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**Heirot****Member**- Registered: 2010-08-08
- Posts: 9

A spherical planet has only one inhabitant that can move freely on the surface of the planet with speed u. A spaceship approaches the planet. Let the maximal speed of the spaceship be 10u. Show that the spaceship can always see the inhabinant regardless of the way he moves.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,726

Am I missing something here? As the spaceship can go lots faster, and assuming it can approach as close as it likes and manoeuver freely, what's to stop it just following inhabitant about.

If my assumptions are not valid, then you'll have to tell me the limits.

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

Show that the spaceship can always see the inhabinant regardless of the way he moves.

If the spaceship approached the present position of the object and never changed course, while the object moved, the spaceship could lose sight of the object. The object could make it to the far side of the sphere.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**Heirot****Member**- Registered: 2010-08-08
- Posts: 9

Assuming that the inhabitant doesn't want to be seen, I've found a way the spaceship can move so to always spot the inhabitant. This gives the lower limit for v/u in this case. I would like to hear your opinion on it. Here it goes...

Set up a spherical coordinate system with the origin at the center of the planet. Let the spaceship be at the distance

from the center of the planet, with angular coordinates and . 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 . Since BOA is a right angle, we have . Now, let the ship circle around the z-axis with the velocity . The time it takes for one complete revolution is . At the same time the ship sweep the strip of the planet of width . 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 should be greater or equal to the time of revolution. So . Using the relation above, one has . Since the relation must hold for all , we take the worst case scenario . Then the only free parameter is and we can choose it so to minimize the left hand side of the inequality. Numerically, one gets .Does this look ok?

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Maybe I'm missing something.

The spaceship approaches the planet, but we're not told how far away it was to start with.

If it's sufficiently far away, then no matter what its movements are, it will always be looking at approximately the same point on the planet.

In this case, the inhabitant can easily move to the other side to avoid its gaze.

Why did the vector cross the road?

It wanted to be normal.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,227

Hi all;

I have you beat, I have been missing the point here since post #3. If you are trying to set up some type of pursuit curve, I don't know if you defined the problem enough. It seems to me that this condition exists:

If the sphere is very small and spaceship very far away, then no matter which direction the spaceship maneuvers, to the sphere it will appear stationary. The inhabitant may then get around to the other side of the sphere.

Say the ship is 100 000 u away and travels at a speed of 10 u. The sphere has a circumferences of 1 u and the inhabitant travels at a speed of 1 u. I think the inhabitant can duck on the other side of the sphere.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**Heirot****Member**- Registered: 2010-08-08
- Posts: 9

I also got the problem not defined well. I' assumed the ship coming form infinity heading towards the planet. In trying to find the optimal algorithm for finding the inhabitant, I assumed it stopped at the distance d and moved the way I described in post #4.

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