﻿ Satellite Orbits - College Physics

# Satellite Orbits

## Introduction

Many satellites move in stable orbits around the earth, some have left the gravitational field of the Earth and move around other planets and a couple are on the way to leave the solar system. An important task of astronomy is therefore to determine the orbits of these satellites.

### Trajectories

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In the calculation of satellite orbits two speeds $$v_1, v_2$$ play a major role. They depend on the starting altitude of the satellite (see below for derivation).

The first velocity $$v_1$$ is the launching speed, which ensures a stable circular orbit. If the satellite is launched at a higher speed its orbit is deformed into an ellipse. If a satellite, however, is launched with a smaller velocity it will probably fall to the ground and explode.

The second velocity $$v_2$$ is the launching speed, which empowers a satellite to escape the gravitational field of the Earth on a parabolic path. If a satellite is launched with even higher speed it will leave the gravitational field on a hyperbolic path.

## Orbit simulation

The following simulation calculates the trajectories of satellites at a given starting altitude and velocity.

Launch parameter
Altitude
$$\rm km$$
Launch velocity
$$\rm \frac{m}{s}$$
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ResetStart

Orbit
Semi-major axis
$$\approx$$ $$\rm km$$
Num. eccentricity

## Stable orbit

### Circular orbit

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A missile theoretically requires at least this speed to remain in an orbit around a celestial body, without falling onto the ground. It then moves on a circular orbit around the celestial body.

$$v = \sqrt{G \, \dfrac{M}{R+h}}$$ $$M$$ = Mass of the celestial body $$R$$ = Radius of the celestial body,
$$G$$ = Gravitational constant, $$h$$ = Altitude of the satellite over the ground

The gravitational force of the earth in this case acts as a centripetal force, which forces the missile onto a circular path.

\begin{aligned} F_\rm{Z} &= F_\rm{G} \\ \\ \dfrac{\cancel m \cdot (v_1)^2}{\cancel{R+h}} &= G \, \dfrac{\cancel m \cdot M}{(R+h)^{\cancel 2}} \\ \\ (v_1)^2 &= G \, \dfrac{M}{R+h} \\ \\ v_1 &= \sqrt{G \, \dfrac{M}{R+h}} \\ \\ \end{aligned}

### Ellipse

If a satellite is launched at a higher speed its orbit deforms into an ellipse.

## Escaping the gravitational field

A missile theoretically requires at least the second cosmic velocity
to escape the gravitational field of the Earth.

$$v_2 = \sqrt{2 \, G \, \dfrac{M}{R+h}}$$ $$M$$ = Mass of the celestial body $$R$$ = Radius of the celestial body,
$$G$$ = Gravitational constant, $$h$$ = Altitude of the satellite over the ground

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In theory to escape the gravitational field of the Earth, a missile has to move infinitely far from the Earth. The required energy can be calculated:

\begin{aligned} \Delta W &= G \cdot M \cdot m_\rm{R} \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right) \\ \\ &= G \cdot M \cdot m_\rm{R} \cdot \left( \dfrac{1}{R+h} - \dfrac{1}{\infty} \right) \\ \\ &= G \cdot M \cdot m_\rm{R} \cdot \dfrac{1}{R+h} \\ \end{aligned}

This work must be available in the kinetic energy of the missile.

\begin{aligned} E_\rm{kin} &= \Delta W \\ \\ \dfrac{\cancel m_\rm{R}}{2} \cdot (v_2)^2 &= G \cdot M \cdot \cancel m_\rm{R} \cdot \dfrac{1}{R+h} \\ \\ (v_2)^2 &= 2 \cdot G \cdot M \cdot \dfrac{1}{R+h} \\ \\ v_2 &= \sqrt{ 2 \,\, G \dfrac{M}{R+h} } \\ \\ \end{aligned}

### Sources

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