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Motion in a plane!

  1. Oct 31, 2009 #1
    1. The problem statement, all variables and given/known data
    Driving a spaceship isn't as easy as it looks in the movies. Imagine you're a physics student in the 31st century. You live in a remote space colony where the gravitational force from any stars or planets is negligible. You're on your way home from school, coasting along in your 20,000 kg personal spacecraft at 2 km/s, when the computer alerts you to the fact that the entrance to your pod is 500 km away along a line 30 degrees from your present heading, as shown in the following figure:
    http://img130.imageshack.us/img130/4743/knightv.jpg [Broken]
    You need to make a left turn so that you can enter the pod going straight ahead at 1 km/s. You could do this with a series of small rocket burns, but you want to impress the girls in the spacecraft behind you by getting through the entrance with a single rocket burn. You can use small thrusters to quickly rotate your spacecraft to a different orientation before and after the main rocket burn.

    You need to determine three things: how to orient your spacecraft for the main rocket burn, the magnitude F-thrust of the rocket burn, and the length of the burn. Use a coordinate system in which you start at the origin and are initially moving aling the x-axis. Measure the orientation of you spacecraft by the angle it makes with the positive x-axis. Your initial orientation is 0 degrees. You can end the burn before you reach the entrance, but you're not allowed to have the engine on as you pass through the entrance. Mass loss during the burn is negligible.

    m = 20000kg
    Vx0 = 2000 m/s
    Vx1 = 0 m/s
    Vy0 = 0 m/s
    Vy1 = 1000 m/s
    2. Relevant equations

    V12 = V02 + 2aS

    3. The attempt at a solution
    The distance in the x direction is 500000cos(30), the y direction is 500000sin(30).
    I use the formula above to solve for the acceleration in the x and y direction, ax = 4.62, ay = 2, so a = [tex]\sqrt{4.62^2 + 2^2}[/tex] = 5.03 m/s/s.
    Thrust force = 20000*5.03 = 100600N
    Then i use inverse tangent to find the angle, which is 23.41 degrees below positive x-axis (where the thruster should point at).
    However, the answer in the book is 103000N and 26.6 degrees below the positive x-axis.
    What did i do wrong?
    Any help is appreciated!
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 31, 2009 #2
    bump, please help!
  4. Oct 31, 2009 #3
    I think they're implying that you want to turn the corner by following an arc that's a quarter of a circle. You would then have to choose what radius that circle would have. Thruster force would be centripetal force. That's the only idea I have.
  5. Nov 1, 2009 #4
    I don't think that's the case. If i turn the corner by following an arc, i would have to use a series of rocket burns, but the problem is asking to do it with a single rocket burn, and i can use small thrusters to change the angle before i apply the main rocket burn, but i'm not sure though.

    Please help! I need to submit this to my teacher by the end of the night!
    Any help is greatly appreciated!!
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