

Because the earth is very slightly flattened at the poles, so the North and South Poles are a bit closer to the center of the Earth than land on the Equator. And the South Pole is in Antarctica, which is a very high plateau, so the gravity there would be a bit less than that at the North Pole, which is in the middle of a deep ocean.

The South Pole is colder than the North precisely because of this high elevation.

Answer 2

Centrifugal force is a purely perceptive force. Which means it doesn't exist. It feels like a force has pushed you, but it's actually your intertial movement. Just like standing on a bus and the bus suddenly stops. There is no force pushing, it's the bus deccelerating. I'm not a planetary physist (i just made that job up it's probbaly in the astronomy field). But the earth is slightly eliptical, I'm not particularly sure why that is, maybe because of it's spin. If the earth was perfectly round, then the net force would be smaller by a tiny tiny tiny fraction. However the earth is slightly elitical so i can only presume there is more mass around the centre of it's spin. And thus there is more gravitatinal pull. So from just deduction about what i hear about earth. No it isn't. And if it was, it wouldn't be due to the 'centrifugal forces' it would be due to the fact of a lighter normal reaction force which in turn doesn't effect teh gravity. Weightlessness is just no normal reaction, you still feel the effects of gravity when you're in the air, it makes not difference if you're on the ground. Hope this helped in some way. Just take into account there are lots and lots of factors. Such as whether or not the moons gravitational force acts upon you and so forth and so on.

Answer 3

Bec ause of the oblateness of the spheroid
the north pole is closer to the centre of the mass - so gravity is stronger.

F = G x M1 x M2 / R^2

M1 being Earth's mass, M2 being the body on which the force (F) is acting R is the separation of their centres and G is the gravitational constant. The polar radius is about 15 miles less than the equatorial radius (i.e 3 times the height of Everest!)

Oh, and by the way, you asked about GRAVITY. Not any other forces, so these people who mention centifugal (centipetal) forces or anything to do with weather temperature etc are barking up the wrong tree.
Gravity is dependent on MASS and DISTANCE. Thats all.

The Earth is shaped a bit like an orange, squashed at the top, fatter round the equator. Then there are lumpy bits like mountains, so there isn't a simple explanation for the shape nor the gravitational variatiuons we can measure. Local density changes can also affect the gravity measurements - if a more dense mass is very close, then the gravitational force is a bit stronger (locally)t. The formula above still works, but must be applied to many more constituents than the simple case of two textbook masses. (e.g. Newton's. apple and the entire planet.) For day to day work, the formula is good enough.

Vulcanologists use very sensitive gravitometers to measure the build up of magma in chambers under volcanoes. I understand that the gravity goes down if the less dense rock is nearer to the observer.

Answer 4

I have had rather a large breakfast, and gravity is stronger here now, so I shall have to lie down for a while for the effects to wear off, and return to normal. The answer is to keep off big breakfasts at the North Pole.

Answer 5

One thing to keep in mind that will allow you dismiss about half the answers is that the inverse square law only applies to the *exterior* of a spherically symmetric object. Since the answer depends on the earth not being a sphere, simply stating that the poles are closer to the center is not an adquate explanation. Dr. Spock is correct about an idealized equipotential earth, but left us with a less than fully satisfying "French scientist Clairaut" said so.

I don't actually know if you weigh more at the poles. As epidavros points out, local perturbation may be more important. Neglecting local perturbations (for academic purposes only), if you're taking about the force due to Newton's universal law of gravity, gravity is stronger at the equator. Only if you subtract off centripetal acceleration at the equator is it higher at the poles. That it, you weigh more at the poles. The reason for this is that earth's surface (again, neglecting mountains and such) is a equipotential surface in the rotating frame. That means the effective potential difference between the surface and the center is the same everywhere on the surface. Since the poles are closer to the center, the mean potential *gradient* is greater at the poles (it has to change by the same amount over a shorter distance). The potential gradient is the effective acceleration (aka strength) of gravity.

Answer 6

It isn't.

Its a common and completely incorrect myth, base don the fact that the Earth is very slightly flattened at the poles (the Eart is, in fact, more spherical than a pool ball).

Gravity depends on the density and make up of underlying rock, and so it varies across the Earth. GRACE is the NASA program to measure this variation. You can see a map at the link

The South Pacific is a high spot. The North Pole is a low spot.

Answer 7

No-one seems to have approached this problem in a well-ordered scientific way.

If the Earth were spherically symmetric, gravity would be equally strong all over its surface.

However, the Earth is rotating, at an angular rate of w radians/s, say. Consequently, its equatorial regions bulge somewhat. Naively, one can think about this in EITHER of two equivalent ways:

(i) "CENTRIFUGAL FORCE" due to the rotation partly counters the attraction due to gravity, effectively weakening it by a fraction of w^2 r / g. That fractional value is about 1 / 289. The resulting g_eff = g(1 - w^2r / g) is the ONLY way that anything "gravitational" such as dropping or rolling things or timing a pendulum can be measured on the SURFACE of the Earth. The g here stands for the local (and now variable) "bare" but separately unmeasurable gravitational g if one is restricted to local gravitational experiments on the surface of the Earth.

(ii) The second way of looking at it is this: Part of g (the fraction ω^2 r / g) is "used up" in providing the CENTRIPETAL ACCELERATION needed at the equator to make that material follow a daily circular path --- consequently, there is a weaker g_effective at the equator, hence it bulges.

No matter which way you prefer to look at it, the consequences result in the equator "bulging" a bit, and therefore the poles being closer to the centre of the Earth than points on the equator are. Therefore g is slightly larger at the poles.

However, that IS a naive approach, as I stated. The more complex truth is that because the Earth is distorted from being spherical, its gravitational field is ALSO distorted from spherical symmetry. Consequently, the form of the surface of the Earth is determined in a more complicated self-consistent way involving its rotation, "centripetal" or "centrifugal" effects AND the consequently nonspherical gravitational field.

With the assumption that the surface of the Earth must ideally be an equipotential in the combined effective force fields due to gravity and the rotational effects, plus certain assumptions about the Earth's run of density throughout its interior, this self-consistent problem can be solved. The first person to do it with reasonable success was the French scientist Clairaut.

Because that combined equipotential surface for g_eff is that of an oblate ("squashed") spheroid, so are other nearby equipotential surfaces. Consequently, one can readily visualize that g_eff (which is INVERSELY proportional to the local distances between successive equipotential surfaces) must be less at the equator than at the poles.

The net result is indeed that both the local g and the more easily measured effective gravity g_eff are somewhat stronger at the poles than at the equator. However, because of the self-consistency requirement, the numerical MAGNITUDE of the effect is different from what one would first naively calculate. One final point is that the Earth is actually slightly pear-shaped as well (this is connected with the fact that ~2/3 of the Earth's continental masses are in the northern hemisphere). This introduces yet one more global asymmetry. However, there are lots of local anomalies due to local mass concentrations. In the end, g at the North pole (and equivalently g_eff which is the same as g, there) is stronger than at most, but not places.

Live long and prosper.

Answer 8

These are all very interesting answers, but they ignore one very important thing. The very latest science, states that the earth has global gravity variations. This is proven by measurements using the very latest instrument. It's a fact. However, no constant factor has yet been discovered to explain these variations in the pull of gravity. So the answer is, all your answers using math and physics are just plain and simply wrong. Sorry about that. These variations are an as yet unexplained phenomonen.

Science is not resting on this one. Even NASA has embarked on a project to explain these variations, and investigating it via satellite under the GRACE programme. As yet even they do not have an answer. The best guesses to date suggest it is caused by ancient fissures and density variations within the earth.

Hope that helps.

Answer 9

Only few individuals can give you the correct answer on that, but if you look at it like this, the Axis of the earth are in the north and south pole, so its bound to have more gravity on these two than any where else and it has to be stronger between this two places to balance the earth.

Any one can object on that, but that's my theory any way!

Answer 10

The Earth is an oblate spheroid, meaning that the it is not quite round, but slightly flattened at the Poles.

SInce gravitational force is related to the mass of the object, and the inverse square of the distance from the centre of the mass, the poles, which are just that little bit closer, have a stronger gravitational attraction.