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Landing problem

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    A 8.73 × 103-kg lunar landing craft is about to touch down on the surface of the moon, where the acceleration due to gravity is 1.60 m/s2. At an altitude of 217 m the craft's downward velocity is 16.5 m/s. To slow down the craft, a retrorocket is firing to provide an upward thrust. Assuming the descent is vertical, find the magnitude of the thrust needed to reduce the velocity to zero at the instant when the craft touches the lunar surface.

    http://edugen.wiley.com/edugen/courses/crs2216/art/qb/qu/c04/EAT_12257263467150_5449420724903514.gif

    2. Relevant equations

    v^2=v0^2+2ax

    3. The attempt at a solution

    Fthrust=Fg? mg=ma?
     
  2. jcsd
  3. Feb 1, 2010 #2

    kuruman

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    What is the needed acceleration to bring the velocity down to zero over the given height?
    How many forces act on the lander as it is coming down?
    Can you write Newton's Second Law?
     
  4. Feb 1, 2010 #3
    Hello onyxorca,

    As a first step you have to figure out which forces act on the rocket. Then you have to use the general equations for x(t) and v(t).
     
  5. Feb 1, 2010 #4
    F=ma

    the forces are gravity and thrust.

    btw the mass is 8.73 × 10^3 kg.

    the acceleration is 0=16.5^2+2a217 and a = -0.627304 m/s^2 ? where do i go from there?
     
  6. Feb 1, 2010 #5

    kuruman

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    Use Newton's Second Law. Find an expression for the net force (the sum of all the forces) and set it equal to mass times acceleration.
     
  7. Feb 2, 2010 #6
    Fthrust - Fg = ma?

    so F - 8730*1.6 = 8730*-0.627304

    F=8491.64 N?
     
  8. Feb 2, 2010 #7
    F_thrust -F_g = ma is correct! :smile:

    Now you need to find out the value of a.

    For that use the general equations for x(t) and v(t):

    v(t) = a*t + v_0

    x(t) = 1/2*a*t^2 + v_0*t + x_0
     
  9. Feb 2, 2010 #8

    kuruman

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    OP found the acceleration in Posting #4. The answer is OK.
     
  10. Feb 2, 2010 #9
    well appearantly 8491 isn't right...
     
  11. Feb 2, 2010 #10

    kuruman

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    I checked the numbers and I checked the method. The answer should be correct. If someone (or a computer program) says that it is not, then you need to take your solution to the person who assigned this problem and ask him/her where the error is.
     
  12. Feb 2, 2010 #11
    V0=V^2+2ax find a

    F=mg-ma=m(g-a)

    i finally got it right, had trouble realizing that the thrust is smaller than gravitational force and i made mistakes when putting in the scientific digits but i finally got it.
     
  13. Feb 2, 2010 #12

    kuruman

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    This is the right answer. What number do you get when you plug in?
     
  14. Feb 2, 2010 #13
    well the numbers in the problem changes every time so i wouldn't know if this is right but i got the right answer for that time.

    in this case though it's probably F = 8.73E3 (1.6+.627304) = 19444.4 N because it's Fg-Fthrust, not the other way around.
     
  15. Feb 3, 2010 #14
    I get the same number.
     
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