Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Featured I Radial movement in a gravitational field

  1. Oct 1, 2017 #1


    User Avatar
    Science Advisor

    Dear all,
    to keep me busy on a Sunday I considered the "1-body radial movement in a (Newtonian) gravitational field problem". I was a bit surprised to find it quite hard finding decent explanations on it. My question is: does anyone have a reference of the explicit solution to the particle's position ##r(t)##? Let me show my calculation:

    We consider a particle with mass m being attracted by a mass M >> m in a gravitational field. We let it go at ##t=0## from a distance ##r(t)=r_b##. Newton's second law states

    [tex]m\ddot{r} = - \frac{GMm}{r^2}[/tex]

    We can integrate this equation to find the total energy,

    [tex]E = \frac{m}{2}\dot{r}^2 - \frac{GMm}{r} \ \ \ \ \ (1) [/tex]

    which is conserved on-shell (by Newton's second law),

    [tex] \dot{E} = (m\ddot{r} + \frac{GMm}{r^2}) \dot{r} = 0[/tex]

    I want to solve for the particle's position ##r(t)## using the energy. For that I rewrite eqn.(1) as

    [tex]\dot{r} = - \frac{\sqrt{2Er + 2GMm}}{\sqrt{mr}} [/tex]

    choosing the minus-sign because the particle's position ##r(t)## decreases. Now we separate variables:

    [tex]\int_{r_b}^{r} \frac{\sqrt{r}}{\sqrt{Er+GMm}}dr = -\sqrt{\frac{2}{m}} \int_0^t dt[/tex]

    The integral can be solved by using the "standard" primitive

    [tex]\int \frac{\sqrt{x}}{\sqrt{x+a}}dx = \sqrt{x^2 + ax} - a \ln{|\sqrt{x} + \sqrt{x+a} |} + C [/tex]

    with ##a## constant and C the integration constant:

    [tex]\Bigl[\sqrt{r^2+ar} - a \ln{|\sqrt{r}+\sqrt{r+a}|} \Bigr] - \Bigl[\sqrt{r^2_b+a r_b} - a \ln{|\sqrt{r_b}+\sqrt{r_b+a}|} \Bigr] = - \sqrt{\frac{2E}{m}} t [/tex]

    where I defined

    [tex]a \equiv \frac{GMm}{E} [/tex]

    This gives us ##t(r)##.

    I must say that, having not done this kind of stuff for a while, I'm a bit surprised that simple radial movement in the 1-body approximation already gives a nasty integral like this. My question is basically this: can I invert my relation to obtain ##r(t)##? Between ##r(t)=r_b## and ##r(t)=0## the function ##r(t)## should be invertible, right? And am I right to be surprised that the result is quite complicated, or did I make a mistake?Thanks in advance!
  2. jcsd
  3. Oct 5, 2017 #2


    Staff: Mentor

    While I can't comment much on your work, the problem did remind me of one we did in Classical Physics.

    We needed to show that a particle initially at rest and then falling from a great height to the Earth took 9/11 of the time to fall the first half of the journey. On the surface, it looks straightforward but then you must relate distance fallen to time fallen and then things get nasty.

    After falling to solve it a prof gave us a hint to use Kepler's laws to simplify it namely the equal areas in equal times law and to consider a very narrow orbit and then the problem fell apart.
  4. Oct 6, 2017 #3
  5. May 18, 2018 #4
    You did all the calculus correctly, but I agree it isn't very satisfying, not being able to invert it to get r as a function of t. However, if you assume that the object falls from an infinite distance and from rest, E = 0. This simplifies the equations considerably leading to the result:

    r(t) = ( rb3/2 - 3(√(GM/2)) t )2/3 .

    I know you posted this several months ago, but I am a physics professor and I thought I'd give you a way to get r(t) if only for a specific case. - Bob
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?