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

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

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)

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)

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

Newton’s Universal Law of Gravitation Force between two masses, M and m: F =GMm r2 e r For Earth GM =  = 3.986 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)