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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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1 Answers1

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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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