1. SVY 207: Lecture 1 Introduction to GPS Theory and Practice
Aim:
Provide you with basic information on this module, including administrative details and scientific background

2. Overview The Module (see handout) Scientific Background Aims
Administrative details
Scientific Background
GPS Positioning: The basic idea
Some basic Physics you’ll need to know

3. The Basic Idea GPS point positioning is similar to trilateration
3 ranges to 3 known points
GPS point positioning
receiver records 4 “pseudoranges” from 4 satellites
estimate 4 parameters:
your receiver’s position (xr, yr, zr) and
your receiver’s clock error (tr)
Questions we need answers to:
How do we know the satellite positions (xs, ys, zs) ?
What are pseudoranges, and how do we model them? r2 r1 r3 P3 P4 P2 P1
(xr, yr, zr, tr)

4. The Basic Idea How do we know position of satellites? Therefore:
Receiver can read the Navigation Message
transmitted by the satellite, encoded on the signal
includes orbit parameters (broadcast ephemeris)
Receiver can then compute satellite coordinates (xs, ys, zs)
using the standard GPS Ephemeris Algorithm
Therefore:
We need to understand these orbit parameters
But first we need to understand satellite orbits
which requires some basic Physics
Newton’s Laws of motion
Newton’s Law of gravitation
Keplers Law’s of orbits (Lecture 2)

5. Newton’s Laws of Motion
Newton I
In the absence of external forces, an object continues to move at constant velocity
Newton II
There exists a reference frame (“inertial frame”) for which: The rate of change of momentum of an object is proportional to the applied force
momentum = mass  velocity p = mv
if mass = constant, F = dp/dt = d(mv)/dt = m(dv/dt) = ma
if F = 0 on an object, a = 0 , v = constant (Newton I)
Newton III
For every force, there is an equal and opposite reactive force
If force is central, the reactive force is also central
can show this implies conservation of angular momentum L = r  p = r  mv
If mass, m, is constant, therefore r  v is constant

6. Newton’s Universal Law of Gravitation
Force between two masses, M and m: F =GMm r2 e r
For Earth GM =  = x 105 km3 s-2 r = distance to geocentre
Force on a satellite of mass m is therefore
F =m r2 e r
Then Newton II gives the equation of motion for a satellite F =ma =m /r2 e r a = /r2 e r
Satellite mass cancels, and is therefore irrelevant (Galileo)