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Engineering LibreTexts Projectile Motion- Very Long Range

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    Suppose that a projectile is launched in the horizontal direction (call it \(X\)) from a high tower, with the initial velocity of \(v_x\). The height of the tower is \(H\). Find the equation of the projectile trajectory, and the spot it hits the ground. It's a standard problem in introductory level physics. The solution is pretty simple: we call the vertical direction the \(Z\) axis, we call the coordinates of the tower top as \(x=0\) and \(z=0\), and assume that the launching takes place a \(t =0\). So, as there is no force in the \(x\) direction, \(v_x\) does not change, and the \(x\) coordinate changes with time as \(x(t) = v_xt\). Downwards, there is a motion with acceleration \(g\), so that \(z(t)= gt^2\). These two equations together constitute the so-called “parametric equations of the trajectory curve''. If one prefers an alternative description of this curve, in the \(z = f(x)\) form, one can readily obtain it, solving the first equation for time: \(t=x/v_x\), and plugging it into the second, to obtain: \(z = -(g/2v_x^2)x^2\), which is the equation of an inverted parabola. And as far as the spot where the projectile hits the ground is concerned, for this spot \(z = -H\), so by solving the equation \(-H = -(g/2v_x^2)x^2\), we obtain \(x=v_x\sqrt{2H/g}\).

    Earlier in the text, general equations for the free fall (Eqs. 1.20) were presented. For the case discussed above, we need to use only two of them, for \(x\) and \(z\). There is no force in the \(x\) direction, and in the \(z\) direction the force is \(F_z = -mg\). Therefore, we obtain a set of two differential equations:

    \[ dfrac{d^2 x}{dt^2}= 0 \;\;\; {\rm and} \;\;\;dfrac{d^2 z}{dt^2}= -g \]

    It can be readily checked by a pretty simple double differentiating that the two parametric solutions for \(x(t)\) and \(z(t)\) obtained above by the “introductory physics method'' indeed do satisfy both those differential equations.

    But will the introductory physics method always work?... Well, note that in such a solving procedure we used an additional “hidden assumption'' -namely, that the Earth is flat. As long as the flight range of of the projectile is relatively short -say, not beyond the horizon visible from the ground level -the “flat Earth'' is pretty good an approximation because the effects of the Earth curvature are still very small. But if the projectile can fly far beyond the horizon, the curvature can no longer be neglected. In WW II, the battleships' most powerful artillery pieces could fire on targets as far away as about 50 km (\)\tilde{30}\)) miles, and at such distance the target is already about 200 m (1/8 mile) below the imaginary “flat Earth level''. And what if we wanted to calculate the trajectory and the range of a projectile with no propulsion, but with the initial speed sufficient to carry it over distances much, much longer than 50 km?

    In such a case there is no other way than to use the general equations. If we put the origin of the \(XZ\) Carthesian system at the Earth center, then the calculation of the force of gravity at any point above the Earth surface of coordinates \(x\) and \(z\) is straightforward: the distance from the Earth center is \(\sqrt{x^2+z^2}\).

    By replacing \(R_{\rm e}\) in the Eq. 2.13 with this distance we get the total force of gravity acting on mass \(m\) at this point:

    \[ F(x ,y) = Gdfrac{M_{\rm e}m}{\sqrt{x^2+z^2}^2}. \]

    In the equations of motion we need not the total force, but its components in the \(x\) and \(z\) directions. To obtain their values and signs, we have to multiply the total force by, respectively, \(-\cos\theta\) and \(-\sin\theta\), where \(\theta\) is the angle defined in Fig. XX. From this Figure, one can readily find that:

    \[ \cos\theta = dfrac{x}{\sqrt{x^2+z^2}}\;\;\;{\rm and}\;\;\;\sin\theta=dfrac{z}{\sqrt{x^2+z^2}}.\]

    By combining Eqs. 1.13, 1.30, and 1.31, we obtain generally valid equations of motion for a projectile launched from an arbitrary point above the surface of \textbf{spherical} Earth:

    \[ dfrac{d^2x}{dt^2} = -GM_{\rm e}dfrac{x}{\sqrt{x^2+z^2}^3} \;\;\;{\rm and}\;\;\; dfrac{d^2z}{dt^2} = -GM_{\rm e}dfrac{z}{\sqrt{x^2+z^2}^3}\]

    To obtain the trajectory of the projectile, one needs to specify the so-called “initial conditions,'' i.e., the \(x_0\) and \(z_0\) coordinates of the launching point, and the velocity components \(v_{x0}\) and \(v_{z0}\) at the moment of launching. And then, find the time-dependent
    coordinates \(x(t)\) and \(z(t)\), which together constitute a set of parametric equations of the trajectory.

    The bad news is that solving these equations using the classical “paper+pencil'' method requires applying sophisticated high-level mathematical techniques. Presenting a step-by-step solution procedure in this chapter would make no sense. However, the good news is that a highly accurate solution can be obtained by using a relatively simple numerical algorithm which takes even a simple today's portable computer only a split second to process. This algorithm and a program in the PYTHON language are described in a greater detail in APPENDIX XX. Here, we will only show a few calculated trajectories in a graphic form in a figure.

    The calculations are done for realistic data, i.e., the Earth radius is taken as 6371 km, the Earth's mass as \(5.9722\times 10^{24}\) kg, and \(G = 6.6741\times 10^{-11}\) m\)^3\)kg\)^{-1}\)s\)^{-2}\). If we wanted to make a plot displaying everything in real proportions, the launching point could not be too low, because the plotted trajectory would “blend'' with the circle representing the planet's contour. Therefore, the projectile is launched from the altitude of 1000 km -you can either thing of a super-giant tower built at the North Pole, or a spacecraft “hovering'' over the North Pole at the altitude of 1000 km, and acting as a platform for a horizontal launch. The value of \(g\) at this altitude is markedly lower than at the Earth surface -from the Eq. 2.16, for \(H=1000\) km one gets \(g(H) = 7.3362\) m/s\)^2\).

    So, the initial coordinates of the launching point are: \(x_0 =0\) and \(z_0 = 6371\) km + 1000 km = 7371 000 m. The projectile will always be launched in the horizontal direction, so that always \(v_{z0}\) -and we will make calculations for several different values of \(v_{x0}\).

    In the Fig. 2.1 one can see a trajectory for the initial projectile velocity \(v_{x0} = 6000\) m/s, plotted with black color.1 The projectile lands at a spot which is about 1/8 of the Earth's circumference from the base of the “launching tower''. For \(v_{x0} = 6850\) m/s (blue curve) the landing spot is close to the Equator. Now, the range becomes quite sensitive for the initial speed value: for \(v_{x0} = 7081\) m/s the projectile lands close to the South Pole. And the most interesting trajectory is definitely that for \(v_{x0} = 7353.6\) m/s (red color): the trajectory is a perfect circle, it never hits the ground -the projectile returns precisely to the spot at the top of the launching tower from which it was launched!

    In the next section it is explained what is the significance of the “7353.6 m/s" speed. The trajectories for higher launching velocities have one in common -they all return to the launching spot, but their shapes get elongated in the “downwards'' direction in the figure (one example plotted with a dashed line is a trajectory for \(v_{x0} = 8000\) m/s). They are no longer circles, but ellipses.

    Figure \(\PageIndex{1}\): The shaded circle symbolizes the Eart, and the sphere bar at its top is a 1000 kilometer high tower erected at the North Pole. The black, blue and green curves are trajectories of projectiles launched horizontally from the tower top with velocities, respectively, 6000, 6850, and 7081 meters per second. They all fall to the ground. But the projectile launched with a velocity of 7353 m/s (red curve) becomes an artificial satellite of Earth -it travels along a trajectory which is a perfect circle and returns to the tower top. A projectile launched with a velocity of 8000 m/s travels along an elliptical trajectory (dashed curve).

    However, for \(v_{x0} = \sqrt{2}\times 7353.6\) m/s = 10399.6 m/s and higher launching velocities there is yet another change of behavior - the trajectories are no longer closed, but open curves -meaning that the projectiles never return to the launching site, but they “leave Earth for good''.

    In summary, there are four different types of behavior of the projectile. For zero launching speed, which is obvious, it's just a vertical fall. Then, for \(v_{x0} > 0\) the trajectories are first “ballistic curves'' ending at the Earth surface. Next, the paths become closed orbits, and the projectiles are now Earth's satellites -and finally, for \(v_{x0}\) larger than the so-called “escape velocity'', the projectile becomes an “artificial asteroid'' and leaves Earth forever.

    And it should be stressed that all these four types of behavior have on thing in common - namely, in all cases the \textbf{only} force acting on the projectile is the force of gravity.

    Everybody agrees that an appropriate term for a purely vertical drop is “a free fall''. Yet, many students react at the first moment th surprise if the same name is used for a ballistic flight, satellite motion, or a spacecraft flight. But physicists insist that as long as there is no engine in the moving object, and it is only the gravity which governs its motion, one should rather invoke the famous “Occam Razor'' and use the same term in each case. One more argument in favor of such thinking is that all types of trajectories are obtained as solutions of the very same set of equations of motion that are the mathematical form of the Second Newton's Law of Dynamics.


    1. 6000 m/s is about 17 times the speed of sound, or about 6 times the speed of a bullet from a powerful military rifle. Projectile Motion- Very Long Range is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Tom Giebultowicz.

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