But is there any gravitational pull from a plane? Although infinite in 2 directions, it’s infinitely non-infinite in that 3rd direction.
I’m curious. Where did you think the force came from that “held up” the shot bullet? From its forward speed? If so, why would forward speed translate to lift in any way?
My guess is this is some sort of instinctive thing from airplanes: they move forward fast and generate lift, so anything moving fast will generate lift. Is this a fair guess? If so the crucial difference is of course the lack of wings on the bullet to translate the forward speed into lift.
If it has a finite area mass density (i.e., x kilograms per square meter), then it has a constant gravitational field. At least, so sayeth Newton-- In general relativity, it can’t exist.
The dropped bullet will hit the ground before the bullet fired from the gun. The bullet fired from the gun has a nose high trajectory which creates aerodynamic lift.
So the gravity acting on an object resting on a halfspace is the same as if that object was 1km away is the same as if it was 1 light year away? I’ll have to grab a piece of paper because this doesn’t sound right to me.
I may have this wrong, but think of it this way: how much mass is involved in the gravitational attraction, no matter where the object is? Now, over what average distance is that mass acting? The answer is the same regardless of where the object is. Another example of why it can be a real mind-bender when you start thinking of what “infinite” means in various contexts… :eek:
I was simplifying the bejeezus out of it for our non-technical OP and may have gone a wee bit too far. Or at least far enough to lose you at the bakery.
My point was that velocity parallel to a (functionally) infinite plane is meaningless in conjunction with any kind of “escape”. The plane’s gravity pulls the bullet downward and no matter how fast the fired one is going, it won’t get to any kind of edge effect on the (functionally) infinite plane before it strikes the surface. And at the same time as the stationary dropped one of course.
But … If the plane is not a true plane, but is instead, say a really, really huge sphere (like say about 8000 miles in diameter), then escape does become conceptually possible.
We normally think of escape velocity as a velocity normal to the surface, i.e. straight up. And (ignoring atmospheres) the concept behind “escape velocity” is that an initial velocity greater than that critical value will result in the projectile’s slowing due to gravity never quite reaching zero as the projectile gets to the “edge” of the launch surface’s gravity well at the “edge” of the universe.
So a projectile fired at a high enough speed tangent to the surface of a very large sphere would, eventually, be far enough from the sphere that for all practical purposes the sphere is directly behind it, or equivalently, the projectile was fired directly away from the center of the gravity well.
Bottom line: So on an atmosphere-free absolutely smooth planet you could fire something tangent to the surface fast enough to enter a stable orbit. You could fire it bunch faster than that and end up with escape.
But that would not work if your surface was a true plane and plenty larger than the distance the bullet travels before it falls to the surface.
The plane doesn’t need to be infinite. It just needs to be larger enough than the range of the bullet that edge effects are negligible enroute to the point of impact. I’m guessing 10 miles square would do for typical rifle ammunition fired from shoulder height. If 10 isn’t enough, let’s try 100. That’s still far enough from infinite that we can ignore all the relativistic reasons that infinite planes can’t exist in the real universe.
I hope that makes some sense.
It was specified in the OP that the bullet is fired in a vacuum.
Infinite mass, infinite distance. And?
IMO this is analogous to an infinite line of charge where the distance from that line definitely does matter. Maybe assuming a halfspace instead of a finite thickness slab that extends infinitely laterally (an exact analogy, I think) messes with it, but I don’t think so.
Hmm, but escape velocity doesn’t depend on the direction… touche.
No, an infinite line of charge is analogous to an infinite line of mass. An infinite plane of mass is analogous to an infinite plane of charge: In either case you get a uniform field. Gauss’s Law applies the same way to gravity as it does to electromagnetism, so you can calculate the field the same way in both cases.
I know this goes against the spirit of the question, but, technically, the bullet is supported inside the barrel of the gun for the amount of time that it is inside the barrel. So, the fired bullet will land a tiny bit after the dropped bullet.
Did I hear someone say it’s in a vacuum on a plane ? Should I be concerned… I have to fly tomorrow. Should I call security ? Where on the plane do they keep the vacuums anyway?
[nitpick] Wings aren’t really necessary to get lift, lifting bodies. [/nitpick]
CMC fnord!
:smack:
Here’s another way to visualize it: When the object is 1 millimeter away, the net gravitational force from the piece of mass *immediately below it *is pretty big (relatively). But as you go out further away on the xy plane (while the object is still 1mm away in the z-direction), all those gravitational forces from all those pieces of mass mostly cancel each other out (as most of them are acting pretty parallel to the ground, in opposite directions).
But as the object moves farther away from the surface(say, 1km in the z-direction), the gravitational force from the piece of mass immediately below it gets weaker, but the net downward gravitational force from the ring of mass further away on the xy plane gets stronger. When you add up the gravitational forces from all those rings of mass (going out to a diameter of infinity), you’ll find the net downward gravitational force adds up to the same at 1km as it was when the object was only 1mm away.
>a large force on one of them would create some difference
This presents a nice opportunity for understanding cosines.
The force delivered during the shooting would make a difference to the degree that it is aligned with the axis of interest. That is, if the shooting force were perfectly aligned with the vertical dimension in which the falling is meaningful, it would be tremendously important. If there were some little angle between the shooting force and the vertical, the importance would diminish a little but not much. However, as you get further from perfect alignment, the force gets way less important, and actually changes sign as you pass through 90 degrees. So, the misalignment dictates a multiplier that is between 1 and -1.
This is what cosines are all about. The cosine of an angle is the multiplier you’d use to account for things not working in the same direction. Cosines are all about dilution.