1. if I were to fire a super fast gun just outside the earth how fast is it if the bullet travel 360 around the globe and hit me in the back of my head?

2.

3. Around 7902 meters per second

4. And for details of how that is worked out, see here: Orbital speed - Wikipedia, the free encyclopedia

5. That would depend entrirely on how far away is 'just outside'. Also there is a large amount of small details one could think of but still.

6. Isaac Newton gave a well known diagram relevant to this:

7. Wow nice question, I think you should buy the energy power of a thermonuclear central to make that shoot lol

8. Then how do you calculate heat and air resistance with this combined?

9. Alright, guys seriously?

First of all, outside the earth the only speeds that are high enough to do the trick lie outside the atmosphere. Second It would take you about 20 times the speed of sound to to it. This doesn't require a very big gun though. As the expansion of gases are much more powerfull and the resistance is much smaller.
Second, if you do use a gun, you will get the same force backwards as the projectile you fire, moving you backwards. Second the idea is really simple. But the Aim is incredibble difficult. You have to fire a target the size of your head over a distance of 72 thousand kilometers away. That is about as accurate as shooting a hair point at the exact cutting edge of a pair of scissors about 10 meters away...

10. You would have to fire this superfast bullet horizontally so that it follows the curvature of the earth right?

Interesting fact: A bullet that is fired horizontally will hit the earth at the same time as one that is dropped from the same height. Gravity will act on a speeding bullet in exactly the same way it acts on a stationary one... they both hit the ground at same time.

Therefor you would need to fire a bullet so fast that it gets around the earth almost instantly... before gravity can act on it so much that it drops and hits you in the neck or torso.

11. Originally Posted by question for you
Therefor you would need to fire a bullet so fast that it gets around the earth almost instantly... before gravity can act on it so much that it drops and hits you in the neck or torso.
Earth will drop before the bullet drop, so it float above ground. It doesn't need to be instanteneous speed.

Surface of Earth is curved downward*

*Post#5 by JonG

12. I was trying to work out how earth curvature would come into play... I saw john's picture but didn't fully understand what it meant. I take it Newton has done all the maths.

I think the thing about gravity having the same effect on the bullet regardless of it's velocity was something Newton said.

13. that's kind of complicated answers. but pumps my interest in science. can anyone here make a formula?

that's kind of complicated answers. but pumps my interest in science. can anyone here make a formula?

What do you want a formula for?

15. for rounds per second ~ curiosity

16. Originally Posted by question for you
that's kind of complicated answers. but pumps my interest in science. can anyone here make a formula?

What do you want a formula for?
The speed of the projectile very much comes into play here. If it's to fast the orbit will increase by the time it makes one revolution and it will miss the back of your head, and if it's to slow the orbit will decrease enough to miss.

Also, if you are in orbit, that will have to be accounted for too. The only chance you have to hit the target is to know all the variables that will help you calculate the exact speed the projectile will need. But if you are not really serious about shooting yourself in the back of the head, not having the formula is a good idea.

17. The speed of the projectile very much comes into play here. If it's to fast the orbit will increase by the time it makes one revolution and it will miss the back of your head, and if it's to slow the orbit will decrease enough to miss.

Also, if you are in orbit, that will have to be accounted for too. The only chance you have to hit the target is to know all the variables that will help you calculate the exact speed the projectile will need. But if you are not really serious about shooting yourself in the back of the head, not having the formula is a good idea.

for rounds per second ~ curiosity
A formula for rounds per second?

You need to fire a certain amount of rounds... such as 10, or for more accuracy 100 and time it in seconds.

Then the formula is: amount of rounds divided by amount of seconds taken to fire them. Say 5 seconds to fire 100 rounds, 100/5. answer 20 rounds per second.

r/s=rps

Maths is useful huh

19. Originally Posted by arKane
Originally Posted by question for you
that's kind of complicated answers. but pumps my interest in science. can anyone here make a formula?

What do you want a formula for?
The speed of the projectile very much comes into play here. If it's to fast the orbit will increase by the time it makes one revolution and it will miss the back of your head, and if it's to slow the orbit will decrease enough to miss.
No. As long as the velocity of bullet is large enough for it to complete a full orbit and is less than escape velocity (~ 1.4 times the velocity required for a circular orbit), the bullet will return to its point of origin. As long as you haven't moved in the meantime, the bullet will return to you

Also, if you are in orbit, that will have to be accounted for too. The only chance you have to hit the target is to know all the variables that will help you calculate the exact speed the projectile will need. But if you are not really serious about shooting yourself in the back of the head, not having the formula is a good idea.
That depends. When you fire the Gun, the bullet enters one orbit and the recoil alters your own orbit. Both orbits will still return to the point where the Gun was fired. The periods of your orbits will be different, but with the bullet mass and velocity you could arrange to have the bullet's period to be exactly twice yours and you'll met up again on your after your second orbit.

If you don't want to go to that much bother, then you'll just have to be patient. Eventually, given enough orbits, your orbit and the bullet's orbit will bring you both back to the same point at the same time again.

20. Originally Posted by Janus
That depends. When you fire the Gun, the bullet enters one orbit and the recoil alters your own orbit. Both orbits will still return to the point where the Gun was fired. The periods of your orbits will be different, but with the bullet mass and velocity you could arrange to have the bullet's period to be exactly twice yours and you'll met up again on your after your second orbit.

If you don't want to go to that much bother, then you'll just have to be patient. Eventually, given enough orbits, your orbit and the bullet's orbit will bring you both back to the same point at the same time again.
I was thinking most guns won't produce a projectile speed capable of making even one orbit. Does anybody know for sure that this is not the case? However I'm thinking some kind of rail gun might get the job done.

21. Originally Posted by question for you

Interesting fact: A bullet that is fired horizontally will hit the earth at the same time as one that is dropped from the same height. Gravity will act on a speeding bullet in exactly the same way it acts on a stationary one... they both hit the ground at same time.
Physics isn't your strongest field is is?

You see, your statement works if the earth is flat. But it isn't. The bullet travels straight, but the gravity field of the earth works towards its core. Try it out, if you stand on the ground. You might see a far away church tower. But there comes a moment where you can't. This isn't because you are too far away, but because the curvature of the earth is in front of it. If you shoot straight then for every kilometer traveled the bullet keeps traveling straight, but the ground beneath it will drop just a little bit.
Hence if you just go fast enough (without too much drag) you can go round quite easily. Now the closer you are to earth the faster this must be, because the gravitational pull is larger.

If you get the right speed (and a lot of patience) I could throw a baseball around the planet (but that would actually be outside the orbit of the moon).

22. Originally Posted by Kerling
Originally Posted by question for you

Interesting fact: A bullet that is fired horizontally will hit the earth at the same time as one that is dropped from the same height. Gravity will act on a speeding bullet in exactly the same way it acts on a stationary one... they both hit the ground at same time.
Physics isn't your strongest field is is?

You see, your statement works if the earth is flat. But it isn't. If you shoot straight then for every kilometer traveled the bullet keeps traveling straight, but the ground beneath it will drop just a little bit.
If the bullet is shot horizontally from a height of 1 metre above the ground. Gravity will pull it 1 metre down from it's projectory in exactly the same time as gravity would make a bullet, dropped from 1 metre above the ground, hit the ground.

Is that better?

This isn't my physics, I just read it somewhere.

23. Originally Posted by question for you
I was trying to work out how earth curvature would come into play... I saw john's picture but didn't fully understand what it meant. I take it Newton has done all the maths.
I'm trying to find it also (how to work out how earth curvature effect bullet trajectory)... Still don't know how.

I don't even know if it even has a formula... but I saw in a computer programs they could calculate this thing very well (ie: "Orbiter" space-flight simulator [1]. I tried this free game and it could even calculate trajectory to other planet!).

[1]http://en.wikipedia.org/wiki/Orbiter_(simulator)

24. I hate to say it, but the dropped bullet gets down first.

25.  A bullet fired horizontally and one dropped from the same height will hit the ground simultaneously. Confirmed Adam and Jamie first carried out two small-scale experiments, one using ball bearings (dropped vs. shot from a spring-loaded launcher), the other using paintballs (dropped vs. fired from a paintball gun). While the first experiment seemed to bear out the myth, the second one contradicted it; Adam attributed this result to imperfections in the paintballs' surfaces that caused them to veer slightly off course. For full-scale testing, they started at a firing range and used a .45 caliber pistol to measure the distance a bullet would travel before hitting the ground. Since the ground there was not level, they set up a second test at Fort Mason. Once they had properly fine-tuned their mechanism to fire and drop the bullets at the same time, they found that the two bullets landed within 39.6 milliseconds of each other. Commenting that this difference was less than the duration of one film frame (shot at 24 frames per second), and thus short enough for the human eye not to notice, they declared the myth confirmed. MythBusters (2009 season) - Wikipedia, the free encyclopedia

26. First of all mythbusters do a great job at promoting experimentation. They are really cool guys but their methods are not scientific. I still have to calculate it myself but i need to include some more factors and I'm in bed now. If you imagine that it is safe to assume a bullet trave travels 400 m/s then in mythbusters experiment the bullet travels about 1,5 meters. Not to mention that bullers rotate around their axes. Second, most importantly. If i'd drop a bullet 1 meter. That takes me approximately 1 second. As any gun manufacturer will tell you at a distant of around 400 meters even the most advanced sniper guns have a certain spread of several centimeters. At this speed a centimeter or 3 higher would make a few hundreds of milliseconds difference. Way larger then the error in the experiment. The curvature doesn't account for much but about meters per kilometer. This is alot. I'll get back to you guys oce i derived a formula

27. Can someone provide a link for the related orbital and projectile physics? (preferably not Wikipedia)

28. I remember a physics class I had in college. It was held in an auditorium. The professor had an apparatus on stage which launched a dart across the stage, and at the same time dropped a target from the ceiling. As long as the dart was initially aimed directly at the target, it would intercept the target as it fell. The dart would drop from its initial trajectory the same amount as the target fell, in the same amount of time.

29. For small velocities and distances the earth is "flat" and in this regime dropped bullets and shot bullets hit the ground at the same time (on average, so we can ignore any imperfections in the bullet, gun etc) for a reasonable definition of same. But when talking about orbital speeds we are no longer in this regime, for example a bullet orbiting the earth at "people" heights would complete its orbit every one and a half hours I think and ignoring air resistance would never hit the ground.

30. Originally Posted by Kerling
First of all mythbusters do a great job at promoting experimentation. They are really cool guys but their methods are not scientific. I still have to calculate it myself but i need to include some more factors and I'm in bed now. If you imagine that it is safe to assume a bullet trave travels 400 m/s then in mythbusters experiment the bullet travels about 1,5 meters. Not to mention that bullers rotate around their axes. Second, most importantly. If i'd drop a bullet 1 meter. That takes me approximately 1 second. As any gun manufacturer will tell you at a distant of around 400 meters even the most advanced sniper guns have a certain spread of several centimeters. At this speed a centimeter or 3 higher would make a few hundreds of milliseconds difference. Way larger then the error in the experiment. The curvature doesn't account for much but about meters per kilometer. This is alot. I'll get back to you guys oce i derived a formula

It's a series of steps to get the answer:

First you find the total orbital energy of the bullet with respect to the Earth by

m is the mass of the bullet, k is 3.987e14 m³/s², and h is the starting height of the bullet.

Then you use

to find "a", the semi major axis.

Then

Gives you the eccentricity of the orbit (e).

Now you use

to find 'q' which the angle from perigee (True anomaly) where this trajectory intersects the Earth

Gives you the eccentric anomaly. And

Gives you the Mean anomaly.

Finally

gives you the fall time for the bullet.

You then compare this to the fall time for a bullet dropped straight down.

If you have a bullet traveling at 400 m/s, fired from a height of one meter, it will hit the ground ~576 microseconds after the other bullet. (under ideal conditions)

31. Let's say the gun can shoot minute-of-angle accuracy, which would be very good accuracy for a hunting rifle, but probably too optimistic for a 45 ACP even firing from a Ransom Rest, as Mythbusters were. A minute of angle is about 1 inch at 100 yards, let's say the error is plus or minus 2 centimeters at 100 meters.
If a bullet is travelling 300 meters per second, it would reach the 100 m target in ~1/3 second. So the gun can impart an error velocity (up, down, left, or right) of ~2 cm/.33 sec = 6 cm per second.
At an acceleration ot 9.8 meters per second per second, an object with zero initial velocity will fall 1 meter in 0.4518 seconds. An object with an initial velocity of .06 meters per second will travel the same distance in 0.4457 seconds, giving an experimental error of 0.006 seconds, or 6 milliseconds.
Mythbusters claimed an error of 39.6 milliseconds, so the bullet dispersion error does not seem to be excessive.

32. As usual folks post a hypothetical question with no relevant conditons or assumptions .l assume there is no atmosphere to contend with,if not your bullet would have to still have velocity after peneitrating 25000 miles of air-good luck. Secondly some one posited a velocity 7902 M/sec which is roughly 24000 ft/sec no gas fired gun could achieve that since the limit at which a bullet can be propelled is hah fast the gas can travel down a barrell (look it up)==rail gun only hope here.
BTW how many angels can really dance on the head of a pin.
Peace

33. is the gravity "just outside" the earth still the same inside earth?

34. The force due to gravity at a point below the surface of the earth is less than it is just above the earth's surface. At a point below the earth's surface, there is less matter between that point and the centre of the earth than would be the case for a point at the earth's surface. The gravitational force due to the spherical shell of matter which is above this point below the earth's surface, and therefore further from the centre, cancels out - actually because of the inverse square law.

35. One variable not mentioned so far is Coriolis force - although the effect of what happens as the bullet travels over the Northern hemisphere would probably be cancelled out by that of the Southern hemisphere (assuming I understand it correctly). I'd be surprised if they balance out exactly - although you could stand on, and take aim along, the Equator. *

Also gravity is not the same all over the globe and varies by as much as 0.5%. It has been mapped but I don't know how accurately it needs to be for this thought experiment. It's been accurate enough to measure the shift in mass of ocean water to land during major flooding that accounted for major (millimetres) up and down variations in average global sea levels. The gravitic influence of Sun, Moon and tidal movements would need to be considered.

And any Earthquakes during the time the time the bullet is in motion can change the Earth's rotation. Even the average of winds interacting with the Earth's surface impact the Earth's rotation, maybe enough to make the difference between hit or miss. Without atmosphere, oceans, tectonic movements and the gravity of Moon and Sun I imagine the problem gets (relatively speaking) easier.

* My Wrong.

36. Coriolis force doesn't matter here as we are doing these calculations as an observer floating above the earth?

37. River Rat, yes. I hadn't really thought about coriolis effect being about the motion relative to a moving curved surface; it only appears to be a change in motion of something because of how it appears relative to the "fixed" surface beneath? Coriolis "force" withdrawn.

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement