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Math Help - Hide and seek on a spherical planet

  1. #1
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    Hide and seek on a spherical planet

    A single inhabinant lives on a surface of a spherical planet on which he can move freely with velocity u. A spaceship approaches the planet with velocity v. Show that if v/u > 10, the spaceship can always observe the inhabitant (sooner or later), no matter how he tries to hide.

    I have no idea where to start. Please help me.
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  2. #2
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    I'll have a bit more of a think, but I'd at least start by defining a polar (EDIT: spherical) coordinate system with the origin at the centre of the planet.

    Is the space ship flying at any height past the planet? And are you taking into account curvature of the planet blocking out the view? If you aren't it's quite simple, but if you are then it's a little more tricky.
    Last edited by Guffmeister; August 23rd 2010 at 09:14 AM.
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  3. #3
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    The problem is not well defined, in my opinion. There's lots of different approaches available. I'm not sure is the spaceship free to cruise around the planet searching for the inhabitant, or is it just passing by the planet. Also, I must the assume the inhabitant is intelligent enough to hide when the spaceship gets closer. Regarding the factor 10, I think is stands for pi^2 because I believe the solution to be a geometric one. That's all I've come up with so far.
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  4. #4
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    I'm not sure that the inhabitant is hiding, but rather it is not visible because it is on the other side of the planet/too far away to be seen. The key is to prove that if the ratio between the inhabitant and space ship is 10, then no matter how far the inhabitant goes the space ship will eventually be able to see it.

    So you have to define geometrically whether the inhabitant is too far round the planet to be seen, in terms of the space ship's position and the inhabitant's position. The problem is, that the space ship can only ever one complete half of the planet, technically speaking, from infinitely far away, and as it gets closer, it will see less and less far around the planet.
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  5. #5
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    If the inhabinant isn't hiding, why would we need to know his speed?
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  6. #6
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    If he is hiding why would you need to know his speed? He'd just hide under a rock or something and sit still. The point of the question is to show that at some point the inhabitant will be in a position where it can be observed by the space ship.

    The inhabitant will move in one direction at a velocity, and the space ship in another at another velocity, and at some point you should be able to prove that independent of size of planet, or distance of the spaceship from the planet, that the inhabitant will eventually be observable at a certain point. And hopefully, that will be the ratio you've been given to prove.

    The path of the two moving objects is interesting though, because if the spaceship was flying in a tight circle above the planet, and the inhabitant was walking in a tight circle on the other side, they'd never see each other. Clearly you need to know more information about how the two things are moving to answer this correctly. My suspicion is that they are either both travelling in a straight line, or the space ship is orbiting.
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  7. #7
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    This seems like a hard problem. If the spaceship is a great distance from the planet then it can observe almost an entire hemisphere. But its maximum angular velocity around the planet will be very small, so it will be easy for the inhabitant to keep moving so as to be always on the far side of the planet from the spaceship. On the other hand, if the spaceship keeps close to the surface of the planet then it will be able to move round the planet very fast. But then most of the planet will be beyond its horizon, so again it only has a low chance of spotting an inhabitant who keeps moving. Presumably there is an optimum height above the planet for the spaceship to operate in order to maximise its surveillance effectiveness. Call this critical height R.

    The next thing is to decide a strategy for the course of the spaceship. My suggestion is that it should initially approach the north pole of the planet. To evade detection, the inhabitant will therefore have to be somewhere in the southern hemisphere. Now suppose that when the spaceship reaches the critical distance R from the planet, it starts to circle the planet in a spiral movement, sweeping round from west to east keeping at the same distance R from the planet, as in the diagram, while gradually moving from north to south until it has scanned the entire planet.

    Hide and seek on a spherical planet-loxo.png

    (This curve is called a loxodrome.) The inhabitant could try to evade detection by sprinting from the southern hemisphere to the north while the spaceship is round the far side of the planet. But this will presumably not be possible if the spaceship's velocity is sufficiently greater than the inhabitant's.

    I won't even try to calculate the ratio of velocities necessary for this to happen, but at least it gives an idea for a possible approach to the problem.
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  8. #8
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    Thank you, Opalg, that's just what I've been looking for! I'll try to do the calculation and let you know if it works.
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  9. #9
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    How's this for a solution...

    Set up a spherical coordinate system with the origin at the center of the planet. Let the spaceship be at the distance d from the center of the planet, with angular coordinates \theta and \phi. If the spaceship is point A and the origin is point O, draw a tangent from the spaceship to the planet and call it point B on the planet - there are two of these, any one will do. Now, denote the angle AOB by \psi. Since BOA is a right angle, we have R=d \cos\psi. Now, let the ship circle around the z-axis with the velocity v=d \sin\theta \dot{\phi}. The time it takes for one complete revolution is t_v = \frac{2\pi d \sin\theta}{v}. At the same time the ship sweep the strip of the planet of width s=2R\psi. So, in order to see the inhabinant, it has to go through all the strips taking care that the inhabitant cannot escape, i.e. the time neccesary for him to cross the strip t_u=\frac{2R\psi}{u} should be greater or equal to the time of revolution. So t_u \ge t_v. Using the relation above, one has \frac{v}{u} \ge \frac{\pi \sin \theta}{\psi \cos \psi}. Since the relation must hold for all \theta, we take the worst case scenario \theta=\frac{\pi}{2}. Then the only free parameter is \psi and we can choose it so to minimize the left hand side of the inequality. Numerically, one gets \frac{v}{u} \ge 5.6.

    Does this look ok?
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