Influence of gravity on ballistic trajectory?

Hi,
Can anybody tell me the influence of gravity on the trajectory of a bullet when the aim is to shoot as far as possible?
To put it simple, if i shoot a bullet from a gun on earth or on the moon (places with different gravity) does the angle, in order to shoot the as far as possible, differ?
If you know any interesting web sites about this subject, i would very much appreciate it!
thanks, bjorn
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far
Yes, the angle would be different because the downward force due to gravity is different.
Look in any physics textbook for a more in-depth explanation.
JD
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bullet
gravity
Not quite: http://scienceworld.wolfram.com/physics/Trajectory.html The range of the projectile is given in equation 7, which depends on gravity, but the *maximum* range occurs at a firing angle of theta = pi/4, regardless of the (non-zero) magnitude of g.
It's 45 degrees on the earth OR the moon - barring air resistance (see my other post in this thread).
Tom Davidson Richmond, VA
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Gravity will pull the bullet down. Horizontal velocity will carry the bullet further. Ideally there is a trade-off between horizontal to make the bullet go *far* and vertical to make gravity take more time to pull the bullet back down to ground level.
In the *absence* of air it turns out that you get maximum range when horizontal and vertical velocities are perfectly balanced at the start - a 45-degree angle.
If you like calculus, this may help: http://scienceworld.wolfram.com/physics/Trajectory.html

far
The 45-degree angle works very well on the moon, but the earth has air, which resists the movement of the bullet. The net effect of air resistance is a force that tries to slow the bullet down no matter which direction it is going.
The bullet will be going slower after the peak of its arch, so for maximum range we need to make better use of the higher horizontal speed it has *before* the peak of its arch.
The exact angle will depend on various details such as the geometry of the bullet, the weight, the density, the wind direction, altitude, humidity, and so on, but a good rough guess will be that you will get your maximum range on earth if you fire the bullet at an angle of 35-38 degrees above the horizontal.

much
For general information, Google "ballistics". For a nifty on-lint calculator for maximum ballistic range: http://www.eskimo.com/~jbm/calculations/maxdist/maxdist.html For other related calculations (including a sunrise/sunset calculator): http://www.eskimo.com/~jbm/calculations/calculations.html
HTH
Tom Davidson Richmond, VA
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If you (a) ignore the effect of air resistance on the bullet's path, and (b) assume a flat world with gravity not changing with altitude (which is reasonably accurate until ranges become quite long), then no, the angle does not change: it's always 45 degrees. Mathematically:
For muzzle velocity V and angle x, the (upward) vertical component of velocity is V*sin(x). The constant (downward) vertical acceleration of gravity, g, reduces that to zero in time V*sin(x)/g; by symmetry, the time taken to reach the ground again is the same, so the flight time is 2*V*sin(x)/g.
The horizontal component of velocity is V*cos(x), and that does not change. So the distance covered in that time is V*cos(x)*2*V*sin(x)/g. A table of trig identities yields sin(x)*cos(x) = sin(2*x)/2, so we simplify and get V^2*sin(2*x)/g as the distance.
Neither V nor g varies with the angle, so distance is maximum when sin(2*x) is maximum. The largest value the sin() function can take is 1, which occurs at 90deg. So maximum range is at 90deg/2 = 45deg, independent of the exact choice of V and g.
If memory serves, moderate air resistance calls for a slightly steeper angle, but the math is ugly.
For long ranges, you cannot ignore the spherical shape of the planet or the reduction in gravity with altitude. (Indeed, if muzzle velocity exceeds the planet's escape velocity -- impractical on any of the planets except possibly Pluto, but easy on the smaller moons -- then surface range is not meaningful because the bullet never strikes the surface.) Then the angle does become a function of both the planet's gravity and its radius. More precisely, you can show that the range is a messy function of just two quantities: the angle, and V^2/(g*R), where R is the planet's radius. I believe a smaller planet, or weaker gravity, or both, means a lower angle for maximum range, although I'd need some effort to confirm that.
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There are a number of artillary games, which allow two players to shoot cannons at each other. The speed and angle can be adjusted, so you can see how these change the trajectory of the shell.
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far
much
This is in many science/physics textbooks. I'd suggest reading one. ;-)
Jeff
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