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Calculating the mass of the moon given radius.

  1. Oct 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Hi guys so i have this problem that i have been trying to solve and i think i might have the right answer but i am a bit confused about something.



    An astronaut conducting experiments on the moon, drops an object from a height of 1.6m and notices that it takes the object 1.4 to reach the moons surface. If the mean radius of the moon is 1.74 * 106m , calculate the mass of the moon.



    2. Relevant equations
    gx = GMx/Rx2

    Mx = gx * R2 / G


    3. The attempt at a solution

    Well this is where i get confused, I know that the gravity on the moon is 1/6th that of earth so its equal to 1.6m/s2.

    So this is what i did

    Mx = 1.6m/s2 * ( 1.74 * 106m )2 / 6.67 * 10-11

    and i got the mass to be 7.26260869 * 1022

    Is this the correct answer?
     
  2. jcsd
  3. Oct 18, 2009 #2

    cepheid

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    Rather than using your existing knowledge about the moon to figure out what g is, I think that you are supposed to use the kinematics data given in the problem (how long it took an object to fall freely from a given height) to *calculate* g. Then you can proceed given the knowledge that g = GM/R2
     
  4. Oct 18, 2009 #3
    Ok so i think this is rigth but i dont know. So i used the 3rd motion eqn to find a.

    d = T +1/2a(T)^2

    and then i found acceleration to be -2.1952 m/s^2

    IS that the correct value i should be using for g?
     
  5. Oct 18, 2009 #4

    cepheid

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    No, because the formula is actually:

    d = v0t + (1/2)at2

    where v0 is the initial velocity. In this case, the initial velocity of the dropped object is zero, so the formula reduces to:

    d = (1/2)at2

    Using this formula will give you an answer close to the 1.6 m/s2 that you remembered.
     
  6. Oct 18, 2009 #5
    Oh yes, that was a silly mistake, i forgot about that. And ya so after your suggestion i got the value fo g to be -1.568 m/s2

    And then i pluged it in to the formula i mentioned in my first post and got the mass to be

    -7.117356522 * 1022
     
  7. Oct 18, 2009 #6

    D H

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    That is not the correct answer, correct in this case meaning the answer the instructor expects you to get using the given information. In particular, you did not use the experimental evidence.
     
  8. Oct 18, 2009 #7

    D H

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    Does that answer make any sense? First off, no units. 7*1022 what? Secondly, when is mass ever negative?
     
  9. Apr 23, 2011 #8
    I have been assigned this question..
    the answer according to my worksheet is given as 7.4E22kg

    I cannot get this answer by finding a through a=d/t^2,

    My attempt:

    Fnet=ma=Fg=Gmm/r^2
    ma=Gmm/r^2
    a=Gm/r^2
    (r^2*a)/G = 3.631304348E22 (its half of the actual answer.. what am I doing wrong??)
     
  10. Apr 24, 2011 #9

    cepheid

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    Look at the posts above. From the kinematics equations for constant acceleration, it should be a = 2d/t2. Perhaps that is the cause of the problem you are having?
     
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