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Gravitational attraction problem

  1. Nov 12, 2007 #1
    1. The problem statement, all variables and given/known data
    An astronaut working in the cargo bay of a space shuttle accidentally released the lifeline when picking up a toolbox. The mission control computers quickly calculated the astronaut was drifting away from the space shuttle at a constant speed of 0.350m/s. The astronaut, toolbox and space suit had a combined mass of 187.5kg and the space shuttle had a total mass of 6.55x10^3 kg

    a) What is the gravitational attraction between the astronaut with equipment and the space shuttle when their centres of mass are 8.00m apart?

    b) Explain how the astronaut could use the toolbox of mass 17.5 kg to get back to the cargo bay within about 20 seconds when she is 10m away.


    2. Relevant equations
    [tex]
    \begin{array}{l}
    F = \frac{{Gm_1 m_2 }}{{r^2 }} \\
    v = \frac{s}{t} \\
    \end{array}
    [/tex]

    3. The attempt at a solution
    a)
    [tex]
    \;F = \frac{{Gm_1 m_2 }}{{r^2 }} = \frac{{\left( {6.67 \times 10^{ - 11} } \right)\left( {187.5} \right)\left( {6.55 \times 10^3 } \right)}}{{\left( {8.00} \right)^2 }} = 1.28 \times 10^{ - 6} \;N
    [/tex]
    Right?

    b) I'm a little confused here.
    The only thing I could think of involving the toolbox is to drop it, but how would that make her drop back to the cargo bay??

    Thanks
    Steven
     
  2. jcsd
  3. Nov 13, 2007 #2
    Have you considered conservation of momentum?
     
  4. Nov 13, 2007 #3
    hmm ok

    so we have the equation;
    [tex]
    \Delta {\bf{p}}_1 = - \Delta {\bf{p}}_2
    [/tex]

    But I'm still a little confused as to how this would help me in solving part b of the question.

    [tex]
    \begin{array}{l}
    m_1 v_1 = - m_2 v_2 \\
    \left( {187.5} \right)\left( {0.350} \right) = - \left( {17.5} \right)\left( {v_2 } \right) \\
    v_2 = 3.75\;ms^{ - 1} \\
    \end{array}
    [/tex]

    Any further help would be appreciated.

    Thanks
    Steven.
     
  5. Nov 13, 2007 #4
    What you have just calculated is the velocity she (so PC!) would have to throw the toolbox to stop her drift away from the space shuttle. What if she threw it harder?
     
  6. Nov 13, 2007 #5
    Right so if 3.75 m/s is the velocity she would have to throw the toolbox to stop her from drifting away, she needs to throw it harder.

    We are told that she needs to drift back within 20 seconds, when she is 10m away.

    This requires a velocity of 10/20 = 0.5 m/s

    So she must throw it with a velocity;
    V - 3.75 = 0.5
    therefore
    V=4.25m/s?
     
  7. Nov 13, 2007 #6
    Not quite. You need to take account of the relative masses of astronut and toolbox
     
  8. Nov 13, 2007 #7
    argh i'm still having trouble.

    The weight of the astronaut is 170kg and the weight of the toolbox is 17.5 kg.
    A minimum velocity of 3.75 m/s is required to stop the astronaut drifting away.

    I don't know what to do to form some kind of equation ... really quite stuck.
    Any further help would be appreciated.
     
  9. Nov 13, 2007 #8
    Using subscript A for astronut and T for toolbox
    [tex]\Delta p_A = - \Delta p_T [/tex]
    Momentum is mass x velocity
    [tex]\Delta(m_{A}v_{A}) = - \Delta(m_{T}v_{T}) [/tex]
    Mass does not change
    [tex]m_{A}\Delta v_{A} = - m_{T} \Delta v_{T} [/tex]
    The question is what [itex]\Delta v_{T}[/itex] is required to change the astronut's velocity from 0.350 away from the shuttle to 0.5 toward the shuttle ...
     
  10. Nov 13, 2007 #9
    Ahh ok.

    So it should be;
    [tex]m_{A}\Delta v_{A} = - m_{T} \Delta v_{T}
    [/tex]

    [tex]
    \begin{array}{l}
    \left( {170} \right)\left( { - 0.35 - 0.5} \right) = - \left( {17.5} \right)\left( {\Delta v_T } \right) \\
    \Delta v_T = \frac{{170 \times - 0.85}}{{ - 17.5}} = 8.26\;ms^{ - 1} \\
    \end{array}
    [/tex]
     
  11. Nov 14, 2007 #10
    Looks good
     
  12. Nov 14, 2007 #11
    excellent

    thanks for your help catkin!
     
  13. Nov 15, 2007 #12
    Hi,

    I have a question about escape velocities, is this a good place to post it?
    (I'm new, so sorry if I'm asking a stupid question, or if this is the wrong place to post it)

    Thanks
     
  14. Nov 16, 2007 #13
    Welcome to PF!

    You are in the right place, but just create a new post in the 'introductory physics' area using the 'new topic' button

    Steven
     
  15. Nov 18, 2007 #14
    Okay, Thanks!

    Jen
     
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