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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?

sammy gerbil
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archit
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  • 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. – alephzero Mar 12 '17 at 17:44
  • "Angular displacement of ground is 2vw/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. – archit Mar 12 '17 at 18:00
  • @alephzero But like an aeroplane it shall the velocity of the earth too – Shashaank Mar 12 '17 at 18:17
  • Possible duplicates: http://physics.stackexchange.com/q/166853/2451 and links therein. – Qmechanic Mar 12 '17 at 19:21
  • Don't see any obvious mistake offhand. But 1.7 km/sec is a very large velocity which is starting to approach escape velocity (11.2 km/sec), and we know that the calculation you outlined breaks down at escape velocity because the projectile never returns. So I wouldn't be surprised to see a significant difference in the ground and projectile impact positions. A 2.34 km difference doesn't strike me as being obviously wrong. Have you tried using smaller velocities or plotting a graph of velocity versus difference in ground-vs-projectile positions? –  Mar 12 '17 at 21:33
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    Possible duplicate of Earth moves how much under my feet when I jump? – sammy gerbil Mar 12 '17 at 23:03
  • @sammygerbil I looked at both of the questions, however one of them is taking into consideration Coriolis Effect and another one Tidal Forces of sun. However in my question the calculations are pertaining to conservation of angular momentum – archit Mar 13 '17 at 02:24
  • Sorry archit, we do not check your work to locate errors of arithmetic or logic. Last 2 answers in Earth moves... give calculations which you can compare with your own. Coriolis Effect is just rotation of Earth beneath the bullet. None of the calculations include tidal effects from Sun or Moon. – sammy gerbil Mar 13 '17 at 02:36
  • You are discussing central forces, so the angular momentum $m\omega r^2$ is conserved--the tangential velocity distinctly slows down with h. Aren't you missing a square? – Cosmas Zachos Mar 13 '17 at 17:07
  • So, then $L=m\omega R^2$ and thus $\omega'=\omega/(1+h/R)^2$, which goes into your integral. You might want to shift your 0 of the time to the highest point of the trajectory, to simplify the integral and symmetrize the limits. Most considerations of Coriolis bunk are prone to mistakes, and only favored by the confusable. – Cosmas Zachos Mar 15 '17 at 13:17
  • This type of question is exactly what we expect students to ask, and we see it quite frequently not just here. Interestingly students rarely bother to either do the experiment themselves or look up to see who else has done it. When you have the answer as to what actually happens in reality THEN ask the question why ! For the benefit of those that do not already know this is typically a question that is often discussed in the Flat Earth forums. Well worth a visit if you have not already been there. – BetterBuildings Mar 28 '17 at 19:41

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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)

Ben L
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