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Gravitational Force/Newton's 2nd Law

  1. Sep 19, 2008 #1
    1. The problem statement, all variables and given/known data
    The problem reads: "If the Earth were suddenly to stop in its orbit, how long would it take for it to collide with the Sun?[Regard the Sun as a fixed point mass. You make use of the formula for the period of the Earth's orbit.]" (The answer is 65 days. Obviously, I'm not interested in the numbers, I want to know where I am going wrong.)

    2. Relevant equations
    Newton's Law:
    [tex]m_{earth} \frac{dv}{dt} = -G m_{earth} m_{sun} \frac{1}{r^{2}}[/tex]
    which implies
    [tex]\frac{dv}{dt} = -G m_{sun} \frac{1}{r^{2}}[/tex]

    A change of variable from t to r:
    [tex]\frac{dv}{dt} = \frac{dv}{dr}\frac{dr}{dt} = v \frac{dv}{dr}[/tex]

    3. The attempt at a solution
    So far, I am getting (plugging in the change of variable)
    [tex]v \frac{dv}{dr} = -G m_{sun} \frac{1}{r^{2}}[/tex]
    Separate the variables.
    [tex]v dv = -G m_{sun} \frac{dr}{r^{2}}[/tex]

    The next step is to integrate. v is integrated over 0 (initial time) to t (t being the time of impact). My question right now is what is r integrated over? If I use R (the "radius" of the circular path) to 0, I end up with a 1/0 term (obviously wrong).

    [tex](1/2) v^{2} - (1/2) v_{0} ^{2} = G m_{sun} [\frac{1}{0} - \frac{1}{R}][/tex]

    Obviously, I am stuck there, but to give you an idea of where I am going, I will set the initial velocity (Vo) to 0 (the Earth stopped). Then I will change v to dr/dt. Separate Variables and integrate again. Then solve for time, t.
  2. jcsd
  3. Sep 19, 2008 #2
    I would integrate v from 0 to t where t is some time BEFORE impact, and integrate r from R to r.
  4. Sep 19, 2008 #3


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    I think you're on the right track. Instead of integrating the vdv equation to the point of impact, integrate to some intermediate distance "r" instead. Then v is the velocity at a distance r from the sun, and the pesky "1/0" term becomes "1/r" instead.

    I'm not positive that the 2nd integral you do (with dt and dr) will be integrable, but it might be so try it and see what happens.
  5. Sep 19, 2008 #4
    I think it will be integrable with a trigonometric substitution.
  6. Sep 19, 2008 #5
    I'm 99% sure you are right about that.

    Thanks for the ideas.
  7. Sep 19, 2008 #6
    I'm 100% sure because I tried it and it worked ;)

    You're welcome.
  8. Sep 19, 2008 #7


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    As a sanity check you can also use Kepler's third law which say T^2 is proportional to R^3 where T is the period and R is the semimajor axis. If the earth stops in its orbit than the orbit changes from circular to a cometary orbit which swoops in close to sun and then back out. So the semimajor axis becomes R/2 and you want half a period to fall in. So the time is 1yr*(1/2)^(3/2)/2. Which is about 65 days.
  9. Sep 20, 2008 #8

    Shooting Star

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    I feel there is no need to consider an elliptical or otherwise orbit. The whole point is that the earth starts to fall towards the point mass M of the Sun from a distance of R which is the radius of the circular orbit of the earth. What Dick has tried to do is perhaps to avoid things at r=0.
    Correctly written, if v is the speed at a disatnce r from the Sun at time t, it should read:

    [tex](1/2) v^{2} = G M_{sun} [\frac{1}{r} - \frac{1}{R}][/tex], since [itex] v_{0}=0[/itex].

    This could have been arrived at directly by conservation of energy. Now you just have to solve this equation by evaluating the integral [itex]\int_o^T {dt} [/itex] directly, as r goes from R to 0. (Put r=Rsin²θ.) If you feel uncomfortable with putting r=0, try taking the limit as r-->0, which will not make any difference. The time is finite even though the speed blows up at r=0. The value of GM will come naturally from equating the centripetal force to the gravitational attraction of the Sun, not by putting the values by hand. The answer should match the one given in the previous post.

    EDIT: Thanks,descendency for pointing out the typo. Correcting it.
    Dick, while appreciating your approach (I'd use it for similar scenarios), I thought it would be perhaps a bit sophisticated for a beginner.
    Last edited: Sep 20, 2008
  10. Sep 20, 2008 #9
    Not to nitpick, but the substitution is r=Rsin2θ. (the r/R term is inside of a square root and has to be simplified to a cos2θ to rid your integral of radicals.)

    that formula for the period is [tex]\tau^{2} = \frac{4\pi^{2}R^{3}}{GM_{sun}}[/tex]

    Once I got past choosing the wrong interval over r, I got the rest of it. Thanks again.
  11. Sep 20, 2008 #10


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