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Homework Help: Earth falling into sun problem

  1. Mar 26, 2014 #1
    1. The problem statement, all variables and given/known data
    If Earth stops rotating Sun suddenly , how long would it take for Earth to collide with Sun ?
    G = gravitational constant
    m_s = 2e30kg
    m_e = 6e24kg
    where m_s is the mass of the sun and m_e , mass of earth
    x = 1au =1.5e11m
    2. The attempt at a solution
    Gravitational acceleration ,
    a = G(m_s + m_e)/x^2

    d^x/dt^2 = a
    dt = sqrt(1/a).dx
    dt = x/sqrt(G(m_s + m_e)) . dx
    integrating over x from x = distance between earth and sun to 0
    t = x^2/(2*sqrt(G(m_s + m_e)))

    The answer i m getting is not right , anyone help me finding out the solution and what i did wrong
    Last edited: Mar 26, 2014
  2. jcsd
  3. Mar 26, 2014 #2


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    According to the OP, the earth is 13 orders of magnitude more massive than the sun.
  4. Mar 26, 2014 #3

    Sorry for that ,fixed it now
  5. Mar 26, 2014 #4
    Your equation for the attractive force is incorrect. It should involve the product of the masses, not the sum. The acceleration should be independent of the mass of the earth.

  6. Mar 27, 2014 #5

    Gravitational forces act on both objects , both are directed towards each other (they finally collide at their centre of mass).
    If we switch our reference frame to an inertial frame of any one body , acceleration of the other is ,
    a = f_1/m1 + f_2/m2
  7. Mar 27, 2014 #6


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    d2x/dt2 is not the same as (dx/dt)2 .
  8. Mar 27, 2014 #7
    Ah. I see what you did. Thanks.

  9. Mar 27, 2014 #8

    OK thanks for that , but how should I solve this problem then.
  10. Mar 27, 2014 #9
    Use Kepler's 3rd law comparing earths actual orbit (semi-major axis = 1 AU) with the earth's orbit in the problem (semi-major axis = .5 AU), and keep in mind that the time to fall into the sun is just half an orbital time.
  11. Mar 27, 2014 #10
    Your starting equation is:


    If this is correct, then dx/dt is an integrating factor for both sides of your equation.

  12. Mar 27, 2014 #11
    The second integration is a bit more challenging though. It is easier to use Kepler's 3rd law and bypass the need to integrate at all (Well the integrations are hidden under the hood of the law).
    Last edited: Mar 27, 2014
  13. Mar 27, 2014 #12
    I hadn't looked at the second integration yet. I'll take a look at it later when I have some time. Is it just a bit more challenging, or is really much more challenging?

  14. Mar 27, 2014 #13
    It can be done fairly easily if you know a couple of tricks of the trade
  15. Mar 27, 2014 #14
  16. Mar 27, 2014 #15
    You forgot to define Θ.
  17. Mar 27, 2014 #16
    It was just a trigonometric substitution to facilitate the integration. But, I'd be interested in your filling me in regarding its relationship to Kepler's 3rd law.

  18. Mar 27, 2014 #17

    D H

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    It's fairly simple, conceptually. Imagine a sequence of orbits all with aphelion distance equal to one astronomical unit but with perihelion distances tending to zero. In the limit, the orbiting object is falls straight toward the Sun from aphelion (r=1 AU), makes a 180 degree turn at perihelion (r=0), and climbs back to aphelion. By symmetry, the time taken in going from aphelion to perihelion and the time taken in going from perihelion to aphelion are both half an orbit.

    That said, I suspect that the OP is supposed to integrate the equations of motion for this radial trajectory.
  19. Mar 28, 2014 #18
    Energy conservation is also a way to solve the problem. You have to solve a first order differential equation if energy conservation is used.
  20. Mar 28, 2014 #19
    Oh, OK, Yes that substitution works. DH already explained the relationship with Kepler's law.
  21. Mar 28, 2014 #20
    That bypasses the 1st integration leaving the second integration to be done by hand.
  22. Mar 28, 2014 #21
    Thanks a lot. Very cute trick.

  23. Mar 28, 2014 #22
    Thank you all for the help.
    I found the solution using Kepler's third law as stated above.
    But I am curious ,how can that integral be solved ?
  24. Mar 28, 2014 #23
    Do the first integration to get the velocity (using either dx/dt as an integrating factor or using conservation of energy). Then use the trig substitution that I recommended, in which x0 is equal to the initial distance.

  25. Mar 28, 2014 #24


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    The differential equation didn't give me the same answer as Kepler's 3rd law. From Kepler I got:
    [tex]t = \pi \sqrt{\frac{r_0^3}{GM}}[/tex]
    Starting with
    [tex]r'' = -\frac{GM}{r^2}[/tex]
    I got:
    [tex](r')^2 = 2GM \frac{r_0 - r}{r_0 r}[/tex]
    Then, using the trig substitution, I got:
    [tex](cos^2(θ))θ' = \sqrt{\frac{GM}{2r_0^3}}[/tex]
    And, finally
    [tex]t = \pi \sqrt{\frac{r_0^3}{8GM}}[/tex]

    Any ideas where I've gone wrong?
  26. Mar 28, 2014 #25
    Nothing is wrong. The r0 of the second method is the distance at aphelium while the r0 of the first method is the semi major axis. One is twice as big as the other for that degenerate orbit. The cube of that factor of 2 explains your factor of 8.
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