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Asteroid Gravity

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data
    An asteroid is discovered to have a tiny moon that orbits it in a circular path at a distance of 113 km and with a period of 41.0 h. The asteroid is roughly spherical (unusual for such a small body) with a radius of 19.3 km.
    a) Find the acceleration of gravity at the surface of the asteroid.

    b) Find escape velocity from the asteroid?


    2. Relevant equations
    Keplers 3rd Law [itex] T^2/R^3=4 π^2/ G M [/itex]
    [itex] g= G M/ R^2 [/itex]


    3. The attempt at a solution
    I started off using the first equation and solving for M. My main question is would R be just the Height of the moon, or the height of the moon, plus the radius of the asteroid?
     
  2. jcsd
  3. Nov 16, 2013 #2

    rude man

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    The law says R is the diistance between the centers of mass of the two bodies.
     
  4. Nov 16, 2013 #3
    Ok. So I put the numbers in and got a mass of 6.2875.... E16.
    So then using the second equation I came up with 14.74 m/s^2 which is totally unreasonable for and asteroid that small.
    I changed all the kilometers to meters so the units all go together, so I'm not sure where I went wrong.
     
  5. Nov 16, 2013 #4

    rude man

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    You'll have to show how you got your numbers if we are to diagnose.
     
  6. Nov 16, 2013 #5
    I have

    [itex] (4 π^2 (19300m+113000m)^3)/((6.674*10^(-11)) (147600s)) [/itex]

    which came out to be
    [itex] 6.28753*10^(16) kg [/itex]

    So then I have
    [itex] g=sqrt(G M/R^2) [/itex]
    or
    [itex] g=sqrt((6.674*10^(-11)) (6.28753*10^(16)))/ (19300^2) [/itex]

    Which came out to be 14.74 m/s^2
    (sorry about the exponents being messed up. Still new to LaTeX)
     
  7. Nov 16, 2013 #6

    rude man

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    a) Rather than find your mistake, why don't you combine the two equations into one for g. Then you get a nice foolproof formula for g.

    b) Compute escape velocity by potential.

    One problem is that they don't make clear what R is. It's either 113 km or 113 km + 19.3 km. Probably they meant the latter so then R = 132.3 km.
     
  8. Nov 16, 2013 #7

    rude man

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    Where did the number 19300 come from? It's certainly not R^2!!
     
  9. Nov 16, 2013 #8
    That's 19.3 km in m.

    So I get
    [itex] g= sqrt((4π^2*(132300)^3)/(19300^2 * 147600^2)) [/itex]
    (132300m is 19.3km+113km.)
    Which comes to .106 m/s^2, but it still says its the wrong answer.
     
  10. Nov 16, 2013 #9

    haruspex

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    It's not clear whether you need to add the 19.km on here. It shouldn't make much difference anyway. More to the point, you forgot to square the time.
    You'd be better off leaving G as a symbol (in fact, calculate GM rather than M). as you can see, the Gs cancel later. I get GM = 23.14 m3s-2

    It's the 19.3km in metres. It doesn't need to be R2 since Psyguy22 wrote 193002.
     
  11. Nov 16, 2013 #10

    D H

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    Of course it's the wrong answer. What are the units of your expression [itex] g= \sqrt{(4π^2*(132300)^3)/(19300^2 * 147600^2)} [/itex] ? (Hint: They are not meter/second squared.)

    Always use units in your expressions and you won't have problems like this. Yes, it's a tiny bit more effort to carry those units around. That little extra effort saves a whole lot of agony.
     
  12. Nov 16, 2013 #11
    Ok so for units I'm getting sqrt(m)/s... which I have no idea what that means...
     
  13. Nov 16, 2013 #12

    D H

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    It means you made a mistake.
     
  14. Nov 16, 2013 #13

    rude man

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    You used the same R in both your formulas which is wrong. It's 132.3 km in the 1st and 19.3 km in the second .... which I only caught now.
     
  15. Nov 16, 2013 #14
    @ D H
    I see... I don't know why I have the sqrt in the g equation...
    But even so, I'm getting .01127 m/s^2 and its still wrong.

    @rude man
    I haven't used the same R.. I used 1323000 m for the top R and 19300 for the bottom R
     
  16. Nov 16, 2013 #15

    D H

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    The question as you wrote it in the original post is a bit ambiguous with regard to what that value of 113 km means. Your updated result is the correct answer if that value of 113 km means the distance from the surface of the asteroid to the small moon. Try interpreting that value of 113 km as meaning the distance from the center of the asteroid to the small moon.
     
  17. Nov 16, 2013 #16

    rude man

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    OK. But a poor 1st post.

    Go with post 15.
     
  18. Nov 16, 2013 #17
    Thank you. The answer was looking for you to say that 113km was the distance between the center of the asteroid.
     
  19. Nov 16, 2013 #18

    D H

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    That illustrates one of the problems with these automated homework tools. Problems are oftentimes stated ambiguously. If this was an old-style, hand in your homework type of problem, a good grader would have seen your addition of 113 km + 19.3 km and looked to see if the problem statement was stated ambiguously. You would have gotten full credit for your answer of 0.01127 m/s2 if that 113 km could legitimately have been interpreted as being altitude rather than radial distance.


    Aside: Learning to deal with ambiguity is a good thing. There are no unambiguous real world problems.
     
  20. Nov 16, 2013 #19

    rude man

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    Agreed on the problem of 'automated homework tools".

    I also see it as the failure of the engineering community to stress good exposition. This problem is exacerbated for our many ESL students. For them especially, clarity is of utmost importance.
     
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