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Asteroid Radius problem

  1. May 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Asteroids have average densities of about 2,660 kg/m3 and radii from 470 km down to less than a kilometer. Assuming that the asteroid has a spherically symmetric mass distribution, estimate the radius of the largest asteroid from which you could escape simply by jumping off. (Hint: You can estimate you jump speed by relating it to the maximum height that you can jump on earth, for this problem assume that height in 1 meter).

    2. Relevant equations

    v = [tex]\sqrt{}(8*pi*G*rho*R^2)/3[/tex]

    v = 2gh



    3. The attempt at a solution

    I derived the above equations to solve for R. I understand "g" is gravity relative to the Earth and not the asteroid. I am struggling to relate the gravity to the asteroid and solve for the largest radius.
     
  2. jcsd
  3. May 9, 2009 #2
    You need to measure the escape velocity you can have (the velocity of your jump on earth). V = 2g(1)

    Then plug that number as your V for the first equation and then solve for R.
     
  4. May 9, 2009 #3

    LowlyPion

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    I'd suggest that you view it as Potential Energy.

    You know for instance what potential energy you have in your jump (m*g*h).

    You can express then the potential energy you would be at on the surface of the asteroid.

    Escape velocity then is where potential energy of the asteroid is 0 and your kinetic energy is 0.

    That looks to simplify to:

    GM_ast*m/r = m*g*h
     
  5. May 9, 2009 #4
    I solved for the escape velocity and the radius and get 11.4 km. The correct answer is 3.63 km. I assume the escape velocity and gravity are relative to the earth and not the asteroid. Wouldn't it be the gravity with respect to the asteroid I need?

    Pion: When it came to the potential energy equation, I wasn't sure how to factor in the Mass of the asteroid as well as gravity with respect to the asteroid.
     
    Last edited: May 9, 2009
  6. May 9, 2009 #5
    Don't forget that the mass of the asteroid is roe*4/3*pi*r^3 and not roe*pi*r^2
     
  7. May 9, 2009 #6
    Right, the volume of the sphere compared to the area of a circle. Every equation I come up with, I end up with 2 unknown variable. Between gravity of the asteroid, R, and mass of the asteroid.
     
  8. May 9, 2009 #7

    LowlyPion

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    You have 2 equations and 2 unknowns.

    They give you the density which relates mass to radius.
     
  9. May 9, 2009 #8
    By taking density = Mass / Volume, and volume = (4/3)*pi*R^3, I can see relating mass to radius but that still leaves me with mass and radius unknown right? i'm still having trouble and not making very much headway.
    The substitution still leaves:

    rho = mass / ((4*pi*R^3) / )

    ....and this is where i still have mass and radius, I am just not making much headway.
     
  10. May 9, 2009 #9

    LowlyPion

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    The point is when you have 2 equations and 2 unknowns then you simply solve. Brute force substitution works just fine.

    GM_ast*m/r = m*g*h

    GM_ast / r = g*h

    2660 = M_ast / (4/3 πr³)

    M_ast = 2660*(4/3 πr³)

    So ...

    G*2660 * (4/3 πr³) / r = g * 1

    2660 * 6.67 * 10-11 * (1.333) * 3.141596 * r2 = 9.8

    Looks to me like 1 equation and 1 unknown.
     
  11. May 10, 2009 #10
    PERFECT!!! I was under the impression you had to solve for the gravitational acceleration on the surface of the asteroid first.

    I kept getting to...

    G*2260*(4/3*pi)*r^2 = g*h

    but i would try to derive another equation to substitute for "r" to try and solve for "g".

    - If the the person is jumping off the asteroid, why is the gravitational acceleration for the Earth sufficient? Why wouldnt you have to solve for the gravity value on the asteroid first?

    Thanks for the help!!!!!
     
  12. May 10, 2009 #11

    LowlyPion

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    The gravitational acceleration is not sufficient. It just turns out to be that way in this case. The g enters the picture because that's the energy budget available. On earth your potential energy in your legs - that you convert to gravitational potential at the height of your jump - is m*g*h.

    We merely substituted that for the kinetic energy that the jumper would have any where at take off. Whether it is on Earth or on an asteroid, the legs can produce that much energy. Since the mass of the jumper cancels out in figuring the Potential energy equation for escape ... it all boils down to how high you can jump ... g*h ... here on earth. Since h in this case is 1, ...
     
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