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The apparent weight of an astronaut in space if

  1. Feb 13, 2005 #1
    what is the apparent weight of a 65 kg astronaut 4200 km from the centre of the moon in a space vehicle a) moving at constant velocity, and b) accelerating toward the moon at 3.6 m/s^2? state the direction in each case.

    [mass of moon: 7.36*10^22 kg]
    [radius of moon: 1.74 * 10^6 m]
    [Period of Moon Revolution: 2.36 * 10^6 s]

    answers: a)18N toward the moon; b)220N away from moon

    i do not exactly know how to do the question, but i think part a) has something to do with this equation: w=mg-m(v^s)/r. and i totally don't get the direction part.

    can somebody give me some hints. thanks in advance.
  2. jcsd
  3. Feb 13, 2005 #2


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    The direction part can be found out by applying the 2-nd law as it should be:in vector form...You'd then have to find a convenient system of coordinates and project on its axis the vector equation...


    P.S.Compute the effective "g" in the first case.
  4. Feb 13, 2005 #3
    actually, i don't even know how to start the question.

    can you tell me that which equation i should use.

    is it the Fgrav.=Fcent. ?
  5. Feb 13, 2005 #4


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    Yes,but you should look at it vector equality.What is then the absolute value of each vector?

  6. Feb 13, 2005 #5
    hmm... after numerous tries, i still dont get it. can you please explain what you meant by "look at its vector quality"?
  7. Feb 13, 2005 #6


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    Well,the second law states that
    [tex] m\vec{a}=\vec{F}_{grav.} [/tex]
    and because the acceleration has a nonzero radial component,u may project the previous relation on the radial diraction and wind up with a scalar egality
    [tex] m\frac{\omega^{2}}{r}=G\frac{mM_{moon}}{d^{2}} [/tex]
    ,where the latter is the distance betweent the astronaut and the center of the moon.The acceleration due to gravity is then found to be
    [tex] g(d)=\frac{GM_{moon}}{d^{2}} [/tex]

    Find it and u'll solve point a).

    As for point "b",things are not that simple,since the astronaut is accelerating...And so,in his noninertial reference system will be subject to inertial forces...

  8. Feb 13, 2005 #7
    now, i finally know how to get the numerical answers for the questions, but i still don't get the direction part. WHY is when the guy is travelling at constant velocity, the direction for his weight points toward the moon; while he is accelerating toward the moon, his weight points away from the moon.
  9. Feb 13, 2005 #8


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    When he's travelling at constant velocity,the only force acting on the astronaut,in his own frame of reference is the gravity pulling him towards the moon,therefore the diraction of this force is POINTING TOWARDS THE CENTER OF THE MOON...
    When he's in an accelerating frame of reference,for his total acceleration to be zero,he finds the inertial force to be acting in opposite direction to the gravity force.So his apparent gravity is the inertial force who's pointing away from the moon and has a magnitude of magnitude of 216N.
  10. Feb 13, 2005 #9
    oh, ok. thanks very much for your help, daniel.
  11. Feb 14, 2005 #10

    Andrew Mason

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    The acceleration at r=4200 km is
    [tex]g = \frac{GM_{moon}}{r^2} = 6.67e{-11}*7.35e22/(4.2e6)^2}[/tex]
    [tex]g = .278 m/sec^2[/tex]

    So if it is moving at constant velocity (0 acceleration) it must be experiencing a force away from the moon equal to gravity: ie. of 65 kg x .278 m/sec. = 18.7 N. Its 'weight' is the inertial reaction to this applied force - it is not a real force. Note that the astronaut's actual velocity (speed and direction) will depend only on its motion at the time this force was applied.

    As for b), the total force is 65x3.6 = 234 N toward the moon, of which 18.7 N is provided by the moon so the ship must be providing another 215.3 N toward the moon. Again, the 'weight' is the inertial reaction to this which appears to be 215.3 N in the opposite direction - radially away from the moon.

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