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Work, energy problem

  1. Apr 2, 2013 #1
    1. The problem statement, all variables and given/known data

    An asteroid (space rock) is 10 m in width and has a density of an average Earth rock of 2000 kg/m^3 (2.0 g/cm^3). (a) Find the mass of the asteroid? (b) Suppose (unrealistically) that the asteroid starts from rest half way between the Moon and the Earth (6.0 x 10^5 km) and accelerates toward the Earth at and average acceleration of (1/10) g = 1.0 m/s^2. What will the velocity of the asteroid be when it reaches the Earth. Note: although the assumptions are unrealistic the result will agree with what is known about asteroid velocities as they pass the Earth. (c) Now calculate the kinetic energy of this asteroid when it reaches Earth. Convert this to tons of TNT (1.0 ton TNT = 1.0 tn = 4.186 x 10^9 J). For comparison purposes note that the atomic (fission) bomb dropped on Hiroshima was about 12 - 14 ktn (kilotons TNT). (d) Finally, how much kinetic energy would an asteroid ten times wider or 100 m wide (still only a medium size asteroid) under the same conditions bring to Earth upon impact?

    2. Relevant equations

    3. The attempt at a solution

    (a) Find the mass of the asteroid?
    I was thinking of using the ol' D = M/V formula...but I have no way of finding V

    (b) I have no idea what to do here...I was thinking using graviational potential energy = mgh being converted to kinetic energy 1/2mv²...but I'm not sure...and still can't figure it out without the mass

    (c) Now calculate the kinetic energy of this asteroid when it reaches Earth
    I thought I would use the results of V (from part (b)) and M (from part (a)) and find K.E here...but the fact that it says "when it reaches Earth" throws me off and makes me think otherwise

    any help here? Thanks in advance
  2. jcsd
  3. Apr 2, 2013 #2
    I'm thinking you can skip part a for later. But for part b, since energy is conserved What do you know about the initial and final energies of the object ? Examine your equations closely.
  4. Apr 2, 2013 #3
    From what I know (just started learning about this in lecture)

    the initial potential energy would be mgh (gravitational p.e) and final p.e would be 0
    and the initial kinetic energy would be...0? because it starts from rest...and final would be kinetic energy 1/2mv²

    the fact that it has an acceleration is totally throwing me off...but here's what I know

    **would I replace g below (and replace it with acceleration??

    but with "g" I have

    mgh = 1/2mv²
    so the m's cancel

    gh = 1/2v²

    rearrange for v² for now

    v² = gh/(1/2)

    or v = √(gh/(1/2))

    anything going right there??
  5. Apr 2, 2013 #4
    That looks about right.

    And yes use the acceleration provided for g
  6. Apr 2, 2013 #5
    since (1/10) g is 1.0 m/s²

    I'm assuming that "g" would be 10m/s² correct?

    so it would look like

    V = √((10m/s² * (6.0 x 10^8m)) / 2
    converted to meters

  7. Apr 2, 2013 #6
    No g would just be as stated, 1.0m/s^2
  8. Apr 2, 2013 #7
    But it states that is what (1/10) of g is....It wouldn't be correct to just simplify that to g = 10m/s² by multiplying both sides by 10?
  9. Apr 2, 2013 #8
    I think you're reading it wrong, 1/10g IS 1.0m/s^2. I don't think its an equation where you have to solve for g.
  10. Apr 2, 2013 #9
    Plus it also wouldn't make sense for g to be 10m/s^2 when it is at that height.
  11. Apr 2, 2013 #10
    That's what I'm thinking but then I'm also reading the

    (B) "Suppose (unrealistically)" part.....it's just really confusing me..I'm not sure it the "unrealistic" part is the fact that it can start from rest there....or that PLUS the odd acceleration

    I guess for the point of doing it....I'll go with your idea that g is the 1.0m/s²
  12. Apr 2, 2013 #11
    Just making sure you're multiplying by 2 not dividing right?
  13. Apr 3, 2013 #12
    Right, sorry about that i'm not sure why I didn't put /(1/2) here...but yes, same thing as multiplying by 2

    Thanks for your help freshcoast!
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