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?


  • $\begingroup$ Try doing the maths if you want to be sure, but otherwise you can just think of it conceptually. If you launched a ball straight upwards on Earth vs. on the Moon, it would go higher and be in the air for longer on the Moon. Thus, since there is no air resistance, it would have more time to travel forward on the moon before it landed. $\endgroup$ – mardat Dec 3 '14 at 20:42

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,

  1. How do you figure out what time the object hits the ground again? (It starts on the ground because of the initial condition I assumed.)
  2. Once you have that time, how do you figure out how far it traveled horizontally?

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."


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