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Homework Help: A asteroid approaches the Sun

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A asteroid is approaching our solar system from a great distance (r->infinity) and is headed straight for the Sun. Because of the gravitational pull of the Sun the asteroid is constantly accelerating.

    Radius of Earth's Orbit around the Sun ([tex]R_{e}[/tex]) 1.5*10^11
    Venus is given as 3/4 of this.
    Sun's mass (M) 2.0 * 10^30kg


    a)What is the speed (v) of the asteroid, which was originally stationary at a very great distance, as it comes as close to the Sun as the Earth orbits?

    b)How long does the asteroid take to travel from crossing the orbit of the Earth around the Sun to crossing Venuse's orbit.

    2. Relevant equations

    [tex]F= \frac{G*mM}{r^2}[/tex]
    [tex]E_{p}=m*g*r[/tex]
    [tex]E_{k}=\frac{m*v^2}{2}[/tex]
    [tex]Work=\int F ds[/tex]

    3. The attempt at a solution

    a) Energy is conserved. The sum of potential and kinetic energy is constant. noting that g of course changes with r (distance from the Sun)

    g=grav. const. m*M/r^2

    [tex]\frac{m*v^2}{2} = \int^{\infty}_{0}\frac{G*m*M}{r^2}dr - \int^{R_{e}}_{0}\frac{G*m*M}{r^2}dr[/tex]


    [tex]v^2 = 2*(\int^{\infty}_{0}\frac{G*M}{r^2}dr - \int^{R_{e}}_{0}\frac{G*M}{r^2}dr)[/tex]

    [tex]v^2 = 2*G*M*(\int^{\infty}_{0}\frac{1}{r^2}dr - \int^{R_{e}}_{0}\frac{1}{r^2}dr)[/tex]

    [tex]v^2 = 2*G*M*r_{E}^-1[/tex]

    v=4,22 m/s?

    I'm posting this because the units don't add up so I must have made a mistake either in the set up or in the integration itself (which I don't think I did). I translated the text from my own language to English so if the instructions or notation doesn't make sense please ask. Thanks for your help and patience! :)

    b) Hm this one proved to be more difficult. Its no trouble at all to see that I more or less basically have a function of v(r).

    [tex]v=\sqrt{\frac{2*G*M}{r}}[/tex]

    r obviously changes from Earth's Orbital radius to the Orbital Radious of Venus (given as 3/4 of Earths).

    So I'm going to mark travelled distance as s, which is of course:
    [tex]s=\int^{t}_{0}v(t)dt[/tex]
     
    Last edited: Sep 12, 2010
  2. jcsd
  3. Sep 12, 2010 #2

    ehild

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    Do you understand the gravitational constant on "grav"? Yes, you get the gravitational potential energy at a distance R from the sun by integrating the force of gravity from R to infinity. If you integrate with respect to r, the limit should be also distance, not some other quantity. And you know what is the integral of 1/r^2, don't you?

    ehild
     
  4. Sep 12, 2010 #3
    Yes, sorry about that I wasn't sure if you guys use G or perhaps Chi for the constant so I just put grav. for gravitational constant.

    I should change the notation. E in the integral limit was supposed to be the radius of Earth's orbit. Which could be confused as energy. I'll edit the opening post and mark it as [tex]R_{e}[/tex].



    The integral of r^-2 is r^-1/-1 or -1/r right?



    Are you saying the solution for a) is basically correct? I tried to check the units twice and they don't match.
     
    Last edited: Sep 12, 2010
  5. Sep 12, 2010 #4

    ehild

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    Yes it is basically correct. But there is no sense to integrate from r=0. The potential energy of a body in a force field is defined at a point P as the work done by the force when the body moves from point P to the place where PE=0. It is infinity for gravity.

    [tex]PE(R_E) = \int^{\infty}_{R_E}{-\frac{GmM}{r^2}}=-\frac{GmM}{R_E}[/tex]

    At infinity, both PE and KE are 0. So the total mechanical energy PE+KE=0 everywhere in the space, far away from other planets.

    The gravitational constant is G=9.6743 E-11 m3 kg-1s-2 . Multiply it by kg2 and divide by meter, the result is Nm = joule, isn't it?


    ehild
     
    Last edited: Sep 12, 2010
  6. Sep 12, 2010 #5
    Thank you. I see now that it was pointless to have two integrals, but my tought process went: "Oh cool the difference between the potential energy at infinity and at Earth distance is equal to the total kinetic energy of the asteroid". And I didn't think about it much further.


    Also It seems that in my notes I had left off ^-1 from kg^-1, which explains why the units didn't match. :blushing:

    Thanks again for you help! :smile:


    After I do some other exercises I'll get to b).
     
  7. Sep 12, 2010 #6

    ehild

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    You did think it well, but the potential is counted from infinity, where it is zero, and not from r=0, where it is infinite.

    As for b), it seems a bit more tricky.


    ehild
     
  8. Sep 12, 2010 #7
    Can anyone give me a hint on b)?
     
  9. Sep 12, 2010 #8

    ehild

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    The asteroid travels straight towards the Sun. How do you define v with time derivative of r?

    ehild
     
  10. Sep 12, 2010 #9
    Did I completely misunderstand you or is this what you are referring to?

    [tex]v=\frac{d(R_{e}-r)}{dt}=\sqrt{\frac{2*G*M}{r}}[/tex]
     
    Last edited: Sep 12, 2010
  11. Sep 12, 2010 #10

    ehild

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    Not quite. r is the distance from the Sun. According to your formula, the distance will increase with time.

    ehild
     
  12. Sep 12, 2010 #11
    Better?

    [tex]v=\frac{d(R_{e}-r)}{dt}=-\frac{dr}{dt}=\sqrt{\frac{2*G*M}{r}}[/tex]
     
  13. Sep 12, 2010 #12

    ehild

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    Yes! A nice separable DE. Solve.

    ehild
     
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