How does latitude affect acceleration due to gravity?

Question

Does the acceleration due to gravity increase as latitude increases?

Answer 1

the answer is no. acceleration of a falling body is the same at any latitude. the is also ZERO weight difference at the poles than at the equator. it doesn't matter that you are spinning at 1000 mph, it is angular momentum that affects the weight of a sitting body, and at one revolution per 24 hours, the effective angular momentum is zero. also, any centrifugal force that WAS exerted by the spinning earth has no effect on a falling body. it only effects bodies that are in contact with the earth

Answer 2

Latitude will effect acceleration! This is because due to the Earths spin the diameter of the Earth is more at the equator than the poles. As a consequence a person at the poles is slightly further away from the Earth's centre of gravity. This means there is slightly less gravitational pull here than at the poles. As a consequence the acceleration due to gravity is slightly less at the equator than the poles.
Having said this local terrain and your altitude (e.g. on top of a mountain) will also have an effect). Iberus's argument has followed the wrong precept in answering this question

Answer 3

Yes, the acceleration increases slightly, because the polar radius of the earth is slightly smaller than the equatorial radius. Here's why:

Newton's law of gravitation, as applied to an object on the earth's surface, is:

F = (G * Me * m) / Re^2

F = force
G = gravitational constant (6.67*10^-11 N-m^2/kg^2)
Me = mass of the earth (5.98*10^24 kg)
m = mass of the object
Re = radius of the earth, (equatorial) = 6,378,135 m, (polar) = 6,356,750 m

And the definition of acceleration is :

F = m * a

F = force
m = mass
a = acceleration

So, the acceleration due to gravity is....

a = F / m = G * Me / Re^2

...at the equator:

a = (6.67*10-11 N-m2/kg2)(5.98*1024 kg) / (6,378,135 m)^2
a = 9.80 m/s^2

....at the pole:

a = (6.67*10-11 N-m2/kg2)(5.98*1024 kg) / (6,356,750 m)^2
a = 9.87 m/s^2

The difference in acceleration due to latitude, and even due to variations in the earth's density (mountains, ice, oceans) is small, but important when computing something that needs precision, like the orbits of satellites, for example.

Answer 4

gravitational acceleration is calculated by g= -MGr^ /r^2.
As the latitude increases, acceleration due to gravity increases.. means as you move towards poles, the weight tends to increase. The increase in gravity might also be attributed to decrease in centrifugal force at poles, which are flattened, compared to the equatorial region.

Answer 5

Yes, the acceleration due to gravity increase as latitude increase. The difference is small, but detectable. The force of gravity at Singapore (on the equator) is 9.781 m/sec^2, and the force of gravity at Stockholm is 9.818 m/sec^2.

Two reasons:

1. The earth is rotating which makes objects at the equator tend to 'spin' off. Technically, the rotation of the earth creates centripetal force which partial negates gravity.

2. The earth bulges at the equator. This means that the objects at higher latitudes are closer to the earth's center and as long as you are on the earth's surface, the force of gravity increases as you move towards the center.

..................................................................... Edit 1
I've added an additional citations since I first answered this question. All confirm that there are two major factors at work, the centripetal acceleration and the Earth's bulge.

...................................................................... Edit 2
Dave, below, is right in his additional comment about local variations in gravity due to density variations below the earth's surface. However, these density variations are not directly related to latitude. But they have to be taken into account in determining the local acceleration of gravity. An interesting map of gravity variations can be found at: http://www.abc.net.au/science/news/stories/s911917.htm

Dave, also points out that the rotational acceleration as you move towards the equator is not gravity but centripetal force, which is technically correct; however, it is one of the two major component in making you weigh less at the equator then you do at the pole. (local density variation being the third most important factor). And if you want to calculate the local acceleration of gravity say in Singapore, you figure the net number (gravity acceleration - centripetal force), because it is this number you will use in all of your work.

(As an aside, if you are working with satellite orbits, you disregard the local centripetal acceleration below you. Why? Because the satellite has its own centripetal acceleration based on its speed and elevation in orbit. This exactly balances the acceleration of gravity and is what keeps the satellite in orbit and its payload weightless.)