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I Falling to a Star With Varying Acceleration

  1. Apr 9, 2016 #1
    Take an object of negligible mass that is dropped from rest 2 kilometers away from a neutron star of mass 1.989*10^30 kilograms (1 solar mass) and radius 7,802 meters. How long will it take the object to reach the surface of the neutron star?

    I'm not terrible at calculus, but I know for a fact that a problem such as this involves some integration that I haven't been able to pin down.
     
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  3. Apr 9, 2016 #2

    Janus

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    [tex]t = \frac{ \cos^{-1} \left ( \sqrt{\frac{x}{r}} \right ) + \sqrt{\frac{x}{r} \left ( 1-\frac{x}{r} \right ) }}{\sqrt{2 \mu}} r^{\frac{3}{2}}[/tex]

    Where x is the radius of your neutron star
    r is the distance of the object from the center of the neutron star
    [itex]\mu[/itex] is equal to G(M+m) with M+m being the sum of the neutron star and object masses. (if m is really small with respect to M, you can ignore it without sacrificing accuracy too much.)
     
  4. Apr 9, 2016 #3
    Thank you! Do you have a derivation for this that I could look at?
     
  5. Apr 9, 2016 #4

    PeroK

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    There was one on here but it was a long time ago and I can't find it. I've got the solution written down if you are interested.

    How much maths do you know? It's really just a solution to a differential equation. It's not too hard but involves a trick or two.
     
  6. Apr 9, 2016 #5
    I'm taking high school level AP calculus right now, but I'd be happy to see the solution! In terms of differential equations, I only know how to solve simple linear differential equations and separable differential equations.

    Also, I just realized. Why couldn't I just use conservation of energy?
     
  7. Apr 9, 2016 #6

    PeroK

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    Yes, you can use conservation of energy, but that's not all you need. I'll show you both methods. I've got slightly different notation:

    ##M## is the mass of the large body, ##R## is the radius of the large body, ##r_0## is the starting position of the falling mass, ##r_1## is the final position of the falling mass, where ##r_1 \ge R##, and ##r## is the variable position of the falling mass. And, ##r'(0) = 0## meaning the falling mass starts from rest.

    First, you have the differential equation for ##r##:

    ##r'' = -\frac{GM}{r^2}##

    The first trick is to multiply both sides by the integrating factor ##2r'## to give:

    ##2r'r'' = -2r'\frac{GM}{r^2}##

    ##\frac{d}{dt}(r')^2 = \frac{d}{dt}(\frac{2GM}{r})##

    ##(r')^2 = \frac{2GM}{r} + C = \frac{2GM}{r} - \frac{2GM}{r_0} = 2GM(\frac{r_0 - r}{rr_0})## (using the initial position to find ##C##)

    This gives us the first equation:

    ##(r')^2 = 2GM(\frac{r_0 - r}{rr_0})## Equation (1)

    The alternative, as you suggested, is to use conservation of energy:

    ##-\frac{GMm}{r_0} = -\frac{GMm}{r} + \frac{1}{2}m(r')^2##

    Which gives you equation (1) more easily.

    The main trick is a non-obvious substitution:

    Let ##r = r_0 cos^2\theta## hence ##r' = -2r_0(cos\theta sin\theta)\theta'##

    Substituting this into (1) gives:

    ##4r_0^2(sin^2 \theta cos^2 \theta)(\theta')^2 = 2GMr_0(\frac{1 - cos^2 \theta}{r_0^2 cos^2 \theta}) = \frac{2GMsin^2 \theta}{r_0 cos^2 \theta}##

    ##cos^4 \theta (\theta')^2 = \frac{GM}{2r_0^3}##

    ##cos^2 \theta (\theta') = \sqrt{\frac{GM}{2r_0^3}}## Equation (2)

    Now we integrate this with respect to ##t##, noting that at ##t = 0, r = r_0, \theta = 0## and at ##t = t_1, r = r_1, \theta = \theta_1##

    ##\int_0^{\theta_1} cos^2 \theta d \theta = \sqrt{\frac{GM}{2r_0^3}}t_1 + C##

    ##\frac{1}{4}[2\theta_1 + sin2\theta_1] = \sqrt{\frac{GM}{2r_0^3}}t_1## (as ##C = 0##)

    This gives us essentially an intermediate solution for ##t_1##:

    ##t_1 = \sqrt{\frac{r_0^3}{8GM}}[2\theta_1 + sin2\theta_1]## Equation (3)

    Where ##r_1 = r_0 cos^2 \theta_1## hence ##\theta_1 = cos^{-1} \sqrt{\frac{r_1}{r_0}}##

    Now, of course, you can use trig identities to give ##sin 2\theta_1## in terms of ##r_0, r_1##:

    ##sin2\theta_1 = 2cos\theta_1 sin\theta_1 = 2 cos\theta_1 \sqrt{1- cos^2\theta_1} = 2 \sqrt{\frac{r_1}{r_0}} \sqrt{1- \frac{r_1}{r_0}}##

    This gives you the formula that Janus posted:

    ##t_1 = \sqrt{\frac{r_0^3}{2GM}}[cos^{-1} \sqrt{\frac{r_1}{r_0}} + \sqrt{\frac{r_1}{r_0}(1- \frac{r_1}{r_0})}]## Equation (4)

    Although, this is actually a more general formula for time to fall from any distance ##r_0## to any other distance ##r_1##.

    There is a special case worth mentioning. If the object falls from a long way away and you model the large mass as a point mass, then ##r_1 = 0## and you get a good estimate for the time it takes the two objects to collide. The exact equation would, of course, have ##r_1 = R##:

    ##t_1 = \sqrt{\frac{r_0^3}{2GM}}[\frac{\pi}{2} + 0] = \pi \sqrt{\frac{r_0^3}{8GM}}## Equation (5)
     
    Last edited: Apr 9, 2016
  8. Apr 9, 2016 #7
    Is it valid to divide the distance r by t above to get the speed of entry?
     
  9. Apr 9, 2016 #8

    PeroK

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    No. You can get the speed at any point simply from conservation of energy.
     
  10. Apr 9, 2016 #9
    Wow. Thank you I actually understand that. That is really awesome.
     
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