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. Ignoring the concepts of "geostationary" as that is not relevant here, this should make sense. When you fire that bullet, you've essentially set that bullet in motion against that curvature. 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. The problem with the thought experiment is that of scale. 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 two will begin to average out, forming a parabola (or bullet drop) over distance, as the two forces work against each other. 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. ![]() 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. 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. this even assumes that the test is conducted over a topological flat ground. 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. The question of a fired vs a falling bullet is an interesting thought experiment. Shoot a bullet fast enough in the horizontalĭirection, ignore air resistance, and you can get it in orbit aroundĪlternatively, if theres a hill nearby, you could get the opposite answer. If the bullet is shot very fast, the curvature of the Earthīecomes important (say if this is an artillery shell being shot at a So the vertical component of the air resistance force will be greaterįor the horizontally shot bullet than for the dropped bullet. Resistance provides a force that increases nonlinearly with speed, and Rapidly, mostly horizontally, but with a small downwards component. Increases with speed and points in the direction opposite to the You to treat the air resistance as being zero and treat the Earth asįlat, both of which we know not to be true.īut more realistically, air resistance provides a drag force which Horizontal and vertical components of the motion. ![]() Kinematics of uniformly accelerated motion separately for the ![]() Well, the simplified physics textbook problem would probably want theĪnswer "yes," since you can treat Newtons second law and the resulting
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