I understand that how helicopter hovers at the same position or a thrown ball falls to the same position even though earth is spinning. This is due to the preservation of angular momentum and my understanding is that as earth is spinning so does every thing residing on it; when a helicopter/ball leaves the surface of earth, it's also spinning with the same angular speed and keeps on doing so. So the helicopter remains at the same position and the ball falls back to the same position. In case of a rocket, as the height increases the angular speed lowers because of the large radius and if we want to keep it stationary (relative to earth's position) we have to apply some horizontal force.

On the contrary, when one throws a ball horizontally while sitting on a spinning merry-go-round the ball curves w.r.t to the thrower while it moves in straight line w.r.t the observer standing stationary besides merry-go-round. This effect is Coriolis effect and can be attributed to the inertia. A sniper also has to consider this effect on spinning earth for long shots.

Now my question is that when a person sitting on the merry-go-round throws a ball then isn't the angular speed of ball same as that of the person when it leaves the person's hand? Shouldn't it go straight w.r.t the person throwing the ball? Also when the bullet leaves the gun then isn't the angular speed of bullet same as that of earth and isn't it ought to go in straight line?

Also (I think) the ball will come to the same position when thrown upward on a merry-go-round because it will have same angular speed, then why not for a horizontal throw?

  • $\begingroup$ "a thrown ball falls to the same position even though earth is spinning" No, it doesn't. A ball thrown vertically upward will land slightly west of the point from which it was thrown. $\endgroup$
    – Sandejo
    Sep 11, 2020 at 13:20
  • $\begingroup$ yeah and that's because the angular velocity will lower as height will increase. But this lead to another question that if angular velocity would lower as height increases (due to conservation of momentum) then wouldn't the angular velocity increase when height will decrease (due to conservation of momentum) $\endgroup$
    – user_3pij
    Sep 11, 2020 at 14:26

3 Answers 3


It seems to me you are suggesting a distinction between the rotating Earth case and the merry-go-round case that isn't actually there.

To emphasize similarity, let's take the case of firing a projectile from the center, towards the perimeter.

On the merry-go-round:
Once the projectile is fired the motion of it is completely unconnected with the merry-go-round. The projectile will travel along a straight line in the direction it was fired at the instant of being fired. The merry-go-round is rotating underneath that non-rotating line. So if you want to hit a particular target located on the rim of the merry-go-round to need to aim at the point where the target will be by the time the projectile arrives at the perimeter.

On Earth:
You are at the north pole, and you fire a rocket, aiming for a target on the equator. It will take that rocket an hour to make it to the equator. The rocket doesn't do any course correction along the way. The rocket gets all of its speed in the first few minutes of the flight.

Here's the crucial thing: for the rocket the journey to the equator is orbital motion. (We are accustomed to thinking of orbital motion as going around all the way, but the point is: during the flight the weight of the rocket isn't supported, that is what makes the motion orbital motion. It's just that the orbital motion is at such a low altitude that the trajectory intersects the Earth after an hour).

During the flight the orbital plane of the rocket doesn't change; it remains the same plane. The Earth is rotating underneath that non-rotating plane. Given that the flight takes an hour you should aim in the direction where you target will be in an hour.

External ballistics

In external ballistics the correction for drop of the bullet has to take the Earth rotation into account.

During its flight the character of the motion of the bullet is in some respects orbital motion. Gravity makes the bullet drop.

For long range firing: the sight is adjusted so that when aiming at the target you are in fact aiming above the target.

When firing in eastward direction the true velocity of the bullet is the muzzle velocity of the gun plus the velocity of co-rotating with the Earth.

Conversely, when firing in westward direction the true velocity of the bullet is the muzzle velocity of the gun minus the velocity of co-rotating with the Earth.

The true velocity of a bullet fired in eastward direction is larger than the true velocity of a bullet fired in westward direction and so you get a different amount of drop.

  • $\begingroup$ Let's say the merry go round is rotating with angular velocity of 5 revolution/sec and there's a canon on its peripheral which is also rotating with angular velocity of 5 revolution/sec. Now if canon fires a ball horizontally then isn't the ball be rotating with 5 revolution/sec and also moving forward. Also if canon fires the ball vertically upward instead of horizontally then isn't the ball come to the same position i.e canon. If so then what's so different for the horizontal case? $\endgroup$
    – user_3pij
    Sep 11, 2020 at 12:05
  • $\begingroup$ It will not, its angular velocity becomes linear velocity as soon as fired $\endgroup$ Sep 11, 2020 at 12:12
  • $\begingroup$ @AdrianHoward I got your point but I'm confused now. When I throw a ball upward from the surface of earth (lets say while standing on equator), then why isn't its angular velocity becomes linear and it lands on rotated earth? I guess if i do the same on merry go round then the ball will surely land on the rotated version of merry go round (not on the original position) $\endgroup$
    – user_3pij
    Sep 11, 2020 at 12:23
  • $\begingroup$ Let's put in other way. Let's say bullet is fired from the equator to another point on the equator (altitude doesn't change in the flight). Then the earth's rotation doesn't matter because the angular speed of bullet is same as that of earth. Now lets say bullet moves from equator to the pole, now the rotation would matter as the bullets angular speed will be more than the angular speed of the target placed near pole and bullet will curve. But still the bullet will have angular speed component along with equator-to-pole direction component and not only equator-to-pole component $\endgroup$
    – user_3pij
    Sep 11, 2020 at 13:03
  • $\begingroup$ If a bullet is fired from equator to pole then it's path would not be a straight line if even seen from the space but in case of merry-go-round the ball moves in straight line when seen from above. $\endgroup$
    – user_3pij
    Sep 11, 2020 at 13:38

The Earth has different angular speeds at different latitudes, so a bullet traveling north or south far enough would have to take this into account. Once a ball is thrown from a moving merry go round gravity and air resistance are the only forces acting on it, so it will not curve with the merry go round.

  • $\begingroup$ So the coriolis effect will be negligible for a small tennis ball? And a bullet travelling from east to west will not show curve?Right? $\endgroup$
    – user_3pij
    Sep 11, 2020 at 11:36
  • $\begingroup$ And why effect of air resistance not observable for a ball thrown from a moving truck where it has same horizontal speed as that of truck while the air resistance could have slowed it down? $\endgroup$
    – user_3pij
    Sep 11, 2020 at 11:44
  • $\begingroup$ Any projectile in the atmosphere can have its trajectory altered by air resistance. $\endgroup$ Sep 11, 2020 at 12:04
  • $\begingroup$ Yeah but lets ignore air resistance for now. Also in this short demo of throwing ball from truck youtube.com/watch?v=j1URC2G2qnc one can see that air resistance is not a notable factor. Put in this way, let's say I am sitting on merry go round and I am holding a ball. My and the ball's angular speed is same as that of merry go round. Now when I throw the ball then isn't the angular speed of ball same as that of merry go round the moment it leaves my hand just like of a helicopter when it leaves the ground? $\endgroup$
    – user_3pij
    Sep 11, 2020 at 12:14

As long. as you hold the balll in your hand, you excert a centripetal force on it, when it leaves your hand this force no longer exists, so it goes forward with the tangential velocity in the horizontal direction, while you on the merry go round continue in a circle, so the ball will land outside it makes a parabola landing at outside, The truck is different since it goes straight, if the truck would describe a circle, the result would be the same as with your merry go round.You are talking about angular velocity, but without centipede force there is no conservation of angular velocity, only tangential velocity is conserved (without friction)

  • $\begingroup$ I think I got your point just last confusion because of the ball throwing example. Does the centripetal force is provided by gravity when I throw ball upward? Let's say there's a fly sitting on a spinning football then would the fly be flown away in a tangential direction as soon as it fly a little upward and loss contact with the football? $\endgroup$
    – user_3pij
    Sep 11, 2020 at 17:34
  • $\begingroup$ A fly ia a bad example, so I see not what you mean. On earth and for satellites gravity provides the centripetal force. On a spinning ball the fly adheres to the ball the way she can walk on walls and the ceiling. $\endgroup$
    – trula
    Sep 11, 2020 at 20:47

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