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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
m\ddot{r} = - \frac{GMm}{r^2}
We can integrate this equation to find the total energy,
E = \frac{m}{2}\dot{r}^2 - \frac{GMm}{r} \ \ \ \ \ (1)
which is conserved on-shell (by Newton's second law),
\dot{E} = (m\ddot{r} + \frac{GMm}{r^2}) \dot{r} = 0
I want to solve for the particle's position ##r(t)## using the energy. For that I rewrite eqn.(1) as
\dot{r} = - \frac{\sqrt{2Er + 2GMm}}{\sqrt{mr}}
choosing the minus-sign because the particle's position ##r(t)## decreases. Now we separate variables:
\int_{r_b}^{r} \frac{\sqrt{r}}{\sqrt{Er+GMm}}dr = -\sqrt{\frac{2}{m}} \int_0^t dt
The integral can be solved by using the "standard" primitive
\int \frac{\sqrt{x}}{\sqrt{x+a}}dx = \sqrt{x^2 + ax} - a \ln{|\sqrt{x} + \sqrt{x+a} |} + C
with ##a## constant and C the integration constant:
\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
where I defined
a \equiv \frac{GMm}{E}
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!
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
m\ddot{r} = - \frac{GMm}{r^2}
We can integrate this equation to find the total energy,
E = \frac{m}{2}\dot{r}^2 - \frac{GMm}{r} \ \ \ \ \ (1)
which is conserved on-shell (by Newton's second law),
\dot{E} = (m\ddot{r} + \frac{GMm}{r^2}) \dot{r} = 0
I want to solve for the particle's position ##r(t)## using the energy. For that I rewrite eqn.(1) as
\dot{r} = - \frac{\sqrt{2Er + 2GMm}}{\sqrt{mr}}
choosing the minus-sign because the particle's position ##r(t)## decreases. Now we separate variables:
\int_{r_b}^{r} \frac{\sqrt{r}}{\sqrt{Er+GMm}}dr = -\sqrt{\frac{2}{m}} \int_0^t dt
The integral can be solved by using the "standard" primitive
\int \frac{\sqrt{x}}{\sqrt{x+a}}dx = \sqrt{x^2 + ax} - a \ln{|\sqrt{x} + \sqrt{x+a} |} + C
with ##a## constant and C the integration constant:
\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
where I defined
a \equiv \frac{GMm}{E}
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!