The earth’s surface gravity varies only a little – about 0.7% over its entire surface.
Why so little, when the surface speed due to rotation at the equator is c. 1000mph and it takes c. 16000mph to reach low earth orbit (i.e. to become weightless). This is why rocket launches are done mostly near the equator, and in the easterly direction.
I would thus expect the gravity at the equator to be 1/16 weaker than at the poles, plus or minus the (much smaller) complication of the earth not being spherical and not being homogeneous.
Peter wrote:
Why so little, when the surface speed due to rotation at the equator is c. 1000mph and it takes c. 16000mph to reach low earth orbit (i.e. to become weightless). This is why rocket launches are done mostly near the equator, and in the easterly direction.This reasoning doesn’t take into account that gravity decreases with the square of the distance from Earth.I would thus expect the gravity at the equator to be 1/16 weaker than at the poles, plus or minus the (much smaller) complication of the earth not being spherical and not being homogeneous.
The actual centrifugal acceleration at the equator is about 0.0034 g. (The square of the angular velocity multiplied with the distance from the centre of rotation. The angular velocity of the Earth in radians/second is 2*pi/(24*60*60). The radius of the Earth is about 6366 km, so the centrifugal acceleration at the equator is about 0,034 m/s2.)
You don’t become weightless by going far from earth, you become weightless by free fall toward it ;)
In case of an orbit you have to be “free fall toward earth” (in whatever shape ir form?) to feel weightless, at international space station 400km hight gravity is 99.6% of what is on the 6400km surface, that is far strong to float at 400km
The same happens on 0G flight on aeroplanes, t is not gravity vs hight it is the shape/speed that give the free fall…
Why you launch at equator is the same as why don’t live in North Pole ;)
My original Q still stands. A 1kg weight travelling along the equator at 16000mph will weigh nothing because the angular acceleration balances the earth’s gravity. Go up a little (the atmosphere is really quite thin relative to the earth’s diameter) and you are in low earth orbit.
I am sure the answer is simple, however 
The answer is already there in https://www.euroga.org/forums/hangar-talk/10207-gravity-puzzle#post_198537
The centrifugal acceleration is a quadratic function of speed.
1000mph is cca 1/16 of orbital speed, hence the reduction of gravity at equator due to centrifugal effect is cca 1/250 of the surface gravity.
(I did not check the numbers … )
You would formulate the question as why take the hassle to orbit from the equator?
It does not seem you have much gain from earth rotation/gravity at equator? Anything to do with upper winds or just convenience?
If you are looking for low-cost energy launch it needs to be like Nasa/Virgin/Tesla ones, sent to low space orbits using TurboProps/Jets that rely on air density to get you higher
I don’t think you need old platforms/rockets, probably if you are going to the moon/mars?
My understanding is that equatorial launches are a great benefit for equatorial orbits (the most common being geostationary), as you do not have to spend fuel changing the plane of your orbit ot match the equator.
Centrifugal force on the equator is about 0.03224m2/s (https://www.omnicalculator.com/physics/centrifugal-force), which is as already mentioned about 1/280th of the normal earth gravity.
Earth is a spheroid, where distance between surface and center is bigger at the equator than that it is at the poles.
Therefore there is more mass between an object and the center at the equator than at the poles, creating a slightly larger gravitational force than at the poles. This compensates for rotational speed and centrifugal force. Rockets aren’t launched at / near the equator because of “less gravity”, but because of the 1000mp/h speed they “already” have when launched, in relation to space – i.e. the moon, venus, or planetary system.
OK; the bit I missed is that the “weight reduction” is proportional to the square of the velocity. So the 1/16 is really 1/256 which is below the noise caused by mass concentrations in the earth.
EuroFlyer wrote:
Earth is a toroid
A toroid is the shape of a doughnut. I think the word you want is “spheroid”.
