Can you answer a physics question discussed by Hooke and Newton?

[Lionel Mandrake]

A ball is dropped from the top of a tower on the equator. Does it land right at the base of the tower, slightly to the east, or slightly to the west? Neglect air resistance, and assume the Earth is perfectly spherical and rotating at constant angular velocity.

Can you prove your answer?

[Xipe Totec]

1. Slightly to the East

My reasoning is as follows.

Assuming the tower was tall enough to reach geostationary orbit, the ball would actually remain stationary above on a spot normal to the surface. If the tower was taller than geostationary then the ball would actually not be able to keep up and would drift to the West as the earth moved under it. Therefore the opposite would happen if the tower was shorter than geostationary orbit and would actually rotate in an orbit faster than the surface of the earth and land slightly to the East.

[juxtaposed]

2. slightly west.The tower is perpendicular to the tangent surface of the earth, as in zenith.

as the height of the tower increases it's speed increases faster than the surface of the earth. I am sure I'm wrong.

[VMA131Marine]

3. Conservation of angular momentum

The Earth rotates west to east. As the ball falls, its angular velocity has to increase to conserve angular momentum (the product of mass times rotation rate times distance from the center of the Earth squared). As distance decreases, linear speed has to increase. Therefore the ball lands slightly east of the tower.

[Lionel Mandrake]

4. Correct.

Here's a slightly different argument: The Earth and the ball start with the same angular velocity. Since the angular velocity of the ball increases while that of the Earth stays constant, the ball will rotate through a greater angle.

[Igel]

5. And if we move away from the equator the same argument holds.

And we get the Coriolis effect. Of course, the movement is to the West as we move towards the equator.

Sniper's equations have to take into account both effects for long-distance shots.

[Lionel Mandrake]

6. It's interesting that you should mention moving away from the equator.

Newton and Hooke also considered this slightly more complicated case and concluded, correctly, that for a tower in the Northern Hemisphere the ball would land slightly to the southeast. The Newton-Hooke correspondence took place before Newton had put the final touches on his laws of motion and gravitation. It's remarkable that Newton and Hooke could get the right answer at that time. We can do so more easily by using conservation of angular momentum in an inertial coordinate system, where there is no Coriolis force.

[Lionel Mandrake]

7. On second thought, using the Coriolis force is a good idea.

Start with local horizon coordinates, i.e., a rotating Cartesian coordinate system with the x, y, and z axes aligned as follows:

x = East; y = North; and z = Zenith.

Let r, v, and a be the position, velocity, and acceleration vectors.
Thus v = dr/dt and a = dv/dt.

The equations of motion are

a = - g - 2 omega cross v,

where the gravitational acceleration is g = -|g|(0,0,1),
the Coriolis acceleration is - 2 omega cross v,
and "cross" is the vector cross product.

The angular velocity of the earth is omega = |omega| (0,c,s), where c and s are respectively the sin and cos of north latitude. Both c and s will be positive for a point in the Northern hemisphere.

The magnitude of omega is approximately |omega| = 2 pi / (24 hours).

The initial position is r = (x,y,z) = (0,0,H), where H is the height of the tower. We can assume H < 1 km, which is a tiny fraction of the Earth radius. The initial velocity is v = (0,0,0).

Without actually solving the equations of motion, we can see that the solution will be like this: The downward acceleration will cause a buildup of downward velocity, such that the Coriolis force will cause an Eastward acceleration, which will cause a buildup of Eastward velocity, such that the Coriolis force will impart a Southward acceleration. Thus the trajectory will veer to the Southeast.