What I'm basically asking is that if a body is projected with sufficiently high velocity so that it doesn't escape from the earth's gravitational field but reaches an appreciable height with respect to the radius of the earth, then when it comes back will it land on the same spot from which it was fired? You can neglect drag force and winds but do consider the rotation of the earth.

Basically what has to be considered is that the net force acts towards the centre of the earth and so I tried conserving angular momentum. That shows that the angular velocity of the object will decrease with increasing height above the earth. So basically the object moves with smaller angular velocity for some time in it's path.

That led me to believe that when the object finally lands back on earth it wouldn't do so at the place from which it was projected. Am I wrong?

  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/48287/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 24, 2015 at 15:35
  • $\begingroup$ In which frame of reference do you want it to move vertically? $\endgroup$
    – kasperd
    May 8, 2015 at 22:01

1 Answer 1


Even if I ignore wind and the drag forces and only consider the rotation of the earth the bullet will not hit the ground at the same place from where it was projected.

There will be Coriolis effect.

Coriolis effect: The Coriolis effect is a deflection of moving objects when the motion is described relative to a rotating reference frame. I suggest this website,here is a derivation of the deflection due to Coriolis effect of a freely falling body. http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node58.html

  • 3
    $\begingroup$ Paul has it right. The only place with no Coriolis effect is at the North and South Poles. If you fired a bullet straight up at either of those two locations, it would indeed fall back in the same spot. $\endgroup$
    – David Rose
    Feb 24, 2015 at 17:13
  • $\begingroup$ There's also no Coriolis effect at the equator, although there is still "centrifugal" force. $\endgroup$
    – user854
    Jan 3, 2016 at 5:43

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