A Miniature Solar System

I can calculate the motion of heavenly bodies, but not the madness of people.Isaac Newton, after a market collapse in 17^th century England

It should be apparent that our Solar system is not a two-body system since the planets themselves exert gravitational forces on one another. Although the planetary forces are small in magnitude compared to the gravitational force of the Sun, they can still produce measureable effects. For example, the existence of the planets Neptune_resize Neptune and Pluto_resize Pluto was conjectured on the basis of a discrepancy between the experimentally measured orbit of Uranus_resize Uranus and its predicted orbit that was calculated from known gravitational forces.

For simplicity, the model considered here is planar and not three-dimensional, and it is also assumed the force interaction is not strictly central since this is clearly not the case. The "AU" is an astronomical unit, which is roughly equal to 93 million miles.

Planetary Perturbations

This program simulates a miniature Solar system of two planets about a fixed Sun-like star.

$\displaystyle{ \frac{\mathrm{d}^2 \mathbf{r}_1}{\mathrm{d} t^2} = -\frac{G M}{r_1^3} \mathbf{r}_1 + \frac{G m_2}{r_{21}^3} \mathbf{r}_{21} }$ $\displaystyle{ \frac{\mathrm{d}^2 \mathbf{r}_2}{\mathrm{d} t^2} = -\frac{G M}{r_2^3} \mathbf{r}_2 + \frac{G m_1}{r_{21}^3} \mathbf{r}_{21} }$

Earth $x$-position: AU1 ≤ $x$ ≤ 5 AU ormi

Earth $y$-velocity: AU/yr1 ≤ $v_y$ ≤ 5 AU/yr ormi/hr

Pluto $x$-position: AU5 ≤ $x$ ≤ 10 AU ormi

Pluto $y$-velocity: AU/yr1 ≤ $v_y$ ≤ 5 AU/yr ormi/hr

Simulation time: yr2 ≤ $t$ ≤ 1000 yr orhr

Notes

gfortran -g -std=f2008 -Wall -Wextra -O2 -o planet2 planet2.f08

The position vector $\mathbf{r}_1$ represents Earth and $\mathbf{r}_2$ represents Pluto. The position of the Sun is fixed at the origin. The variables ratio1 = $+m_2 G$ and ratio2 = $-m_1 G$ for the interaction masses of Pluto and Earth, respectively. To further simplify the code I've adopted these ratios to be in terms of Solar mass units, or ratio1 = $+(m_2/M) G M$ and ratio2 = $-(m_1/M) G M$.

Period of a Satellite in Low-Earth Orbit

Say we have a satellite with the attributes of mass $m$, tangential velocity $v$, altitude $r$, and period of rotation $T$. This satellite happens to be orbiting about a planet with mass $M$, volume $V$, mean radius $R$, and mass density $\rho$. Let's compute the period of the satellite for the case of low Earth orbit , or $r \simeq R$.

A schematic of a satellite in low Earth orbit. As shown here, orbital motion is the result of a continual deflection from a tangential path by the action of a force directed toward the center.

Start with the net force (gravity) on the satellite by the planet:

$\displaystyle{ \begin{align*} F_\text{g} & = G \frac{m M}{R^2} = G \frac{m}{R^2} \left( \frac{4}{3} \pi R^3 \rho \right) \\ & = \frac{4 \pi R G m \rho}{3} \text{,} \end{align*} }$

(N.

where $G$ is the gravitational constant. The satellite experiences a centralized (centripetal) force that acts to convert the straight-line motion of Newton's first law into a circular orbit with circumference $2 \pi R$:

$\displaystyle{ F_\text{c} = m \frac{v^2}{R} = \frac{m}{R} \left( \frac{2 \pi R}{T} \right)^2 = \frac{4 \pi^2 R m }{T^2} \text{.} }$

(N.

Newton's third law allows us to set Equations (N.1) and (N.2) equal to each other. (Note that invoking Newton's second law is also valid.) After a lot of cleaning-up, we get a prediction for the orbital period of the satellite:

$\displaystyle{ \boxed{ T = \sqrt{ \frac{3 \pi}{G \rho} } \text{.} } }$

(N.

Equation (N.3) shows that the orbital period of a satellite in a low, circular orbit is independent of the planet's radius but is inversely proportional to the square root of the planet's density. The reader should recognize that Equation (N.3) is a rewritten form of Kepler's third law.