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Complicated Asteroid problem

  1. Feb 11, 2007 #1
    Hi everyone, this problem is truly pain in the a**. I understand that Newton's third law applies here, but whatever I have tried nothing worked out. I know that the force applied is the same in magnitude, but opposite in the direction and accelerations of these two masses are different.....but no luck. Can anyone please help me out?

    At a time when mining asteroids has become feasible, astronauts have connected a line between their 3090-kg space tug and a 5770-kg asteroid. Using their ship's engine, they pull on the asteroid with a force of 510 N. Initially the tug and the asteroid are at rest, 550 m apart. How much time does it take for the ship and the asteroid to meet?

  2. jcsd
  3. Feb 11, 2007 #2
    For starters can you find the accelerations of each of the two objects?
  4. Feb 11, 2007 #3
    Hi, yes, as a matter of fact I did. I assumed down direction to be negative so F=ma, a=F/m for the first the tug a (tug)=510N/m For the asteroid a=-510N/m (asteroid). From the formula Xf=Xo+Vot+1/2at^2 it became clear that Vo=0 m/s they started from rest and hence Xf-Xo=1/2at^2 or simply 550m=1/2at^2. Now, what the hell is acceleration here? Of what object? At this point I have no clue!
  5. Feb 11, 2007 #4
    try forming an equation for each object that describes its position, and then use simultaneous equations
  6. Feb 11, 2007 #5
    it did not work, I'm not sure about one thing while tug is pulling asteroid, is tug moving? Ok I'm completely lost, I need someone's help badly, please!
  7. Feb 11, 2007 #6


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    Are you assuming they are talking about some kind of winch? If the tug is pulling the asteroid, of course the tug is moving! Earlier you said "accelerations of these two masses are different". No, they are not! As long as the tow line is taut (and assuming it can't stretch), the two masses (tug and asteroid) have exactly the same motion.
  8. Feb 11, 2007 #7
    ok if they have the same motions they must have the same accelerations which means they will never meet? I thought about winch, but there is nothing in the problem shows it. Can you please tell me if Xf=Xo+Vot+1/2at^2 is the correct equation to use for it because honestly I have no clue what to do right now and the problem is due pretty soon ((
    Last edited: Feb 11, 2007
  9. Feb 11, 2007 #8
    HallsofIvy I re-read you message again and I have a question about your point: when you said that they have the same accelerations which is reasonable assuming taut's stretching is trivial, how in the world they suppose to meet? I mean we can neglect gravitational forces between then since the setting of the problem is in space. Damn that leaves no room for thought.
  10. Feb 11, 2007 #9
    May well be wrong, but the way I interpret the question is that the motor is fired long enough to establish tension in the line, then cut out, so that the force that established the tension is no longer operative. At that point, the tension will exert a brief acceleration on both objects toward each other. Any velocity achieved during the motor firing can be ignored.
  11. Feb 11, 2007 #10
    There is not enough info to calculate that...moreover there is no gravity present from what I understand. Considering if two objects are moving, especially one is pulling the other: there are to scenarios, either one is stationary and using some mechanism to pull the other towards itself (winch) or one has greater velocity. I have no damn clue
  12. Feb 11, 2007 #11
    I think sometimes questions are worded deliberately vaguely to make you think thru all the scenarios, do it both ways, as I suggest and with a winch. See what you come up with.
  13. Feb 11, 2007 #12
    I have done it and computer says that all answers are wrong....now I'm trying to write two separate equations for displacement and make then equal and solve for time we'll see...
  14. Feb 22, 2010 #13
    When the tug pulls on the asteriod, both objects move (accelerate) toward each other. So, find the acceleration for both objects, add them, and plug the value into the formula, X= Vot + .5a(t^2). The Vo here is zero, so just X = 550m, and you are solving for t.
    **Don't assume one of the accelerations to be negative; the objects are moving towards each other, so relative to each other, you have to add them.**
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