# Projectile motion with non-uniform gravitational acceleration [duplicate]

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Out of Curiosity, I asked myself what would be the equation of motion of projectile motion (Time Period, Height, Range) if the velocity of the projectile is sufficiently large so that the value of gravitational acceleration cannot be assumed constant. It is launched at angle theta

## marked as duplicate by valerio, Martin, stafusa, Jon Custer, Kyle KanosJan 21 '18 at 12:18

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## 1 Answer

I assume you are asking about a study of a projectile's motion similar to what is done in a mechanics course, so I will simply decompose the motion on the vertical and horizontal axis.

The acceleration of a body of mass $m$ at the ground level due to the gravitational attraction is $g_0={Gm}/{r^2}$, where $r$ is the radius of the planet. The original meaning of $r$ is the distance between two masses, the falling body and the planet itself (its center), so for a generic height $h$ one should write $g(h)={Gm}/{(r+h)^2}$. Let's write down the ratio $g(h)/g_0$:

\begin{equation} g(h)/g_0=r^2/(r+h)^2=\frac{1}{[1+(h/r)]^2}\approx(1-2\frac{h}{r}). \end{equation} Where, in the last equality, I applied the binomial approximation, since $h/r$ should be a small quantity. Then: $g(h)=g_0[1-(2h/r)].$

Since $g$ only appears on the vertical part of the projectile's motion, let's look at $y(t)$, which indicates the vertical position of the bullet at the time $t$.

$y(t)=-\frac{1}{2}g(y(t))t^2+{v_0}_yt$: we now have to keep in mind that $g$ depends on $y(t)$ since it indicates the height of the bullet.

\begin{equation} y(t)=-\frac{1}{2}g_0[1-2y(t)/r]t^2+{v_0}_yt\end{equation}

\begin{equation} y(t)=-\frac{1}{2}g_0t^2-g_0[y(t)/r]t^2+{v_0}_yt\end{equation}

\begin{equation} y(t)+g_0[y(t)/r]t^2=-\frac{1}{2}g_0t^2+{v_0}_yt\end{equation}

\begin{equation} [1+(g_0/r)t^2]y(t)=-\frac{1}{2}g_0t^2+{v_0}_yt\end{equation}

\begin{equation} y(t)=\frac{1}{1+(g_0/r)t^2}(-\frac{1}{2}g_0t^2+{v_0}_yt)\end{equation}

It can be seen that the new motion equation for $y(t)$ is identical to the original one, with the addition of the multiplicative term $\frac{1}{1+(g_0/r)t^2}$. Since $g_0/r \approx 10^{-6} s^{-2}$, this new term is pretty identical to $1$ for any reasonable kind of bullet motion. It differs from $1$ only for really big values of ${v_0}_y$ which can make the bullet to fly for a great (and impossible) amount of time. Though in this very last case we should not talk about "projectile": it would be the case of the motion of a satellite around the planet and the approximation we did at the start would be valid no more, nor will be the result we obtained.

Finally, the equation of $x(t)$ and $y(t)$ should be: \begin{equation} \begin{cases} x(t)={v_0}_xt\\ y(t)=\frac{1}{1+(g_0/r)t^2}(-\frac{1}{2}g_0t^2+{v_0}_yt) \end{cases} \end{equation}

The fly time, which can be obtained solving $y(t)=0$ would be the same of a constant-g problem, as one could multiply both members by $(1+(g_0/r)t^2)$. From that, follows that, in our approximation, also the range would be the same of the original problem.

• in general it will be an ellipse – Sahil Chadha Jan 20 '18 at 16:30
• Sahil Chadha - How it would be an ellipse? I am not able to figure out the equation of ellipse from here – Computer Guy Jan 21 '18 at 7:14