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Gravitational field conceptual question

  1. Nov 21, 2011 #1
    Suppose a case where two masses are essentially far away form all other bodies so that we can assume the two masses are not affected by any other interactions going on beside them and they do not have the property of gravitation until a time( they are perfectly isolated{in theory everything is possible}). If suddenly at t=0 the gravitational field starts how do we find the position of the two masses as a function of time due to gravitational interaction ( and nothing else)??
  2. jcsd
  3. Nov 21, 2011 #2

    Simon Bridge

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    For the sake of the thought experiment, we can imagine a universe with only two masses in it.
    In theory, not everything is possible, that is why we have theories. However, we can imagine that the objects have particular positions at time t=0 when
    You mean apart from finding their mutual acceleration by Newton's gravitation?
  4. Nov 21, 2011 #3
    Unless I am missing something in the question, this is a textbook example of a classical mechanics problem. You simply apply Newton's laws, F = m a where the force is given by Newton's law of gravitational. Break vectors into components and integrate the acceleration, which is the second derivative of position with respect to time, to get equations of motion, being careful to remember that the force of gravity depends on position as well. This is all pretty basic stuff you could find in any textbook. Which part don't you understand?
  5. Nov 21, 2011 #4


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    You don't even need to decompose vectors, since there are only two masses in this universe and I assume gravity attracts them "toward" each other.
  6. Nov 21, 2011 #5


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    The resulting differential equation d^2x/dt^2 = K/x^2 (where K is some constant) is not trivial to solve though since it's a non-linear ODE. Unless I'm forgetting something basic.

    One can use conservation of energy to get the velocity at any point in the trajectory, but I forget if this greatly simplifies the solution for the full equation of motion.
  7. Nov 21, 2011 #6

    Simon Bridge

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    [tex]\ddot{x} = \frac{K}{x^2}[/tex]

    multiply both sides by [itex]\dot{x}[/itex]
    [tex]\dot{x}\ddot{x} = \frac{K\dot{x}}{x^2}[/tex]

    Now you can integrate both sides, rearrange to get:

    [tex]\dot{x} = \sqrt{ \frac{2K}{x} +C }[/tex]

    which you can do. It's not pretty.

    [tex]t = \frac{\dot{x}}{C} - \frac{K\ln|2K + 2C\dot{x}|}{C^{3/2}} + D [/tex]

    More generally:
  8. Nov 22, 2011 #7
    I meant setting up the problem is straight-forward. Solving the differential equation is not always doable analytically. When not, you an always solve differential equations numerically, either by using per-packaged software or writing your own code.
  9. Nov 22, 2011 #8

    Simon Bridge

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    Oh boy - IRL: differential equations can get very nasty very fast. Real life is messy.

    The nice quadratic ballistic equations we learn in school only work under special conditions ... linear gravity and no other forces.

    Actual accurate satellite orbits are determined numerically and cannot be expressed analytically like this.

    OTOH: the 2-body problem described was worked out quite a while ago.
  10. Nov 24, 2011 #9
    Can't the two body problem be reduced to a similar one body problem? Taking the Lagrangian of the system, assuming the origin to be the COM ([itex]r_1m_1 = -r_2m_2[/itex]) and defining [itex]r = r_1 - r_2[/itex]you have
    [tex] L = \frac{1}{2}m_1\dot{r_1}^2 + \frac{1}{2}m_2\dot{r_2}^2 - U(r)[/tex]
    We can express the position vectors as [itex] r_1 = \frac{m_2r}{m_1+m_2} [/itex] and [itex] r_2 = \frac{m_1r}{m_1+m_2} [/itex], then define [itex]m = \frac{m_1m_2}{m_1+m_2} [/itex] which then reduces the Lagrangian to
    [tex] L = \frac{1}{2}m\dot{r}^2 - U(r) [/tex]
    [tex]U(r) = -\frac{m_1m_2 G}{r} [/tex]

    I guess this would be more of a question than an answer though? :)
    Last edited: Nov 24, 2011
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