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Decending Lunar lander velocity

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Lets imagine we are back in 1969 and the Apollo Lunarlander is on approach to the moon. It has a mass of 200 kg and initial velocity of v1 = 6 m/s. The Astrofolks push the big red button in the cabin and engine burst of energy to break the spacecraft where time = 2 sec the resulting Force of the engine is 800 Newton and time = 4 sec the Force produced of the engine is 800 Newton. Whats the velocity of the decending lunar lander at t = 5 sec?

    My professor did a fancy calculation where he included the mass of spacecraft and the resulting force of the engine bursts. But I tried something else :)

    Lets imagine that after the they used the engine to break the spacecraft that its free-falling towards to the moon then.

    3. The attempt at a solution

    [tex]v_{final} = v_{initial} - g_{moon}\cdot t[/tex] where t = 5 sec and [tex]g_{moon} = 1.62 m/s^2[/tex]

    Then I plug this into the formula I get:

    [tex]v_{final} = 6 m/s - 1.62 \cdot 5 = - 2.1 m/s[/tex]

    Since the engine isn't given bursts at t = 5 then can't the LL be see as free falling and we can disregard its mass? And since the LL is decending the the final velocity is negativ?

  2. jcsd
  3. Mar 18, 2010 #2

    Filip Larsen

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    Gold Member

    Your calculation would be correct, if the speed in question is the vertical speed, with positive up, and if there were no thrusting taking place between t = 0 and t = 5.

    But, even though I have trouble understand the exact details in the problem text as you have quoted it, it do read like there is thrusting taking place and this will make the resulting speed come out different. No matter what method you use it must include the effect of the specified thrusting somehow.
  4. Mar 18, 2010 #3
    The only equation I can come up with

    [tex]W_1 + W_2 - m\cdot g = 1/2 \cdot m \cdot v^2[/tex]

    where W_1 and W_2 is the thrusting of the engine at t = 2 and t = 4 in KWh.

    But if I plug this into this formula I don't get the right result :grumpy:

    which is v = 2,1 m/s

    What am I doing wrong?

  5. Mar 18, 2010 #4
    You cannot add a work and a force, they are of different units.
    You have an initial speed.
    You have two forces acting on the spaceship: gravity and engine thrust.
    So what is the acceleration?
  6. Mar 18, 2010 #5
    I found another interesting formula

    Which states that if object of mass m is effected by a Net Force T then if the object moves vertically


    [tex]\sum F_y = T = m \cdot a_y [/tex] which is implies [tex]a_y = T/ m [/tex]

    Why is that these physics problems aren't formulated so one can use the off the shelf formulas in the book :(
    Last edited: Mar 18, 2010
  7. Mar 19, 2010 #6
    Okay. You have the acceleration. Now you can come up with the speeds in different points of time.
    You can use off the self formulas in the book, just have to know which ones and in what order:)
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