Numerical Methods

Outline

Introduction

Linear
Equations

Interpolation

Numerical
Solutions

Computer
Measurement

      

Satellite orbits

A satellite orbits the earth. The graviational force on the satellite is,

$\large \vec{F} = -\frac{Gm_e m_s}{r^2} \hat{r}$,

where $G= 6.6726 \times 10^{-11}$ N m²/kg² is the graviational constant, $m_e = 5.97219 \times 10^{24}$ kg is the mass of the earth, $m_s$ is the mass of the satellite, and $\vec{r}$ is the position of the satellite measured from the center of the earth.

As long as the earth is much heavier than the satellite, the mass of the satellite does not matter. The satellite mass does not appear in the expression for the acceleration. Only the intitial position and velocity of the satellite determines its orbit.

The initial conditions at $t=0$ are:
$x=$  m  $y=$  m  $z=$  m  $v_x=$  m/s  $v_y=$  m/s  $v_z=$  m/s

If the orbit falls below about 6400000 m, the satellite will crash into the earth. There are various kinds of orbits such as geosynchronous orbits, geostationary orbits, low earth orbits, elliptical orbits, and graveyard orbits. The difference just depends on the initial conditions of the satellite. A long time step should be used to calculate satelite orbits.

 Numerical 6th order differential equation solver 

$ \large \frac{dx}{dt}=$

$v_x$

$ \large \frac{dv_x}{dt}=$

$ \large \frac{dy}{dt}=$

$v_y$

$ \large \frac{dv_y}{dt}=$

$ \large \frac{dz}{dt}=$

$v_z$

$ \large \frac{dv_z}{dt}=$

Initial conditions: 

$t_0=$

 

$\Delta t=$

$x(t_0)=$

 

$N_{steps}$

$v_x(t_0)=$

 

Plot:

vs.

$y(t_0)=$

 

$v_y(t_0)=$

 

$z(t_0)=$

 

$v_z(t_0)=$

 

 

 $t$   $x$   $v_x$   $y$   $v_y$   $z$   $v_z$