Suppose a bullet were to be fired with a velocity $v$ from the surface at equator. Considering the spin of Earth will it land at the same spot?

I tried to solve this assuming constant acceleration due to gravity.

Since the horizontal velocity of the bullet remains constant at height $h$ the angular velocity will be in (angular velocity at ground is $\omega$ and radius of Earth is $R$) $$ \omega' = \frac{\omega \times R}{(R+h)} $$ Now $\omega'$ can be written as $d\theta/dt$ and $h$ as $v\times t-0.5\times g\times t^2$ so after rearranging we get $$ d\theta = \frac{\omega\times R}{(R+vt-0.5\times g\times t^2)}dt $$ Integrating this wrt to $t$ from $0$ to $2\times v/g$ gives me the angular displacement of the bullet. Angular displacement of ground is $2\times v\times \omega/g $and if we subtract the two and further multiply by $R$ we should get the distance traveled further by the ground.

However after calculating everything I'm getting very high distance. For example, a bullet fired at $1700m/s$ will land $2,340m$ away which seems absurdly wrong. Can anyone point out where my mistake is?

  • $\begingroup$ You have the right idea, but you probably forgot that the surface of the earth has also rotated by nearly 2,340m while the bullet was in the air. For an observer "standing still" on the earth, the position where the bullet lands is the difference between those two nearly equal quantities. $\endgroup$
    – alephzero
    Mar 12 '17 at 17:44
  • $\begingroup$ "Angular displacement of ground is 2*v*w/g and if we subtract the two and further multiply by R we should get the distance traveled further by the ground." I didn't forget doing that. Maybe my calculations are wrong but I've checked everything multiple times now. $\endgroup$
    – archit
    Mar 12 '17 at 18:00
  • $\begingroup$ @alephzero But like an aeroplane it shall the velocity of the earth too $\endgroup$
    – Shashaank
    Mar 12 '17 at 18:17
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/166853/2451 and links therein. $\endgroup$
    – Qmechanic
    Mar 12 '17 at 19:21
  • 1
    $\begingroup$ Possible duplicate of Earth moves how much under my feet when I jump? $\endgroup$ Mar 12 '17 at 23:03

Angular velocity of earth: $\Omega=\frac{2\pi}{24*60*60}\frac{rad}{s}$

Radius of earth: $R\approx 6371km$

The velocity and height of the bullet in radial direction:$$v_{radial}=v_0-gt+R\Omega \sin(\Omega t)$$ $$h=\int{v_{radial}} \ dv=v_0t-\frac{1}{2}gt^2+R\cos(\Omega t)$$

Setting $h=0$ gets the point in time $t_2\approx1323.5s$ where the bullet hits the ground.

Distance travelled in tangential direction: $s_{tang}=R\sin(\Omega t)$

Distance travelled by earth: $s_{earth}=-R\Omega t$

Net distance: $s_{net}=s_{tang}+s_{earth}\approx -946m$ (ridiculously high)

  • $\begingroup$ When the surface is moving, the bullet also has a horizontal component $\endgroup$
    – Andrei
    Mar 12 '17 at 18:52
  • $\begingroup$ In the first equation, shouldn't it be -gt^2/2? Also the bullet has a horizontal component. $\endgroup$
    – archit
    Mar 12 '17 at 18:55
  • $\begingroup$ you re obiously right, compete rework $\endgroup$
    – Ben L
    Mar 12 '17 at 22:32
  • $\begingroup$ Still too high, I tried looking over it many times but can't find the mistake. Also 1700m/s isnt too high as it's the muzzle velocity of a modern AR. $\endgroup$
    – archit
    Mar 19 '17 at 8:01

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