# Thread: Effects of Earth's rotataion on a simple experiment

1. I am not sure if this question even makes any sense, it just popped in my mind while reading a book on special relativity.
Suppose an observer throws a ball upwards, now at the time of throwing, both the observer and the ball have a horizontal velocity same as the instantaneous velocity of earth at that moment (considering earth's rotation only).
Now the ball, after falling back will have no vertical displacement (relative to the observer), but if the ball is thrown with sufficient velocity, shouldn't it land ahead of the observer, as the path of the observer is somewhat an arc, due to earth's rotation, so its displacement should be less than the horizontal displacement of the ball?  2. The ball does come down in a slightly different spot from where it was thrown, but not ahead of the spot, but behind it. The ball also follows an arc due to the rotation of the Earth (actually a segment of an ellipse), but unlike the person, who's speed remains fixed, the ball looses speed as it climbs, and gains it back as it falls. In addition, depending on your latitude, the ball can also fall a bit to one side. The person would continue to travel along the same latitude, but the path of the ball would not as the ellipse its path is a section of has its focus at the center of the Earth.  3. Long distance rifle shots will drop differently when shooting in the direction of earth's rotation (west) compared to counter to earth's rotation (east).

Demonstrating how the Coriolis effect affects bullet drop at 1000 yards

This important because bullet drop is something that can be modified by adjusting powder loads.
It must be added that bullet drop is reduced when shooting up or down hill.

The coriolis effect also causes left and right drift (east or west drift) which is most severe but still small when shooting in the north-south directions (and vice versa)  4. Originally Posted by amitSingh95
I am not sure if this question even makes any sense, it just popped in my mind while reading a book on special relativity.
Suppose an observer throws a ball upwards, now at the time of throwing, both the observer and the ball have a horizontal velocity same as the instantaneous velocity of earth at that moment (considering earth's rotation only).
Now the ball, after falling back will have no vertical displacement (relative to the observer), but if the ball is thrown with sufficient velocity, shouldn't it land ahead of the observer, as the path of the observer is somewhat an arc, due to earth's rotation, so its displacement should be less than the horizontal displacement of the ball?
Dear amitSingh95,

Welcome to the forum! I'm sorry for not getting to this earlier. I thought I posted a response but it appears that when I posted my response it never got through. Now I have to write it all over again. This time I'm saving it! Ideally, no. It will land at a different place. The difference is so small for the typical ball through speed then it will land almost exactly where it left from.

However, there are certain forces acting on the ball called inertial forces which effect the path the ball takes. You can read all about them at

Fictitious force - Wikipedia, the free encyclopedia
Accelerated Frames of Reference

During World War II the deck guns on battleships were so powerful and the projectiles propelled so far that the Coriolis effect had to be taken into account. However if this is about something more closer to people size and speeds then we can neglect it. For example; in the text Classical Dynamics of Particles and Systems - Third Edition by Jerry B. Marion and Stephen T. Thornton they provide a worked out example on pages 345 to 347. It works out to about two pages of calculations to solve this problem so it's not something that's simple to do. The question states on page 345
Find the horizontal deflection resulting from the Coriolis force of a particle falling freely in the earth's gravitational field.
....
An object dropped from a height of 100 m at a latitude of 45 degrees will be deflected 1.55 cm.
1.554 cm is about the width of a finger. This ignores air resistance. Mind you that 100 meters is very high! It's about 1/5 th the height of One World Trade Center!!

I hope that helps.  5. Sincere thanks for that paper.

For my amusement I reduced this to a simple geometry and elementary physics problem.
The difference between the tangential velocity at earth's sea level and 100 meters above that is 7 mm/s at the equator and 5 mm/s st 45 degree latitude.

Assuming a vacuum and thus zero boost and zero braking by air (laterally and vertically) and assuming no moment of inertia effects and a sort of "flat earth landing zone",
A 4.5 second drop would be affected by 32 mm at the equator and 23 mm at 45 degrees.
Obviously I am only within a order of magnitude of a more refined calculation.  6. Originally Posted by pikpobedy Sincere thanks for that paper.

For my amusement I reduced this to a simple geometry and elementary physics problem.
The difference between the tangential velocity at earth's sea level and 100 meters above that is 7 mm/s at the equator and 5 mm/s st 45 degree latitude.

Assuming a vacuum and thus zero boost and zero braking by air (laterally and vertically) and assuming no moment of inertia effects and a sort of "flat earth landing zone",
A 4.5 second drop would be affected by 32 mm at the equator and 23 mm at 45 degrees.
Obviously I am only within a order of magnitude of a more refined calculation.
In a no air environment, I get 24mm for the Equator scenario under "real" conditions. Here's the basic steps:

Upon release, the object follows an elliptical trajectory with the Earth's center at a focus. (Basically is enters a highly eccentric orbit around the center of the Earth. The full orbit cannot be followed because the Earth gets in the way, but we can still use the parameters of the full orbit to work out the problem as the part of the trajectory that is above the Earth's surface is not affected).

From the object's tangential velocity and starting distance from the Earth's center, we get the orbital energy of the object.

This in turn, allows us to find the semi-major axis of the orbit and, in turn, it's eccentricity and period.

We use this information to find the eccentric and mean anomalies, which let's us get the time from periapsis (closest approach to the Earth's center) to Earth surface distance. Subtracting this from 1/2 the period of the orbit gives us the "fall time" from when the object is released until it hits the surface.

The orbital information from above also allows us to determine the true anomaly (the angle from periapsis) of the orbit at the point where it intersects the surface of the Earth. Subtracting this from 180 degrees give the angle between the release point and intersection point.

Taking the difference between the arc formed by this angle along the surface of the Earth and the distance the Earth's surface moves during the "fall time" gives us the final answer as the the "drift" for the falling object.

For the 45 degree latitude problem, it's a bit more complicated.

Here's the thing, in the above example, both the plane of the object's trajectory and the plane for the path for the point on the equator below where it was dropped passed though the center of the Earth and remained parallel to each other. In the 45th latitude example, this is no long the case. The the plane of the object's path still cuts through the center of the Earth, but that for the path of the point on the Earth's surface does not. The two paths do not remain parallel, and you will get a North-South drift as well as a East-West one. I'll leave it at that without working out the whole problem.   