I am currently stuck on the following question.

A stone is dropped from a stationary helicopter 500m above the ground, at the equator. How far from the point vertically below the helicopter does it land and in what direction? Solve this problem by conserving angular momentum.

My current interpretation is that when the stone falls in order to converse angular momentum the angular velocity must increase. This leads to the stone travelling to the east.

I equated the angular momentum at the start L0=*(re+h)mVo

where re is radius of earth, h is the height above that, m is mass and Vo is inital velocity. To the angular momentum at any time


where the 1/2at^2 term comes from gravitational acceleration and Ve is the additional velocity.

I rearranged to find Ve and then integrated over the time it takes to fall the 500m in order to work out the extra distance the stone traveled. However, my answer is 1/2 what the answer should be. I have checked the maths and it seems to be fine. So I believe my set up is wrong. Any help would be appreciated.

  • $\begingroup$ I think the error is in this statement L=(re+h-1/2gt^2)m(Ve+Vo) where you assume the height lost is equal to 1/2gt^2.Presumably because we are always taught you can separate velocity components, but I don't think this is actually true. If, say, the helicopter was instead at 36000 km (the height of geostationary orbit) the stone would not actually move toward earth at all, in fact it would just enter orbit. In this case, the height does not change at all, whereas you assume it always changes by 1/2gt^2. I say this but I don't know how to rectify.Solved a similar problem once, but using ellipses $\endgroup$
    – Joshua Lin
    Dec 25 '14 at 22:54

From conservation of angular momentum we can derive angular velocity as a function of height:

$$\omega(h) = \omega_0 \left(\frac{R + h_0}{R+h}\right)^2$$

where $\omega_o$ is the angular velocity of the earth, $R$ is the radius of the earth, $h$ is the current height and $h_0$ is the initial height.

The horizontal velocity (in the frame of reference of the earth) is

$$v(h) = (\omega_0-\omega(h))(R+h) =\\ \omega_0(R+h) - \omega_0\frac{(R+h_0)^2}{(R+h)}$$

A bit more manipulation gives

$$v(h) = \frac{\omega_0}{R+h}\left((R+h)^2 - (R+h_0)^2 \right)\\ \approx\omega_0\frac{2R(h-h_0)}{R+h}\\ \approx2\omega_0(h - h_0)$$

(The approximations are valid because $R>>h$).

In other words, the velocity scales directly with the difference in height. As a function of time, we write

$$v(t) = 2\omega_0((h_0-\frac12gt^2)-h_0)\\ =-omega_0gt^2$$

Now we integrate:

$$x = \int_0^T - \omega_0 gt^2 dt\\ = -\frac13\omega_0gT^3$$

where $T$ is the time taken for the fall, roughly $T=\sqrt{\frac{2h_0}{g}}$

Substituting, we get

$$x = -\frac13 \omega_0g\left(\frac{2h_0}{g}\right)^\frac32=24 cm$$

Since you did not tell us what the "correct" answer was, or what the details of your math were, I can't tell whether this solves your problem...

  • $\begingroup$ For your first equation, shouldn't the height part be squared? The way I did it I used angular momentum, so L = Iw, but the moment of inertia for a stone can be considered (since it is small compared to the earth) as simply mr^2 right? Where did the square factor go? $\endgroup$
    – Joshua Lin
    Dec 26 '14 at 2:14
  • $\begingroup$ You're right... and that explains why my answer is off by a factor 2. Let me take another crack. $\endgroup$
    – Floris
    Dec 26 '14 at 2:16
  • $\begingroup$ Also how do you get the 'horizontal velocity' in the second equation? It seems to be using v = rw (where w is the net angular velocity) but isn't that assuming circular orbit, whereas this follows an elliptical trajectory? Or can it be used in all cases? $\endgroup$
    – Joshua Lin
    Dec 26 '14 at 2:18
  • $\begingroup$ For small angles / displacements etc, a little bit of an elliptical path is well approximated by a circle. Pretty sure the solution is now correct. $\endgroup$
    – Floris
    Dec 26 '14 at 2:26
  • $\begingroup$ Pretty sure it's insanely elliptical. If the Helicopter were at 35km altitude, then it would be perfectly circular, but the greater the deviation the less 'circular' it is, in this case it's 500 metre altitude which is ellipse to the max. I think. $\endgroup$
    – Joshua Lin
    Dec 26 '14 at 2:31

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