Might be an appallingly easy question to some I was wondering how does difference in gravity affect distance travelled by a projectile? Assuming no air resistance or other external forces and that height thrown/velocity/launch angle remains the same.
In lower gravity, does distance travelled increase or decrease and why?
Thanks
The equation of motion for something under constant acceleration $\vec{a}$ is
$$\vec{x}(t) = \vec{x}(0) + \vec{v}(0)t + \frac12 \vec{a}t^2,$$
where $\vec{x}, \vec{v}, \vec{a}, t$ are displacement, velocity, acceleration, and time, respectively.
If you launch at an angle $\theta$ with respect to horizontal, then the $x$ and $y$ components of the initial velocity are $v_x(0) = v \cos\theta$ and $v_y(0) = v \sin\theta,$ where $v = |\vec{v}(0)|.$
Let's take the directions to be $+x$ to the right and $+y$ up.
Let's also define $\vec{x}(0) = \vec{0}.$
If your constant acceleration is gravity, then $a_x = 0$ and $a_y = -g$.
Then, your two equations for $x$ and $y$ become
$$x(t) = v(\cos \theta)t, \\ y(t) = v(\sin \theta)t - \frac12 g t^2.$$
So, having these,
The second answer will depend on $g$, so this will tell you how gravity affects the range.
There is a much easier way to see how changing $g$ works to change the distance travelled $L$ (assuming the geometry of the problem is in all other ways the same):
$G$ has dimensions $\frac{L}{T^2}$ (acceleration is a length per time squared). The only other input to the problem that has dimensions is the initial velocity $V$, which has dimensions $\frac{L}{T}$.
Therefore, the distance traveled must be some constant (that depends only on the non-changing geometry such as the angle of launch) times that combination of $g$ and $V$ that has dimension of length. That combination is (uniquely) $$ \frac{V^2}{g}$$
Therefore, the length depends inversely to $g$; if you halve $g$ you double the length.
And the length goes as the square of the initial speen of throwing, $V$.
"Units (and dimensional analysis) are your friend."
