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
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).
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
If you like calculus, this may help:
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
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
For general information, Google "ballistics".
For a nifty on-lint calculator for maximum ballistic range:
For other related calculations (including a sunrise/sunset calculator):
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
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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