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A physics (calculus) problem that I can't set up.

  1. Dec 16, 2007 #1
    I first want to say that this isn't a problem from school or anything, I just thought of it one day and when I tried to do it, I couldn't!

    1. The problem statement, all variables and given/known data
    If the earth suddenly stopped orbiting the sun in its circular path, it would immediately begin the accelerate toward the sun in a straight path. From a classical kinematic point of view, how long will it take the earth to reach the sun if r(0)=ri (distance from earth to sun), v(0)=0, and a(0)=0.

    I understand classical kinematics (a=dv/dt=d^2x/t^2), but in a macroscopic case like this, acceleration isn't constant; its a function of position, according to Newtons Law of gravitation a=G*m/r(t)^2.

    2. Relevant equations
    Newton's law of gravitation: A smaller object will accelerate towards a larger object with an acceleration = G*m/r(t)^2, where G is the gravitational constant, m is the mass of the bigger object, r(t) is the distance between the two objects.

    3. The attempt at a solution
    The first thing I thought to do was integrate a=G*m/r(t)^2 twice with time to get s as a function of t. => v=G*m*t/r^2 => s=G*m*t^2/(2*r^2) and s(ti)=r and s(tf)=0. I don't know where to go from there because of I have position as a function of time and position (if that makes sense?)

    So r(t)=ri - s. => s=ri - r(t) => ri - r(t) = G*m*t^2/(2*r^2).

    Can anyone help me out with this one? Thanks!
  2. jcsd
  3. Dec 16, 2007 #2
    Since the earth orbits in an elliptical path it would vary depending on where the earth is in its orbit.
  4. Dec 16, 2007 #3
    No, ignoring complications like that. With ri= mean distance from earth to the sun.
  5. Dec 16, 2007 #4
    Well you'd have an easier time using energy, at t=0 there is no kinetic energy of the system only potential
  6. Dec 16, 2007 #5


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    The 'time to fall integral' is a little difficult (but not undoable). But you can use Kepler's third law to get an estimate. The cube of the semimajor axis is proportional to the period squared. If the earth's velocity suddenly falls to almost zero then it's orbital path will be one that passes very close to the sun and then returns. That means that the semimajor axis is cut in half. What does that do to the period? Time to fall to the sun is then 1/2 of that new period.
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