# Acceleration of free fall at different parts of Earth

“free fall is any motion of a body where gravity is the only acceleration acting upon it.” -Wikipedia

Question 1 - If an object is falling freely, will it follow the Earth’s rotation too? Or will the earth rotate whilst the object falls in a straight line (in its own reference frame). Or more simply put, does it have centripetal acceleration?

Question 2 - Assuming that the earth IS spherical AND has uniform density. Weight of the same mass is different at the Pole and at the equator; thus, the gravitational field strength, $$g$$ is different at those two places. Are these different values of $$g$$ also the same as acceleration of a body falling freely at those different places?

• Q2: if the earth is spherical and of uniform density, then the same mass has the same weight at the poles and at the equator. Why do you think they would be different? If you consider it as not spherical, but as an oblate spheroid, then the weight is slightly different. Aug 5, 2019 at 3:39
• @NickD I believe it’s because at the equator there is centripetal acceleration while at the poles there isn’t (at least less than at the equator). According to Wikipedia, in order to have a net centripetal force at the equator, the magnitude of restoring force (of the spring of the balance) must be less than the force of gravity. en.wikipedia.org/wiki/Centrifugal_force (Weight of Object at the poles and on the equator) Aug 5, 2019 at 3:50

Question 1. No. A projectile will not follow the Earth's rotation. This phenomenon results in very similar to the Coriolis effect. Let it freely fall, but launch it perpendicular to the equator toward one of the poles and you'll see this effect. There is, however, the effect of the wind in Earth's atmosphere, which complicates this answer.

Question 2. The weight is the same at the pole and the equator, i.e. if you consider weight as the gravitational force on a mass. True, the centrifugal 'force' will cause a scale to read a slightly lower result, but that is because part of the acceleration towards the center of the Earth now includes the object's centripetal acceleration. The centrifugal force is directed away from the Earth. The reading on a scale is the object's weight minus the centrifugal 'force' so that the net force is zero (assuming the object is not in freefall and resting on the surface of the Earth).

1.The object is attracted to the center of gravity of the earth, attracted to the center of the earth. The gravity on the object is unaffected by the spinning of the earth. The object does not in any way “know” the earth is spinning or what direction etc. So if it is held far above the earth and let go, then it will fall straight toward the center of the earth and the earth can spin under it.

If instead of just letting it go, you take it to some elevation and throw it, then it will still be in free fall - that’s because no other force is acting on it. It will also still be attracted to the center of the earth, but will travel in an arc. It still only cares about the center of the earth and which way you threw it though. You could throw it to move along with the spin of the earth or the opposite way, or perpendicular, etc.

2.Poles don’t matter for gravity. In fact the earth doesn’t even have to be round; it could be a cube. The object will still always be attracted to the center of the earth. If the earth isn’t round, then elevation alone won’t tell you how far it is to the center. For example whenever we are above point A on the earth, then a mile above the surface means 1,000 miles from the center of the earth. But above point B, a mile above means 1,005 miles from the center (if and only if not round).

Finally, it doesn’t usually come into play, but the distance to the center tells you how strong gravity is. Further from center and gravity gets weaker. This doesn’t usually matter because even if we are thousands of feet above the surface, we are still around the same distance (thousands of miles) from the center.

Consider a mass hanging on a length of string. The upper end of the string is attached to a spring scale. At the equator the force of gravity must exceed the reading on the scale in order to provide the centripetal acceleration. The mass has the same angular velocity as the earth. but a larger linear velocity than points below it on the earth. If the string breaks, the mass will land very slightly to the East of a point directly below its starting position. At a middle latitude, the component of gravity toward the axis of rotation must be greater than the opposing component from the string. The top of the string leans slightly toward the pole. This lean is enhanced by the extra mass in the equatorial bulge, and is perpendicular to the surface of a still body of water. The direction of the string is usually taken as the definition of “up” (or “down”) and it does not point exactly at the center of the earth.