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Rotational Motion of a rocket

  1. Mar 16, 2009 #1
    1. The problem statement, all variables and given/known data

    You are working on a project with NASA to launch a rocket to Mars, with the rocket blasting off from earth when earth and Mars are just aligned along a straight line from the sun. As a first step in doing the calculation, assume circular orbits for both planets. If Mars is now 64.7 degrees ahead of the earth in its orbit around the sun, when should you launch the rocket?

    Give your answer in days to the nearest whole number (i.e. 45.6 = 46)

    Note: For this problem you need to know the fact that all the planets orbit the sun in the same direction, and the year on Mars is 1.88 earth years.



    2. Relevant equations

    The rotational kinematics equations, but I can't seem to figure out which one to use. If you don't have them handy, here's a link:
    http://bama.ua.edu/~jharrell/PH105-S03/exercises/rot_mot_eqs.htm [Broken]
    Sorry, I don't know how to retype those on here.

    3. The attempt at a solution

    I honestly don't know where to start. I had a few ideas but none of them came close. The answer is 140 days. Any help is appreciated.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 16, 2009 #2

    Doc Al

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    Staff: Mentor

    What have you tried? How would you describe the motion of the planets? Accelerated, non-accelerated, uniform, non-uniform?

    Hint: Angle corresponds to distance.
     
  4. Mar 16, 2009 #3
    I didn't write anything down; I just thought out some concepts and what I was doing in my head. The motion is uniform with no radial accelaration but with centripital accelaration, although I don't think centripital accelaration comes in to play here. My guess would be that the angle has something to do with the time it takes to complete one orbit, I guess I'm having troubel relating the orbit's of Earth and Mars in roder to get one answer.
     
  5. Mar 16, 2009 #4
    I've tried a few more things since my last post, still nothing, does anyone have any other suggestions?
     
  6. Mar 17, 2009 #5

    Doc Al

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    Staff: Mentor

    Unless you show us what you're trying, we can't tell if you're on the right track.
    Both planets move uniformly in a circle at different speeds (find those angular speeds). They are centripetally/radially accelerated, but you're right--that doesn't matter here. All you care about is their tangential motion around the circle.

    If you had two cars, A and B, with B ahead of A by 100m but A moving faster. Could you figure out how long it takes A to catch up with B? It's the same problem, only with angle instead of distance.
     
  7. Mar 17, 2009 #6
    I had tried to regard Mars' orbit as a constant 1 (since I don't know actual speeds, only relative ones), that way Earth's would be 1.88^-1 which was .56 or something like that, I don't have my calculator with me at the moment. And I divided 64.7 by that number and got 121 plus a few decimal places. My logic was that if I regard Mars' orbit as constant, and Earth's orbit as changing, ignoring the fact that they are both orbiting a larger body, I can just use that number and the amount of space (degrees) Earth needs to cath up. Other slight variations of that technique didn't work either.

    Also, about teh car analogy: I've actually had several problems just like that before except we were given the speeds or accelarations of the cars and coudl relatively easily solve the problem using two or so f the kinematics equations. Here, all I'm supposed to use is the number 1.88 and 64.7, which is what made me think I had to toy with relative velocities.

    I've also tried plugging in the actual answer to see where it came from, like a sort of reverse solution, but I couldn't get it either.

    Does any of that sound remotely correct?
     
  8. Mar 17, 2009 #7

    Doc Al

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    Ah, but you do have the speeds of the planets.

    The speed of Earth = 1 orbit/earth year = 360 degrees/365 days

    What's the speed of Mars? (Use the given information to figure it out.)
     
  9. Mar 17, 2009 #8
    The rocket can be launched anytime you please. There are some constraints missing.
     
  10. Mar 17, 2009 #9

    Doc Al

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    The launch time (with respect to "now") is specified:
     
  11. Mar 20, 2009 #10
    Then I shall have to learn to read question not as they are written, but as they are intended to mean.
     
  12. Mar 27, 2009 #11
    Hey thanks for all the help guys, I ended up getting caught up with some other work and hadn't had a chance to look at this problem again until last night, but as I'm sure some of you know, all it takes is some time off (of a specific problem) to make you realize how simple it really was. I honestly want to slap myself. (joking, but not really). I sat down, and in about a minute got the right answer and thought, "why did I think that was so hard?"

    To anyone who's also trying to get this problem: read the thread, then find the angular velocities in rad/sec for both planets. Angular accelaration=0 and then you establish an arbitrary point (probably Earth's initial position) for position=0 and use whatever angle Mars is ahead of Earth as it's initial position. Use these values for the angular kinematics equation relating final position to initial position, initial velocity times time, and half of the accelaration times the time squared. Since you want both final positions to be the same (planets should be aligned as said in the problem) set the two equations equal to each other. You now should have an equation with just one unknown: final position. Solve for this, then do some unit conversions and you have your answer.

    Thanks again for everyone's help in this thread.
     
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