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Suppose a projectile is launched from the Earth's surface with initial velocity $v_0$ well below speed of light and initial angle $\theta_0$ with respect to the vertical line perpendicular to the Earth's surface. Omitting Earth's rotation, but knowing that Earth is not flat (as in the real world), what is the maximum height of the projectile with respect to center of the Earth? (Suppose air resistance is negligible.)

Is it possible to solve this problem using only the material in the first $8$ chapters of Fundamentals of Physics; D. Holliday, R. Resnick & J. Walker? I'm eager to see a various number of solutions!

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I do not have the book you mentioned, but I could find something in my notes from last year's Classical Mechanics lecture.

Firstly, if your projectile reaches escape velocity, then it will of course travel away from the Earth forever, which answers your question with $\infty$. When does this happen? The potential energy of the projectile is given by:

$V(r) = -\frac{GMm}{r}$

We will escape the Earth's potential, if we have enough energy to reach the potential at $r=\infty$, which obviously corresponds to $V=0$. So we need at least the potential at the Earth's radius in kinetic energy:

$K = \frac{1}{2}mv_0^2$



Where $G$ is the gravitational constant, $M$ the mass of the Earth, $m$ the mass of your projectile and $R_E$ the radius of the Earth (which is the current distance of your projectile from the Earth's center). So if the projectile's velocity is higher than this we have an unbound orbit, and there is no maximum height. If it is less than this we have a bound orbit, whose trajectory we need to figure out.

Now gravitation is a radial inverse-square law force of the form $\textbf{F}=\frac{k}{r^2}\textbf{r}$ where $\textbf{r}$ is a unit vector pointing radially outwards. Since it is attractive $k<0$ (in fact $k=-GMm$). For these forces a bound orbit will be elliptical, with the center of the Earth at one of the ellipse's foci. The equations I could turn up let you determine the eccentricity $e$ and the semi-latus rectum $h$, which define the ellipse, from the angular momentum and the total energy of the projectile.

$E = K + V = \frac{1}{2}mv_0^2-\frac{GMm}{R_E} \\ \textbf{L} = m\textbf{R}_E\times\textbf{v}_0=mR_Ev_0\sin(\theta_0) \\ h = -\frac{L^2}{mk} = \frac{L^2}{GMm^2} \\ e = \sqrt{1+\frac{2EL^2}{mk^2}}=\sqrt{1+\frac{2EL^2}{G^2M^2m^3}}$

Using $e$ and $h$ we can figure out $r_A$ and $r_B$, the distance from the Earth's center to the "perigee" and "apogee" (the points that are nearest and farthest from the Earth's center, respectively). We do this using standard ellipses formulae:

$r_A = \frac{h}{1+e} \\ r_B = \frac{h}{1-e} \\$

Since $e$ is between $0$ and $1$, we obviously have $r_B > r_A$ as expected. So the largest height above the Earth's surface would then be $r_B - R_E$. We would be done here... if there was not the possibility that the projectile hits the Earth's surface before reaching the apogee (for example because $\theta_0>\pi/2$). I am honestly at a loss when it comes to figuring out whether the projectile reaches the apogee, the perigee or both, but I have a feeling that if it reaches any of the two, it would be the apogee as required. But I cannot figure it out properly right now. When I do, I will edit the answer.

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If you're taking into account the curvature of the Earth you presumably also have to account for the change in gravitational acceleration with distance from the Earth. In effect your object is in orbit and travelling in an ellipse. It just happens that the ellipse intersects the surface of the Earth. So your problem reduces to calculating an orbit given the velocity at a known position (i.e. the launch point).

I'm not familar with the book Fundamentals of Physics, but a quick Google gives lots of hits for calculating orbits.

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You probably don't mean "gravitational constant" but "magnitude of gravitational force" or something. ;) – Martin Ender Nov 7 '12 at 18:12
Oops, yes, well spotted. If I'd discovered the gravitational constant changed, that would be worth a Nobel or two :-) – John Rennie Nov 7 '12 at 18:14
I'll probably get scolded for posting this on a stackexchange site, but I can't help it: – Martin Ender Nov 7 '12 at 18:16

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