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Problem using integral form of Work K.E. Thm

  1. Sep 24, 2013 #1
    1. The problem statement, all variables and given/known data

    I have this problem where the earth immediately loses all orbital velocity and begins to fall towards the sun, and I need to find the time it takes for the earth to hit it.
    Seemed straight forward enough.

    2. Relevant equations

    Started with the work k.e. theorem,
    .5mv(x)2-.5mvo2=∫[from xo to x]F(x)dx.
    Where m=the mass of the earth, vo=0, and F(x)=F(r)=-GMm/r2 where M is the mass of the sun.

    3. The attempt at a solution

    So I made the bounds translate from xo→x to rAU→r(t). Solved the integral and got
    v(x)=sqrt(-2GM((1/rAU)-(1/r(t))))
    Seeing as how r(t) is never going to get bigger than 1 AU, this doesn't make any sense. The answer is already imaginary, and I haven't even gotten to the integral for solving for t(r) yet. Anybody know what I did wrong?

    note: In case it wasn't that obvious, I'm using the initial position of the Earth as rAU and the final point I'm trying to get the Earth to is the radius of the sun, ill just write as ro. The center of the sun is the origin of the coordinate system.
     
  2. jcsd
  3. Sep 24, 2013 #2

    ehild

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    Is r(t) smaller or larger than rAU during the motion of the Earth towards the sun? So what is the sign of -(1/rAU -1/r(t))?
    You got the magnitude of the velocity from conservation of energy. How do you define the vector v in terms of dr/dt?

    ehild
     
    Last edited: Sep 24, 2013
  4. Sep 25, 2013 #3
    Okay I looked at that wrong, however i still end up running into a problem later.
    I eventually get that
    v(r)=sqrt(2GM([1/r(t)]-[1/rAU])).
    I need to find t(r), which would be
    t(r)=∫(from rAU to rsun)dr/v(r).
    →=1/(sqrt(2GM))∫(from rAU to rsun)dr/(sqrt([1/r(t)]-[1/rAU])
    This is a very gross integral that I plugged into wolframalpha, and got a very long answer for, and when I plugged in the bounds I got imaginary numbers again.
     
  5. Sep 25, 2013 #4

    ehild

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    You miss a minus sign. dr/dt=-v=-sqrt(2GM([1/r(t)]-[1/rAU])), as r is decreasing.
    You can find the same problem here in Physicsforums. Try to search. The integral can be done with suitable substitution.

    ehild
     
  6. Sep 25, 2013 #5
    Maybe I'm missing something, but putting a negative sign before all of that wouldn't help me avoid getting an imaginary answer.
     
  7. Sep 25, 2013 #6

    ehild

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    You should not get imaginary answer. Using AU as unit of distance, and approximating the final distance as 0, type in int _1^0 (1/sqrt(1/x-1)dx) in Wolframalpha. As I see, Wolframalpha does integration by parts.

    ehild
     
  8. Sep 25, 2013 #7
    Okay, that's somewhat satisfying. But how come if you don't approximate the final distance as zero, and instead make the bounds from 2 to 1 (since I was planning on going from 1 AU to the radius of the sun), you get an imaginary number?
     
  9. Sep 25, 2013 #8

    ehild

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    The bounds are from 1AU to the radius of the Sun, which is much less than 1 AU.
    Use indefinite integral in wolframalpha and see what you get. The solution is real for x<1AU

    I have to leave now.

    ehild
     
    Last edited: Sep 25, 2013
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