# How to derive the period of non-circular orbits?

• TheMisterOdd
TheMisterOdd
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.

Have you tried to (re)search what the orbital period of an elliptical orbit is?

TheMisterOdd said:
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.
I didn't see anything wrong with what you did, but there is a simpler approach based on Kepler's second law. The body sweeps out area at a constant rate, so the period is the area of the ellipse divided by the rate.

PeroK
From the Kepler equation you get [EDIT: corrected in view of #5]
$$T^2=\frac{4 \pi^2 a^3}{GM},$$
where ##M=M_{\text{sun}}+M_{\text{planet}}##, which for our solar system is approximately ##M_{\text{sun}}##, because even Jupiter has a mass much smaller than that of the Sun (about 1/1000 of the mass of the Sun).

Last edited:
vanhees71 said:
From the Kepler equation
The ##4\pi## should be ##4\pi^2##.

vanhees71
Thanks. I've corrected the typo.

Filip Larsen

## How do you start deriving the period of a non-circular orbit?

To start deriving the period of a non-circular orbit, you begin with Kepler's laws of planetary motion, particularly the third law, which relates the orbital period to the semi-major axis of the ellipse. You also need to understand the geometry of the ellipse and the gravitational forces acting on the orbiting body.

## What role does the semi-major axis play in determining the orbital period?

The semi-major axis is crucial because Kepler's third law states that the square of the orbital period is proportional to the cube of the semi-major axis. This relationship can be expressed as $$T^2 \propto a^3$$, where $$T$$ is the orbital period and $$a$$ is the semi-major axis.

## How do you incorporate eccentricity into the period calculation?

Eccentricity does not directly affect the period of the orbit; the period depends solely on the semi-major axis for elliptical orbits. However, eccentricity affects the shape of the orbit and the speed at which the orbiting body travels along different sections of the ellipse.

## What is the role of the gravitational constant in the derivation?

The gravitational constant $$G$$ is essential in the derivation because it appears in the formula that relates the period to the semi-major axis. The formula $$T = 2\pi \sqrt{\frac{a^3}{GM}}$$ shows that the period $$T$$ depends on the gravitational constant $$G$$ and the mass $$M$$ of the central body.

## Can you derive the period using energy considerations?

Yes, you can derive the period using energy considerations. By equating the total orbital energy (kinetic plus potential) and knowing that the average distance over one period is the semi-major axis, you can derive the same relationship for the period. The total energy of an orbiting body is $$E = -\frac{GMm}{2a}$$, and using this, you can arrive at the same result for the orbital period.

Replies
26
Views
3K
Replies
17
Views
2K
Replies
4
Views
1K
Replies
16
Views
722
Replies
1
Views
909
Replies
0
Views
703
Replies
1
Views
1K
Replies
1
Views
2K
Replies
10
Views
1K
Replies
11
Views
2K