Bullets, Distance and acting forces

[Andreee]

So me and my friend are having an argument on how far it is possible for a bullet to travel on earth, with gravity. Under the conditions we just want to see how the farthest a bullet could travel under the best circumstances. We looked up some windspeeds and found that 372 Km/h was the fastest on earth ever recorded. We also found that the fastest bullets travel at around 2km/s. so taking into account this info. if a bullet was shot from a 45 degree angle, how far could it possibly travel? If anyone has an idea or can work this out it would be a great help

[dr432]

well wouldn't it circle the earth and hit you in the back of the head? some useful calculations:

 

2km/s = 7200km/hour

Edited August 7, 2009 by dr432

[iNow]

well wouldn't it circle the earth and hit you in the back of the head?

 

There's a little thing called gravity that will prevent that from happening. Also, probably a mountain range or two... Which brings up the interesting question... how far above the ground is the person who pulls the trigger?

[CharonY]

Also the Coriolis force and friction.

[DJBruce]

Here's how I did it:

 

First I broke the vector into its components:

[math]v_{y}=2000Sin(45)=1414 m/s[/math]

[math]v_{x}=2000Cos(45)=1414 m/s[/math]

 

Then I used the equation for a parabolic projectile to find out how long it is in the air:

 

[math]d(t)=-\frac{1}{2}at^{2}+V_{0}t[/math]

[math]d(t)=-\frac{1}{2}(9.8)t^{2}+1414t[/math]

 

I graphed this and found the x-intercept. I found the projectile is in the air for 288s.

 

I assumed that the wind only effected the horizontal velocity. So:

 

[math]v_{x}=1414-103=1311 m/s[/math]

 

None I simply multiplied time by the bullets horizontal velocity to get its horizontal distance:

 

[math]d_{x}=(1311 m/s)(288 s)=377,600 m[/math]

 

That seems to big for me, but that's the process we used in physics. Also I know that is the probably not the proper way to deal with drag, but I assumed that wind would only affect the horizontal velocity and as I said this is the way my basic physics course dealt with a similar problem.

[dr432]

there's a little thing called centrifugipal force that's why satellites spin around you know?

[iNow]

there's a little thing called centrifugipal force that's why satellites spin around you know?

 

Actually, satellites fall to earth at the same rate that the curve of the earth falls away from the satellite. If you want to keep making up your own answers, then you'll keep being told you are wrong and you will keep being corrected.

[dr432]

i think what you're forgetting is that the bullet isn't going nearly fast enough to become a satellite at such a low height, and even if it was it would quickly get slowed down by air resistance.

[DJBruce]

i think what you're forgetting is that the bullet isn't going nearly fast enough to become a satellite at such a low height, and even if it was it would quickly get slowed down by air resistance.

 

No one, other than yourself, said it had the needed velocity to do so. If the bullet had the needed velocity, 11.2 km/s, once it reached space its orbit would decay slowly not quickly as you say because the air resistance is very small in space.

Edited August 7, 2009 by DJBruce
Addition/Spelling Correction

[dr432]

oh i was assuming you're shooting it parallel to the slope at the closest point so it would stay at a certain height

[DJBruce]

oh i was assuming you're shooting it parallel to the slope at the closest point so it would stay at a certain height

 

Even if you shoot the gun "parallel to the slope" it would not stay at a certain height because Earth's gravity would be pushing it downwards.

[dr432]

listen to inow:

 

the bullet falls to earth (gravity) at the same rate that the curve of the earth falls away from the bullet

 

like earth has gravity right? but the sun keeps circling around it it doesn't fall onto earth. because its moving fast enough and in space where there's no air resistance so its not slowed down/

[iNow]

listen to inow:

 

the bullet falls to earth (gravity) at the same rate that the curve of the earth falls away from the bullet

 

Satellites, not bullets.

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Merged post follows:

Consecutive posts merged

So me and my friend are having an argument on how far it is possible for a bullet to travel on earth, with gravity. Under the conditions we just want to see how the farthest a bullet could travel under the best circumstances. We looked up some windspeeds and found that 372 Km/h was the fastest on earth ever recorded. We also found that the fastest bullets travel at around 2km/s. so taking into account this info. if a bullet was shot from a 45 degree angle, how far could it possibly travel? If anyone has an idea or can work this out it would be a great help

 

Andreee,

 

Lots of the posts above are not helpful in the least. Sorry for that. You might learn more by looking to sites like this:

 

 

http://www.ehow.com/how_5185428_calculate-bullet-trajectory.html

Here's how to calculate a bullet's trajectory, specifically, the time aloft, the point of landing and the highest point in its path. In this example, certain assumptions have been made, for computational simplicity: negligible air resistance, no wind and insufficient firing distance for the Earth's rotation to have effect.

 

First...

[DJBruce]

listen to inow:

 

the bullet falls to earth (gravity) at the same rate that the curve of the earth falls away from the bullet

 

like earth has gravity right? but the sun keeps circling around it it doesn't fall onto earth. because its moving fast enough and in space where there's no air resistance so its not slowed down/

 

This only happens once the projectile has the escape velocity needed to become a satellite, which as I showed is 11.2 km/s not 2 km/s. Until the bullet has escape velocity the bullet is falling to Earth faster thane the curve of the Earth is falling away.

 

Also the currently accepted model of our solar system heliocentric model not a geocentric one, so the point about the sun is just plain wrong.

 

Also if anyone would carry to check my answer it would be appreciated.

Edited August 7, 2009 by DJBruce
added last line

[insane_alien]

a pretty major point here, 11.2km/s is escape velocity not orbital velocity.

 

satellites only circle the earth at ~7km/s depending on altitude(orbital speed decreases with altitude).

 

with 11.2km/s youjust wouldn't be coming back to earth, you'd enter a solar orbit rather than an earth orbit.

[swansont]

A bullet shot parallel to the surface, neglecting air resistance, would need a speed where the centripetal acceleration were equal to the gravitational acceleration in order to orbit.

 

[math]\frac{v^2}{r} = g[/math]

 

Or, about 7.9 km/s

 

This neglects aiming to miss any obstacles. You can get about a .5 km/s boost by shooting with the rotation at the equator. Remember to duck after shooting.

[swaha]

i didnt get the question. was it how long the fastest bullet can travel in air?

then i think the answer would depend on 1. gravity. 2. centrifugal force. 3. air resistance. 4. latitude for we would need the velocity of earth also.

[iNow]

It didn't say anything about "fastest bullet," but you're pretty correct in your paraphrase of the question. The question seems to ask, "What is the farthest distance a bullet can travel on Earth?" I assume it's implicit that this is a relatively normal bullet fired from a relatively normal gun under relatively normal atmospheric conditions.

[Sisyphus]

It's more fun on the Moon. There's no air resistance, and orbital velocity at the surface is only about 1700m/s. So, you know, careful with those high-powered rifles...

[Janus]

Here's how I did it:

 

First I broke the vector into its components:

[math]v_{y}=2000Sin(45)=1414 m/s[/math]

[math]v_{x}=2000Cos(45)=1414 m/s[/math]

 

Then I used the equation for a parabolic projectile to find out how long it is in the air:

 

[math]d(t)=-\frac{1}{2}at^{2}+V_{0}t[/math]

[math]d(t)=-\frac{1}{2}(9.8)t^{2}+1414t[/math]

 

I graphed this and found the x-intercept. I found the projectile is in the air for 288s.

 

I assumed that the wind only effected the horizontal velocity. So:

 

[math]v_{x}=1414-103=1311 m/s[/math]

 

None I simply multiplied time by the bullets horizontal velocity to get its horizontal distance:

 

[math]d_{x}=(1311 m/s)(288 s)=377,600 m[/math]

 

That seems to big for me, but that's the process we used in physics. Also I know that is the probably not the proper way to deal with drag, but I assumed that wind would only affect the horizontal velocity and as I said this is the way my basic physics course dealt with a similar problem.

 

Ignoring air drag and wind, there is another thing to consider. At such a range, the curvature of the Earth comes into play.

 

Using your example with no wind and no drag, you get a time of flight of 288s, and a horiz speed of 1414m/s for a distance or 408,040 m

 

To take curvature in to account you have to do it somewhat differently.

 

You have to treat the trajectory as a highly eccentric orbit which intersects the Earth's surface at two points.

 

To do this, you need to find the semi-major axis(a), period(P) and eccentricity(e) of said orbit.

 

You can find "a" by noting that the total energy of the bullet at launch is:

 

[math]E = \frac{mv^2}{2}-\frac{GMm}{r}[/math]

 

Where m is the mass of the bullet, G the gravitational constant, M the mass of the Earth, and r the radius of the Earth(from where the bullet is fired.)

 

And also is:

 

[math]E=-\frac{GMm}{2a}[/math]

 

Combining these two equations eliminates "m". So using 5.97e24kg for M, 2000 m/v for v, and 6.673e-11 for G we get:

 

a=3294489m

 

P can be found by

 

[math]P= 2 \pi \sqrt{\frac{a^3}{GM}}[/math]

 

thus P= 1882. sec

 

e can be found from the expression:

 

[math]\frac{Ar}{2}= \frac{ \pi a^2}{P} \sqrt{1-e^2}[/math]

 

where A is the Areal velocity, which in this case is the horizontal velocity of the bullet.

 

solving for e gives us:

 

e = 0.96852

 

The focus of this orbit is the center of the Earth.

 

We can find the radial vector (distance from the focus) at any point of the orbit from:

 

[math]R = a \frac{1-e^2}{1+e\cos \theta}[/math]

 

Where theta is the angle between perigee and the point of the orbit.

 

From the information we have, we can rearrange and solve for theta when R = 6367000m (the radius of the orbit) . Subtracting this from 180° and doubling it gives us the arc it covers during its trajectory.

 

This comes to 3.79°.

 

There are 111317 m to every degree on the Earth's surface, which gives a total distance traveled of:

 

421,798 m

 

compared to the 408,040 m you get neglecting the Earth's curvature.

 

Note that the above doesn't take into account the Earth's rotation. To do so, you have to adjust v and A accordingly.

 

For instance, for firing Eastward at the Equator, you adjust A to 1414+464 =1878 m/s and v to

 

[math]\sqrt{1414^2+1878^2} = [/math]2351 m/s

 

Which increases the range to 574,509 m.

 

Firing Westward decreases it to 278,192 m