Q:

If you shot a bullet parallel to the Earth’s surface and dropped a bullet at the same time from the same height, would both bullets hit the ground at the same time?

- David

Clear Lake High School, Texas

- David

Clear Lake High School, Texas

A:

Well, the simplified physics textbook problem would probably want the
answer "yes," since you can treat Newton's second law and the resulting
kinematics of uniformly accelerated motion separately for the
horizontal and vertical components of the motion. They probably tell
you to treat the air resistance as being zero and treat the Earth as
flat, both of which we know not to be true.

But more realistically, air resistance provides a drag force which increases with speed and points in the direction opposite to the velocity of the bullet through the air. The bullet is traveling very rapidly, mostly horizontally, but with a small downwards component. Air resistance provides a force that increases nonlinearly with speed, and so the vertical component of the air resistance force will be greater for the horizontally shot bullet than for the dropped bullet.

If the bullet is shot very fast, the curvature of the Earth becomes important (say if this is an artillery shell being shot at a target many miles away). Shoot a bullet fast enough in the horizontal direction, ignore air resistance, and you can get it in orbit around the Earth.

Alternatively, if there's a hill nearby, you could get the opposite answer.

Tom J.

But more realistically, air resistance provides a drag force which increases with speed and points in the direction opposite to the velocity of the bullet through the air. The bullet is traveling very rapidly, mostly horizontally, but with a small downwards component. Air resistance provides a force that increases nonlinearly with speed, and so the vertical component of the air resistance force will be greater for the horizontally shot bullet than for the dropped bullet.

If the bullet is shot very fast, the curvature of the Earth becomes important (say if this is an artillery shell being shot at a target many miles away). Shoot a bullet fast enough in the horizontal direction, ignore air resistance, and you can get it in orbit around the Earth.

Alternatively, if there's a hill nearby, you could get the opposite answer.

Tom J.

*(published on 10/22/2007)*

Q:

So what is the answer to the question??

Which bullet will hit the earth first?? I’ve been arguing this with a mate for years. I think the one dropped will hit the earth before the bullet that is shot.

write back to (deleted)

- Dombowski (age 27)

Hong Kong

Which bullet will hit the earth first?? I’ve been arguing this with a mate for years. I think the one dropped will hit the earth before the bullet that is shot.

write back to (deleted)

- Dombowski (age 27)

Hong Kong

A:

Before one can answer a yes/no, black/white kind of question that you have posed you must be extremely precise in specifying all the environmental conditions involved. Do you want to neglect air resistance, air currents, the earth’s rotation, geographical position, direction of firing, and on and on and on...? The classic high school text answer is that they both fall to the earth at the same time. However if you are at the earth’s equator the answer depends on whether you are firing toward the east or toward the west. Figure that one out... you can have more arguments with your mate.

LeeH

LeeH

*(published on 10/22/2007)*

Q:

The question of a fired vs a falling bullet is an interesting thought experiment. It would seem to me that a bullet fired with significant velocity at a perfectly horizontal angle to the immediate plane of the earth (in other words perfectly perpendicular to the earth's gravitational force) would hit the ground after the other bullet, being of identical size and mass, which is dropped from the same exact height above ground. this even assumes that the test is conducted over a topological flat ground. Bullets falling to the ground, whether by dropping or firing, are set into motion on a linear path toward the gravitational center of the earth. In this example we can ignore the slight gravitational variations the fired bullet would travel since the increase of gravitational force would be too small to appreciably affect the result. The main issue here is that by firing the bullet we have applied energy to the bullet that has acted on it in contradiction to its original linear path; we have set it on a new linear path. The two will begin to average out, forming a parabola (or bullet drop) over distance, as the two forces work against each other. But, since we have applied energy to the mass in a linear direction different than that of the linear direction of the gravitational energy, we have acted on it, which as Newton so aptly pointed out, will affect its tendency to "stay in motion". The problem with the thought experiment is that of scale. If you step back and look at it from a larger scale, like that of the bullet's relationship to the entire planet (and the spherical curvature of space-time surrounding our planet's gravitational center) it makes more sense. When you fire that bullet, you've essentially set that bullet in motion against that curvature. Satellites are a good example. Ignoring the concepts of "geostationary" as that is not relevant here, this should make sense. The way that a satellite stays in orbit is by getting it up to the desired height and applying a linear momentum to the satellite perpendicular to the earth's gravitational field. In this way, the satellite will want to travel in a straight line, conserving linear momentum, but the earth's gravity will "tether" the satellite as it exerts energy to pull the satellite along its linear path. The result is the two forces fighting each other: The satellite can't fly off into deep space because of the gravitational "tether", and the satellite won't just fall straight to the earth due to its desire to travel in a linear path perpendicular to this other force. If you balance these forces, the satellite maintains orbit. If, however, you launch the same satellite up to the same height and fail to give it any linear momentum contradicting the gravitational attraction of the two bodies, the satellite will simply fall back to earth, following its linear path along the curvature of space-time towards the gravitational center. The bullets are the same way, only on a much smaller scale. Theoretically, if you fired the bullet with sufficient energy as to have equal energy in both perpendicular paths, take out obstacles and drag coefficient, and could theoretically maintain the linear energy by adding additional energy to the fired bullet as it travels, it would essentially orbit the earth at 3 feet from the ground. It's a matter of energy, of which mass is only a portion of the equation. Velocity is the other portion of that equation. I can imagine that if the test were conducted with the use of a 50 caliber bullet or a large rail gun then the time differences would be such that it would be easier to see the differences between a dropping a bullet with no energy or momentum in the perpendicular linear path or dropping a bullet with vast quantities of energy and high velocities in the perpendicular linear path. I'm by no means a physics expert or professor so if my terminology is off then I apologize, but the concept is evident. Perhaps this angle (pun partially intended) is the one that has been overlooked in the thought experiment because it's hard to imagine a tiny bullet in scale to the entire earth and the space-time in which the earth sits and is traveling.

- Michael (age 38)

Greer SC

- Michael (age 38)

Greer SC

A:

Hello Michael,

You are right; a bullet fired horizontally will take a longer time to fall to earh. If fired with enough velocity it can orbit the earth of even go off into space. Your problem was described by Isaac Newton in the early 18th century. It goes by the name "Newton's Cannon Ball". See: https://en.wikipedia.org/wiki/Newton's_cannonball for a full explanation.

LeeH

*(published on 12/30/2015)*