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In theory, yes. In practice, no.

In theory, You could fire the bullet with such a great speed that it would orbit the Earth, even if it was only at a height of 1 meter from Earth's surface.

In practice, however, their is air resistance. This would cause the bullet to lose speed, and eventually it would fall the the ground.

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Q: If you fired a bullet and it never dropped would it orbit the earth and hit you?
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When a bulltet is fired from a rifle it is the bullet which causes damage but nit recoiling gun answer you according to momentum or third law of motion?

I believe that you are asking if the bullet is what causes so much damage to something it strikes, then why doesn't the shooter suffer an equivalent amount of damage? The answer does have a lot to do with momentum. I think that there may be some confusion between the conservation of momentum, which is the essence of Newton's Third Law, and the definition of kinetic energy.The magnitude of the momentum of the bullet just as it leaves the barrel of a gun is equal to, but in the opposite direction of, the magnitude of the momentum of the gun plus any momentum imparted to the shooter's body.I apologize if you already know much of my answer, but I will show the equations for momentum and kinetic energy. Momentum is a vector because it has both direction and magnitude, and the equation for momentum is p = m•v, where p is momentum, m is the object's mass in kg, and v is the object's velocity in meters/sec. (The vector quantities are in bold. I'm using the units prescribed in the kms system.)The reason the bullet causes damage to whatever it strikes is due to the bullet's kinetic energy, which is defined by the equation Ek=1/2mv2, where m is the mass of the bullet in kg and v is the bullet's speed in meters/sec. The fact that the kinetic energy is proportional to the square of the velocity is of paramount importance. Just like momentum, energy is also conserved. Energy (or the combination of energy and mass in the cases of nuclear fission and nuclear fission) is never lost; it can only change forms. All the energy released from every bullet, bomb, and shell used in World War II, or any other war, is still present somewhere in the universe, although almost all of it is still present on the Earth in some form. Thus, the kinetic energy of a bullet fired from a gun is initially exactly equal to the kinetic energy imparted to the backward-moving gun, the gun's report, the flash of light from the muzzle, the heating of the gun's receiver, chamber, and barrel, the heating of the bullet and of whatever it hit, the frictional heating of the air through which the bullet passed, and the energy expended in damaging what the bullet struck.The reason the shooter of the gun is not severely damaged (although anyone who has ever fired a very powerful rifle such as a 375 H&H or a 460 Wetherby Magnum might disagree) is because the energy of the bullet is proportional to the square of the velocity. Because the mass of the gun is so much greater than the mass of the bullet, the rearward velocity of the gun is very much less than the velocity of the bullet such that the gun cannot cause significant damage to the shooter.


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