how does the gravitational force change with latitude?

Question

since the force on a body on the surface of the earth is f=(m2*g)/d^2 and the force varies with the distance from the center of the earth ,our weight at different latitudes should vary with the least at the equator and the maximum at the poles since the earth is slightly flattened at ther poles,what is this variation?

Answer 1

The formula for gravity is based on a gravitaional constant (g) times mass 1(in this case the Earth's mass) times mass 2 (the person's mass) all of which will be divided by the disance between the 2 center of masses.

The Earth's equatorial radius, is the distance from its center to the equator and equals 6,378.135 km (≈3,963.189 mi ).(Wikipedia)

The Earth's polar radius, is the distance from its center to the North and South Poles, and equals 6,356.750 km (≈3,949.901 mi ) (Wikipedia)

The average radius (R) at various latitudes can be determined using the formula for the radius of an ellipse:

R(θ) = √ [(a2cos(θ))+b2sin(θ))]/[(acos(θ))2+bsin(θ))2]

If the earth were a truely sphereical body, there would be no chang in radius from one latitude to another. So, in that case, there would also be no change in gravity.

Since the earth is flattened at the poles, the rate of radius change is not a constant.

If you draw a circle (as a model) representing the Earth and the flatten the top and bottom you could measure the radial differences moving from its pole to its equator. You will probably find that there is a point when the raduis length jumps sharply and then levels off. However, to accurately draw a model you would already have to know the raduis of the Earth at each latitude. (catch 22)

Of course these are average values for the radius and a person's weight could vary greatly if the were in an ocean trench or on a mountain top.

http://www.nasa.gov/vision/earth/lookingatearth/earthshape.html This nasa page offers more insight into the shape of the Earth and lends more confusion to determining gravity. It states that the shape of the Earth (Its oblateness) changes constantly from phenomenon.


Or do you simply want to know the gravity at the equator and at the poles?

g (0 degrees - at equator) = 9.78039 m/sec^2

g (90 degrees - at North pole) = 9.83217 m/sec^2

Answer 2

I believe there is an important component one might want to consider in the above formulation, if one were interested in the entire net force diagram on a body at a given latitude. That is, the NORMAL force acting on a point varies with latitude (even if one assumed a perfectly homogeneous spherical Earth). This is caused by the requirement for an object to be undergoing centripetal motion (at any place where latitude =/= 90 degrees) about the axis of the Earth's rotation.

Assuming a perfectly homogeneous spherical Earth, I formulated that:

|Fn(t)| = (A^2 + (Zcos(t))^2 - 2AZ * (cos(t))^2 ) ^ (1/2)

where t = angle of latitude (ie. 0 to 90 degrees), A= GMm/R^2 (ie. the gravitational attractive force of Earth on a body with mass m), and Z = R4m(Pi^2)/P^2 where P is the period of Earths rotation in standard metric units, and R is the mean radius of the Earth (for the assumed spheroid model of Earth).

Hopefully someone finds this interesting- I found the idea interesting, although in practice it leads to relatively negligible differences in the magnitude of the normal force.

(please comment if you find errors in the formula).

Answer 3

The gravitational stress with the aid of an merchandise of mass M on an merchandise of mass m is proportional to the made from the masses and inversely proportional to the sq. of the gap between them. Mathematically it somewhat is F = GMm/r^2 This stress is a stress of attraction, and in accordance to Newton's third regulation there's a reaction stress with the aid of mass m on the object of mass M. on condition that the stress is conservative and acts alongside a radial course it somewhat is asserted to be an important stress.

Answer 4

The gravitational pull IS least at the equator. That is why many countrys along the equator are starting to launch space vehicles from there. It takes less energy and therefore allows more payload to be put on space vehicles,and therefore can use less fuel to launch.

Answer 5

earth is not a perfect sphere, it is slightly ablong at equator & flattened at poles. the other reason is, density and consequently, mass of earth is not evenly distributed. this is another factor that we ignore but plays important role in force of gravity. besides it is slightly affected by spin motion of earth.

Answer 6

who cares you're a loser