# What is the meaning of change of angular momentum of a ballistic object during its flight?

In a 2D world, three stones, whose magnitude of initial velocities are 5000m/s, are thrown from the North pole towards the Equator with horizontal initial angles of 15o, 30o, 45o and 60o angles. Their trajectories are as below:

Their respective angular momentums untill they fall to the ground are as below:

I'm not a physicist, I'm a grad student in Electronics Engineer and I'm very unfamiliar to the term "momentum". This homework is about controlling flight of aerial objects. I only know angular momentum as a quantity that is product of mass and speed. I did my assignment up to this point. And now, I have to make some comments on it. But those curves of angular momentum doesn't mean anything to me.

What should I write in my report? Why is angular starting from a different positive value for each stone, and then decreasing in time, finally rising up to its initial value when the stone falls on the ground? What is the meaning of this physical phenomenon?

EDIT:

I iterated these two differential equations:

$\ddot{r} - r\dot{\theta}^2 + \frac{GM}{r^2} = 0 \\ \frac{d}{dt}(r^2\dot{\theta}) = 2r\dot{r}\dot{\theta} + r^2\ddot{\theta} = 0$

With initial conditions:

$r = R_0 \\ \theta = \frac{\pi}{2} \\ \dot{r} = 5000 sin(\gamma) \\ \dot{\theta} = -\frac{5000 cos(\gamma)}{R_0} \\$

Where;
$R_0 = 6378000 m$ (radius of the Earth)
$G = 6.6742 \times 10^{-11} N (\frac{m}{kg})^2$ (the gravitation constant)
$M = 5.9722 \times 10^{24} kg$ (mass of the Earth)
$\lambda = 15^o, 30^o, 45^o, 60^o$ (launch angles for each stone)

At each iteration step, I calculated the angular momentum per unit mass by the following formula:

$\mbox{Angular Momentum per Unit Mass} = r \dot{\theta}^2$

EDIT 2:

I realized that I was wrong with the formula of angular momentum per unit mass. I must have been:

$\mbox{Angular Momentum per Unit Mass} = r^2 \dot{\theta}$

When I modified the formula, I obtained these curves below in which angular momentum stays constant:

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The people who are most familiar with the term "momentum" are not physicians (doctors who cure other people) but physicists. An ex-student leader of the 1989 Velvet Revolution made the same mistake in a talk he gave to a conference of mathematicians and physicists in Prague – he admired the mathematicians and physicians over there. – Luboš Motl Nov 20 '12 at 8:06
The angular momentum is constant in any system with a central force, so I don't understand how you get a variation in the angular momentum. How did you calculate the lines in the second graph? – John Rennie Nov 20 '12 at 10:05
@Prathyush Thank you for the correction. But I realized a few minutes before you. :) – hkBattousai Nov 20 '12 at 18:38
correct if im wrong but $L=p\times r=mr^2 \dot θ$. Which is infact your second equation – Prathyush Nov 20 '12 at 18:41