- #1
TheMisterOdd
- 1
- 0
- Homework Statement
- How to derive the period of non-circular orbits?
- Relevant Equations
- $$
T = \sqrt{\frac{4 r_0^3}{\tilde m}} \int_{1}^{r_{min}/r_0}\frac{\rho}{\sqrt{\left ( 1 - \rho \right) \left (\rho - \frac{\ell^2}{\tilde mr_0} \left (1 +\rho \right) \right )} } \; d\rho
$$
By conservation of mechanical energy:
$$
E(r_0)=-\frac{GMm}{r_0}+\frac{1}{2}\mu \left (\dot{r_0}^2+r_0^2 \omega_0^2 \right)
$$
where R0 =Rmax. Because our body is located at the apoapsis the radial velocity is 0. Hence:
$$
E(r_0)=-\frac{GMm}{r_0}+\frac{1}{2}\mu (r_0\omega_0)^2
$$
By the conservation of angular momentum we have:
$$
\omega_0 =\frac{L}{\mu r^2}
$$
$$
E(r_0)=-\frac{GMm}{r_0}+\frac{L^2}{2 \mu r_0^2}
$$
Then, we can relate the radius r at any given time, to the initial energy:
$$
-\frac{GMm}{r} + \frac{1}{2}\mu \left (\frac{dr}{dt} \right)^2 + \frac{L^2}{2 \mu r^2}= -\frac{GMm}{r_0}+\frac{L^2}{2 \mu r_0^2}
$$
$$
\left (\frac{dr}{dt} \right)^2 = \frac{2GMm}{\mu} \left ( \frac{1}{r} - \frac{1}{r_0} \right) + \frac{L^2}{\mu^2} \left ( \frac{1}{r_0^2} - \frac{1}{r^2} \right)
$$
Let
$$\tilde m = \frac{2GMm}{\mu}, \; \ell^2 = \frac{L^2}{\mu^2}$$
$$
\left (\frac{dr}{dt} \right)^2 = \frac{\tilde m}{r} \left ( 1 - \frac{r}{r_0} \right) - \frac{\ell^2}{r^2} \left (1 - \frac{r^2}{r_0^2} \right)
$$
Rearranging a little bit:
$$
\frac{1}{\sqrt{\tilde m}} \int_{r_0}^{r_{min}}\frac{\sqrt{r} \; dr}{\sqrt{\left ( 1 - \frac{r}{r_0} \right) \left (1 - \frac{\ell^2}{\tilde mr} \left (1 + \frac{r}{r_0} \right) \right )}} = \int_0^{T/2} dt
$$
In the radial integral, the limits of integration go from, R0 to Rmin, covering half of the elliptic path of the orbit. This would take T/2 seconds. Because of this, our limits of integration, for t, are 0 and T/2.
Let's define $$\rho = r/r_0$$
$$
dr = r_0 \; d \rho
$$
We finally get:
$$
T = \sqrt{\frac{4 r_0^3}{\tilde m}} \int_{1}^{r_{min}/r_0}\frac{\rho}{\sqrt{\left ( 1 - \rho \right) \left (\rho - \frac{\ell^2}{\tilde mr_0} \left (1 +\rho \right) \right )} } \; d\rho
$$
Solving this integral yields a formula for the period of a body orbiting a star, planet or other celestial body. Is this line of reasoning right? Is this integral even possible to solve? I would appreciate any hint.
$$
E(r_0)=-\frac{GMm}{r_0}+\frac{1}{2}\mu \left (\dot{r_0}^2+r_0^2 \omega_0^2 \right)
$$
where R0 =Rmax. Because our body is located at the apoapsis the radial velocity is 0. Hence:
$$
E(r_0)=-\frac{GMm}{r_0}+\frac{1}{2}\mu (r_0\omega_0)^2
$$
By the conservation of angular momentum we have:
$$
\omega_0 =\frac{L}{\mu r^2}
$$
$$
E(r_0)=-\frac{GMm}{r_0}+\frac{L^2}{2 \mu r_0^2}
$$
Then, we can relate the radius r at any given time, to the initial energy:
$$
-\frac{GMm}{r} + \frac{1}{2}\mu \left (\frac{dr}{dt} \right)^2 + \frac{L^2}{2 \mu r^2}= -\frac{GMm}{r_0}+\frac{L^2}{2 \mu r_0^2}
$$
$$
\left (\frac{dr}{dt} \right)^2 = \frac{2GMm}{\mu} \left ( \frac{1}{r} - \frac{1}{r_0} \right) + \frac{L^2}{\mu^2} \left ( \frac{1}{r_0^2} - \frac{1}{r^2} \right)
$$
Let
$$\tilde m = \frac{2GMm}{\mu}, \; \ell^2 = \frac{L^2}{\mu^2}$$
$$
\left (\frac{dr}{dt} \right)^2 = \frac{\tilde m}{r} \left ( 1 - \frac{r}{r_0} \right) - \frac{\ell^2}{r^2} \left (1 - \frac{r^2}{r_0^2} \right)
$$
Rearranging a little bit:
$$
\frac{1}{\sqrt{\tilde m}} \int_{r_0}^{r_{min}}\frac{\sqrt{r} \; dr}{\sqrt{\left ( 1 - \frac{r}{r_0} \right) \left (1 - \frac{\ell^2}{\tilde mr} \left (1 + \frac{r}{r_0} \right) \right )}} = \int_0^{T/2} dt
$$
In the radial integral, the limits of integration go from, R0 to Rmin, covering half of the elliptic path of the orbit. This would take T/2 seconds. Because of this, our limits of integration, for t, are 0 and T/2.
Let's define $$\rho = r/r_0$$
$$
dr = r_0 \; d \rho
$$
We finally get:
$$
T = \sqrt{\frac{4 r_0^3}{\tilde m}} \int_{1}^{r_{min}/r_0}\frac{\rho}{\sqrt{\left ( 1 - \rho \right) \left (\rho - \frac{\ell^2}{\tilde mr_0} \left (1 +\rho \right) \right )} } \; d\rho
$$
Solving this integral yields a formula for the period of a body orbiting a star, planet or other celestial body. Is this line of reasoning right? Is this integral even possible to solve? I would appreciate any hint.
