1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.

    Chet
     
  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

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

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

    Chet
     
  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:

    [tex]\frac{d^2x}{dt^2}=-G\frac{(m_s+m_e)}{x^2}[/tex]

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

    Chet
     
  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?

    Chet
     
  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
    [tex]x=x_0\cos^2θ[/tex]
     
  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.

    Chet
     
  18. Mar 27, 2014 #17

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Earth falling into sun problem
  1. Sun and Earth Problem (Replies: 10)

Loading...