Bullet Velocity Matching Gravity on a High Platform?

[InquiringMind]

Every once in a while, I wonder about something off the wall. Here's the latest:

If you were on a hypothetical stationary platform, very high above the Earth (x number of miles), and you fired a gun downward, would the speed of the bullet eventually slow down to the velocity of gravity? Maybe "slow down" isn't the right term, maybe "match" the velocity of gravity would be a better term.

 

[daniel_i_l]

What is the "velocity of gravity"? Gravity is a force. (in classical mechanics)
The speed of the bullet would eventually reach the terminal velocity with is the velocity where the force of air-resistance on the bullet equals the force of gravity. If air resistance is -R*V and gravity is Mg then
R*V_terminal = Mg
V_terminal = Mg/R
Where R is some constant and M is the mass.

 

[InquiringMind]

What is the "velocity of gravity"?

oops! Sorry I used the wrong term. It's been a long time since I've been in school, and it was late when I posted . I think I would have more correctly phrased the question if I'd asked - would the "velocity of the bullet" at some point match the "the rate of acceleration of gravity" (10m/sec squared)? I believe you are saying it will, due to the effect of air resistance on the bullet.

 

[nrqed]

oops! Sorry I used the wrong term. It's been a long time since I've been in school, and it was late when I posted . I think I would have more correctly phrased the question if I'd asked - would the "velocity of the bullet" at some point match the "the rate of acceleration of gravity" (10m/sec squared)? I believe you are saying it will, due to the effect of air resistance on the bullet.

The question is meaningless. It's comparing apples to oranges. The *acceleration* due to gravity is in m/s^2. The speed of the bullet can never "match" that value since a speed can never be equal to an acceleration!

 

[arildno]

It is certainly true that if the bullet gets close enough to the Earth, the force it experiences from Earth will become approximately constant.

 

[arunma]

Every once in a while, I wonder about something off the wall. Here's the latest:

If you were on a hypothetical stationary platform, very high above the Earth (x number of miles), and you fired a gun downward, would the speed of the bullet eventually slow down to the velocity of gravity? Maybe "slow down" isn't the right term, maybe "match" the velocity of gravity would be a better term.

I think what you meant to ask is, "would the bullet eventually reduce its acceleration to 9.8 m/s²? First of all, it's important to note that if you fire a bullet in space, it won't have any acceleration. Acceleration is a change in velocity. The bullet will accelerate while it's still in the gun barrel. But once it leaves, it's acceleration is 0. As it gets close to the surface of the earth, it will accelerate at about 9.8 m/s² (ignoring the fact that the gravitational acceleration outside the atmosphere is about 30% less than surface gravity). However, as soon as it enters the atmosphere, the force of air resistance will oppose the force of gravity. The bullet's acceleration will eventually reach 0, at which point it will fall at a constant velocity. This velocity is called the terminal velocity.

Well, I hope that helps.

 

[russ_watters]

The bullet will accelerate while it's still in the gun barrel. But once it leaves, it's acceleration is 0.

This is only true if the bullet and gun are in orbit, and per the OP's statement about a stationary platform, I would say they are not. Immediately after leaving the barrel of the gun, the bullet will accelerate at whatever g is at that altitude. A few hundred miles up, it is still a good 9 m/s/s.

 

[arunma]

This is only true if the bullet and gun are in orbit, and per the OP's statement about a stationary platform, I would say they are not. Immediately after leaving the barrel of the gun, the bullet will accelerate at whatever g is at that altitude. A few hundred miles up, it is still a good 9 m/s/s.

You're right of course. Though if we want to get technical (just for fun), then I must object to the term "orbit." If the gun were truly in a circular orbit, then the gravitational force would be locally transformed away, and the acceleration really would be zero. But I know what you mean, and you are correct. Therefore I'll make the following note.

My previous post makes the assumption that the gun is initially a sufficient distance from the Earth that the gravitational force is effectively zero. But the result holds true anyway.

 

[InquiringMind]

This is only true if the bullet and gun are in orbit, and per the OP's statement about a stationary platform, I would say they are not. Immediately after leaving the barrel of the gun, the bullet will accelerate at whatever g is at that altitude. A few hundred miles up, it is still a good 9 m/s/s.

Actually, my orginal thought, which I didn't post, had to do with standing on the edge of a VERY high cliff, and throwing a ball straight down. At first the ball would travel faster than if you had just dropped it, but I wasn't sure if it would continue to travel faster than if dropped, or if at some point it would merly be falling at a rate determined by the force of gravity. I think this is the same as the gun analogy without the issues of space and orbits.

 

[nrqed]

This is only true if the bullet and gun are in orbit, and per the OP's statement about a stationary platform, I would say they are not. Immediately after leaving the barrel of the gun, the bullet will accelerate at whatever g is at that altitude. A few hundred miles up, it is still a good 9 m/s/s.

Even if an object is dropped from the space shuttle in orbit around the Earth, its acceleration is NOT zero! What do you define as being "in orbit"? Even at the altitude of the space shuttle the value of g is far from being zero (it's smaller than 9.8 m/s^2 but is still far from zero). The astronauts in the space shuttle do NOT have a zero acceleration! (otherwise they would not be in orbit!)

 

[Ki Man]

I don't understand the OP. what the question here

 

[Integral]

The ball thrown from a cliff is a much simpler situation.

If the initial velocity of the ball is greater then the terminal velocity, then the ball will slow down to the terminal velocity. If the initial speed of the ball is less then the terminal velocity it will accelerate at g until it reaches terminal velocity.

In a vacuum, ie no air Resistance, the ball will accelerate at g. Its speed when it hits the ground will be the initial velocity plus the speed gained due to the acceleration of gravity.

 

[kendr_pind]

do u guys mean to say that a meteorite will hit at the terminal velocity...!
in this case the initial velocity is much much greater than terminal velocity...
simply throw a piece of iron at a magnet with whatever force u can..when it hits the magnet obviously it will hit with a velocity more than if u had just let the magnet attract the iron piece by itself...

 

[Danger]

If the atmosphere is thick enough and dense enough, relative to the initial speed of a meteoroid, the latter will indeed decelerate to its terminal speed. That is assuming that it doesn't just burn up as most do due to compression and friction heat. Keep in mind that they can be moving at over 100,000 mph relative to Earth, which introduces entirely different conditions than the OP was asking about.

 

[russ_watters]

Keep in mind that they can be moving at over 100,000 mph relative to Earth, which introduces entirely different conditions than the OP was asking about.

Actually, the maximum you'd typically see except for a very unusual case would be about 40,000 mph. But the same idea still applies. If we say the atmosphere starts at 150 miles, that's only 13 seconds for the meteor to reach the ground for a vertical impact. It would need to decelerate at 1500 g for 26 seconds to reach terminal velocity before hitting the ground.

 

[russ_watters]

Even if an object is dropped from the space shuttle in orbit around the Earth, its acceleration is NOT zero!

You're right, of course, but it's not relevant for the OP's purpose of dropping the object off a platform. I could have worded it better.

Even at the altitude of the space shuttle the value of g is far from being zero (it's smaller than 9.8 m/s^2 but is still far from zero).

Right - which is what I said.

 

[dimsun]

moving in the direction away from Earth g becomes smaller.
Doesn't the value of g becommes a little little bit bigger towards earth?
Mayby it's a small effect, but is stil there.

Also the gravity field strength of the object which is falling towards earth, is atrackting the Earth itself. (but that is a really really small effect.)

(is there an official name for a change in acceleration?
I call it myself the escalation).

Dimsun.

 

[Danger]

Actually, the maximum you'd typically see except for a very unusual case would be about 40,000 mph.

Yeah, I was using an extreme example. I like to scare newbies.

(is there an official name for a change in acceleration?

Hmmm... I don't know. As far as I know, a change of acceleration is still just an acceleration (since acceleration itself is a change of speed or direction). I guess that you could refer to it as delta a if you want.

 

[DyslexicHobo]

(is there an official name for a change in acceleration?
I call it myself the escalation).

I believe I remember my physics teacher calling it "jerk", and I guess would be measured in m/s^3... not really sure if that's the official name for it or not. A quick google search could probably help out here.

The answer to what I THINK the OP was really trying to ask:

The ball dropped and the ball thrown down will eventually be traveling at the same velocity. When the ball reaches this velocity, their acceleration will be zero (hence the same, constant velocity). Acceleration due to gravity for the Earth (for all intents and purposes) is equal to 9.8 m/s^2. This means that gravity will always be applying the force of weight (F=ma) just enough to make its downward acceleration 9.8. This force is constantly being applied, but at higher velocities the ball will begin to experience drag. This force will counter the force of gravity, and they will eventually cancel out. When this happens, the acceleration is zero (F=ma, the F is zero due to the gravity/drag canceling)-- this is called an object's terminal velocity. However, the ball thrown will always have traveled more distance total at any time (besides t=0 and when they hit the ground).Hope this helps.

 

[disregardthat]

I have actually asked the very question before on this forum about change in acceleration. Change in acceleration is indeed 'Jerk', and change of Jerk is 'snap' (or Jounce). 'Crackle' and 'pop' follows.. To keep a constant velocity you need no force. To keep a constant acceleration you need a constant force. To keep a constant Jerk you need a constant change of force. If you want snap you need a constant change of the change of force... and so on..

In a serie: Velocity, Acceleration, Jerk, Snap (Jounce), Crackle and Pop

Want to know why the names are like that? http://en.wikipedia.org/wiki/Jounce

 

[Danger]

Well now... you've just made my life a hell of a lot more complicated than it needed to be.

 

[xAxis]

I think the OP thinks that when you push an object towards the Earth with some significant velocity, it might become "too fast" for gravity to accelerate it further, and therefore it might slow down to

... maybe "match" the velocity of gravity ...

The thruth is that gravity will accelerate the object regardless of it's speed.

 

[dimsun]

I have actually asked the very question before on this forum about change in acceleration. Change in acceleration is indeed 'Jerk', and change of Jerk is 'snap' (or Jounce). 'Crackle' and 'pop' follows..

Maybe it's of topic, but I have to ask it:
There also is the serie: mass, momentum, energy. What's the next in this serie? (again I made a word myself, calling it "valention").

Dimsun

 

[americanforest]

Maybe it's of topic, but I have to ask it:
There also is the serie: mass, momentum, energy. What's the next in this serie? (again I made a word myself, calling it "valention").

Dimsun

These quantities aren't simply derivatives of the previous quantity. It's not really a series of anything. There is a connection, but it's not as simple as the connection between velocity/acceleration/jerk, etc... I like your word though. Can you explain what it's supposed to be?

 

[Integral]

This thread seems to be done, it is now wandering off topic. LOCKED