Wiploc

Both, sort of. You weigh a tiny bit more at the equator because of the shape of the earth. And circling at a thousand miles an hour gives you a tiny bit of lift. I have no idea which effect is stronger.
crc

SULPHUR

gravity is the more important factor. YOU weigh less at the equator becausr you are a at greater distance to the centre of gravity than at the poles where distance is less. This follows from the law which states the force is inversely proportional to the square of the distance. The greater force at the poles increaces the "weight"than at the equator where the force is less and therefore the "weight" is less.

Wiploc

This doesn't work. It's akin to saying you would weigh less on a larger planet because the center would be farther away.

Let's exagerate the earth's equitorial bulge by way of illustration: suppose that a giant squoze the earth between thumb and forefinger, reducing the pole-to-pole distance to 1000 miles, and increasing the equitorial diameter to 16,000 miles. As a first approximation, a guy standing at the equator would weigh sixteen times as much as much as a guy at the pole. That's because there's sixteen times as much matter in the downward direction from him. Yes, the effect would be reduced by the more distant center of mass; but it should be clear that if a planet is shaped like a quarter, the guy standing on the rim will weigh more than a guy on Washington's nose.
crc

Silent Acorns

The best way is to use a sextant. After all this is what navigators used for hundreds of years. It's cheap, simple, and very accurate with a little practice. If there was something simpler they would have used it. Just find the North Star (or whatever the South Star is), measure the angle, and voila: you have your angle of latitude.

Alternatively, at night and far from city lights, point a camera in the general direction of a pole (the south pole if you believe you may be too far south, and the north pole if you may be too far north). Make sure the camera is perfectly level (i.e. not pointing up or down). Set the camera for a long exposure - about 4 hours should be plenty. The picture you take will show the stars tracing out circular arcs around a common centre. Assuming that you are within 10 degrees of the equator, this common centre should be in the photograph. If you're on the equator, the centre will be right in the middle of the picture. This method works best if you are on the top of a hill and if there is nothing obstructing your view of the horizon.

Defiant Heretic

Simpsons episode.

SULPHUR

I was referring to variation of weight on earth.From the equation on gravitational force F=Gmm/d^2 obvioslly the greater mass the greater the force or weight There are two different issues here.Hopefully I have made it clearer.

SULPHUR

If the mass remains the same,the distance becomes greater so dividing by d^2 the force becomes less hence your weight at the equater is less.

Jesse

That rumor has been around for a while, I heard it well before the Simpsons episode.

Bumble Bee Tuna

yeah but gravity from the Earth is really an integral of the gravity from each individual particle. Taking Wiploc's extreme example of a disk-planet, the edge has stronger gravity than standing on the center of the plane. The gravity vectors from the edge would be mostly in the downward direction. Standing on the plane, the vectors would be mostly horizontal, and all the horizontal vectors would cancel each other out. Sure, the distance d would be tiny for the small amount of mass at the center of the disk, but the vertical components of the vectors would be tiny anywhere outside of the close proximity. Or at least, that makes sense to me, maybe the integrals do work out with that crazy math to be equal but I doubt it.

Also, are the poles actually closer than the equator? Sure, there may be an equatorial bulge but I'll bet that makes the planet a bit more cylindrical and the poles stick out farther.

-B

SULPHUR

All i'm stating is basic physics which some how is being questioned. Why come up with these side issues Do you under stand the basic eqation for gravitational attraction.If so ,what are your problems with it.