1
$\begingroup$

“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?

$\endgroup$
4

4 Answers 4

1
$\begingroup$

For simplicity consider the earth a sphere with the acceleration of gravity on an object the same anywhere on the surface of the earth. The earth rotates about an axis through its poles. Define the "weight" of the object as the force the object exerts on a support attached to the earth when the object is sitting stationary on the support. The weight is equal in magnitude and opposite in direction to the force of the support on the object. Consider an object dropped from a small height above the surface of the earth at the equator.

Consider an object initially at rest on the support at a height $h$

Consider an inertial reference frame "watching" the earth's rotation. At a pole there is no rotation and no centripetal force; the magnitude of the force of gravity equals that of the force of the support on the object. In contrast, at the equator the magnitude of the force of gravity exceeds that of the force of the support on the object just enough to provide the centripetal force to keep the object rotating with the earth. The force of gravity on the object is the same both at the pole and the equator. Therefore, the force of the support on the object is greater at the pole than at the equator, so the weight of the object is greater at the pole than at the equator.

Consider a non-inertial frame rotating with the earth with origin at the center of the earth. The object is at rest in this frame. At the pole the magnitude of the force of gravity equals that of the force of the support on the object. At the equator, in this frame the object also experiences a fictitious centrifugal force outwards. Due the centrifugal force, the force of the support on the object is less at the equator than at a pole, hence, the weight of the object is less at the equator.

The difference in weight is very small.

Consider an object initially at rest at the equator, then dropped

Once dropped, the only force on the object is that of gravity. Using polar coordinates, in the inertial frame, the radial and tangential equations of motion are $-mg\hat n = m(\ddot r - r\dot {\theta^2})\hat n$ and $0 = m(r\ddot \theta + 2\dot r \dot \theta)\hat l$ where $r$ and $\theta$ are the radial and angular positions, with unit vectors $\hat r$ in the increasing radial direction and $\hat l$ in the increasing angular direction, respectively. $g$ is the acceleration of gravity. The radial equation of motion has a centripetal acceleration term: $-r\dot {\theta^2}\hat n$ so the object does have centripetal acceleration when dropped, but this is a small correction for this case and the radial motion can be accurately evaluated as $-mg\hat n = m\ddot r\hat n$. The object initially had the same angular velocity as that of the earth before it was dropped. Once dropped, the object is only subject to the force of gravity, a central force, and its angular momentum must be conserved. This means that as the object is dropped its angular momentum is constant and as it is dropping (decreasing in height) its angular velocity exceeds that of the surface of the earth, so when it contacts the earth, the object is displaced relative to a vertical from the initial height of the object to the surface of the earth. See the answer by @Ricardo Ochel to Deviation of free falling objects (Coriolis effect) using conservation of angular momentum for a detailed evaluation in the inertial frame.

In the non-inertial frame rotating with the earth with origin at the center of the earth, the object was initially at rest, but when dropped it now experiences an additional fictitious force besides the centrifugal force, the Coriolis force, that deflects it sideways as it drops. See the text Analytical Mechanics by Fowles for a detailed calculation in the non-inertial frame.

The deviation from the vertical is the same whether evaluated in either the inertial or non-inertial frame.

So regarding your questions:

(2) g is not different. Sometimes it is said that the "effective" g is different where effective g means g minus the centrifugal effect.

(1) Consider an object at the equator that is dropped.

Viewed in the inertial frame, the object initially followed the earths rotation. Once dropped it maintains constant angular momentum and its angular velocity is not the same as that of the surface of the earth, so it has a displacement from the vertical on impact with the earth.

Viewed in the non-inertial frame, the displacement from the vertical is due to the Coriolis fictitious force.

$\endgroup$
0
$\begingroup$

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).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.