Global Positioning

Systems

Mathematics on the Battle elds: GPS 1 2023-09-28 13:22 Mark Lau

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 Introduction
• The Global Positioning System (GPS) is a satellite-based navigation system that
was developed by the U.S. Department of Defense (DoD) in the early 1970s.

• Initially, GPS was developed as a military system to ful l U.S. military needs.

• However, it was later made available to civilians. (We will talk about this in the
last page.)

• GPS consists, nominally, of a constellation of 24 operational satellites. The

satellites are specially arranged to ensure continuous worldwide coverage.

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 • The 24 satellites are arranged such that at least 4 of them is “visible” to any
location on Earth: https://upload.wikimedia.org/wikipedia/commons/9/9c/
ConstellationGPS.gif

• GPS consists of three segments: the space segment, the control segment,
and the user segment.

• If the distances from a point on Earth (a GPS receiver) to three GPS

satellites are known along with the satellite locations, then the location of
the point (or receiver) can be determined by simply applying the well-known
concept of trilateration.

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 • When a GPS receiver (e.g., your phone) is switched on, it will pick
up the GPS signal through the receiver antenna. Once the receiver
acquires the GPS signal, it will process it using its built-in software.
The partial outcome of the signal processing consists of the
distances to the GPS satellites through the digital codes and the
satellite coordinates.

• Theoretically, only three distances to three simultaneously tracked

satellites are needed. In this case, the receiver would be located at
an intersection of three spheres. From the practical point of view,
however, a fourth satellite is needed to account for the receiver
clock o set.

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 Trilateration

Lying in outer space

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 Only 3 are needed, if no time adjustment is needed.

The 4th satellite for time adjustment

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 • The procedure of the GPS range determination, or pseudo-
ranging, can be described as follows.

• Let us assume for a moment that both the satellite and the
receiver clocks, which control the signal generation, are perfectly
synchronized with each other.

• When a “precision” code is transmitted from the satellite, the

receiver generates an exact replica of that code at the same
time. After some time, equivalent to the signal travel time in
space, the transmitted code will be picked up by the receiver.

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 • By comparing the transmitted code and its replica, the receiver can
compute the signal travel time. Multiplying the travel time by the
speed of light (≈299,729,458 m/s) gives the range between the
satellite and the receiver.

• A GPS satellite has an atomic clock on board, which is the most

advanced and accurate clock.

• Unfortunately, the assumption that the receiver and satellite clocks are
synchronized is not exactly true. In fact, the measured range is
contaminated with synchronization error between the satellite and
receiver clocks along with other types of errors. For this reason, this
quantity is referred to as the pseudo-range, not the range.

• A good overview of GPS: https://youtu.be/ozAPGnr-934 (Length: 7:41)

• About atomic clocks: https://youtu.be/dS3MkLamhI8k (Length: 4:00)

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 • Let (x, y, z) be the receiver’s (e.g., your phone) location.

• Let (Ai, Bi, Ci) be is the location of the i-th satellite, which is constantly
tracked by the ground control and therefore is known. Let ti be the time
on the i-th satellite’s clock.

• Let c be the speed of light. Finally, d represent error, which includes the
time di erence between a satellite’s clock and the receiver’s clock.

• Then, we have (each equation for one satellite):

(x − A1)2 + (y − B1)2 + (z − C1)2 = c(t1 − d)

(x − A2)2 + (y − B2)2 + (z − C2)2 = c(t2 − d) (Ai, Bi, Ci)

.
(x − A3)2 + (y − B3)2 + (z − C3)2 = c(t3 − d)
(x, y, z)
(x − A4)2 + (y − B4)2 + (z − C4)2 = c(t4 − d)
Mathematics on the Battle elds: GPS 9 2023-09-28 13:22 Mark Lau
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 • The equations can be solved to nd out the receiver’s location
(x, y, z) and the timing error d.

• A GPS measures not only position but also altitude (it’s called
geometric altitude), since it gives a point in the 3D space.

• The quantity d can be used to calibrate the clock on your phone

to match the atomic clock on a satellite.

• Solving these equations is not an easy task by a hand-and-paper

method. It requires advanced algorithms and is performed by the
GPS chip on your phone.

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 Time Dilation
• To calculate a receiver's location, the receiver needs to know the
time required for radio wave to travel from a satellite to the receiver.

• Suppose the signal is sent from the satellite at time ts and is

received by the receiver at time tr. Then, the time required is tr − ts.

• However, the clocks on the satellite and the receiver are not
perfectly synchronised, because

1. Receiver’s clock is less advanced and hence less accurate

than that on a satellite.

2. The physical phenomenon of time dilation.

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 • Time dilation was proposed by Albert Einstein
in the theories of special and general relativity.

• It happens in two situations:

• (Velocity time dilation) Suppose that an Clock 2

observer is at rest on Earth. A clock that is
moving relative to the observer will be
measured to tick slower than the
observer’s clock.

• (Gravitational time dilation) A clock subject Clock 1

to greater gravitation pull (say, near a

massive planet) will be measured to tick
slower than the observer’s clock.

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 • In an ending scene of Interstellar, Cooper is younger than his daughter
after visiting the interstellar space and a black hole. It is because his
daughter was on Earth but

• he was on a really fast spaceship (velocity time dilation);

• the black hole had huge gravitation pull acting on him (gravity time
dilation).
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 • Velocity time dilation can be illustrated (not a formal mathematical
proof) by the following imaginary clock:

• The clock is formed by two mirrors. A light beam hits the top
mirror, which is then re ected to the bottom mirror, and so on.
Every time the beam hits the bottom mirror, the clock ticks.

• The clock on the left is at rest, while the clock on the right is
moving.

• The beam on the right clock takes a longer path. If the speed of
light is a constant, then the time required is longer.

To see the animation, visit here:

https://web.pa.msu.edu/courses/2000fall/phy232/lectures/relativity/dilation.html

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 • Let ts be the time measured by a stationary observer and t
be the time measured by another clock moving at a
velocity v relative to this observer. Let c be the speed of
light. Then,

Stationary clock t Moving clock

ticks faster ts = ≥ t. ticks slower
v2
1−
c2

• This is explained in the following two video clips:

• https://youtu.be/mU04-vJB6gc

• https://youtu.be/l_QoEUmwVPU

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 -4 m/s 4 m/s

-6 m/s 6 m/s

• In the above equation, v is the “relative velocity”.

• For example, if an observer is moving at 4 m/s, and another

object is moving at 10 m/s in the same direction, then the
relative velocity v = (10 − 4) m/s = 6 m/s. (Or,
v = ((−10) − (−4)) = − 6 m/s, depending on which
direction you call it positive.)

• Hence, the velocity time dilation equation still holds, even

when the observer is not absolutely stationary. It is because
v in the time dilation formula is the relative velocity
between the two observers. (That’s why Einstein’s work is
called the theories of relativity.)

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 Example
• A GPS satellite moves at 10,000 m/s relative to a receiver on Earth. How much slower does a clock
on the satellite ticks due to velocity time dilation? (The orbital speed of a typical GPS satellite is
about 14,000 km/hr ≈ 3,889 m/s.)

• Let t be the time recorded on the satellite, and te be the time recorded on Earth. The time di erence:

v2 v2 v2
t = te 1 − 2 = te 1 − 1 + 1− 2 = te 1 − 1 − 1− 2
c c c

( 2.998 × 108 )
3
10 × 10
= te 1 − 1 − 1−

≈ te (1 − 5.56 × 10−10) .

• For every second on recorded on Earth, the satellite’s clock is

5.56 × 10−10 = 5.56 × 10−1 × 10−9 = 0.556 nanoseconds slower.

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 Example
• What will be the error in estimating the distance from the
receiver to the satellite, if the time dilation considered in the
previous page is not adjusted?

• We saw that, for every second on the receiver, the satellite’s

clock is Δt = 5.56 × 10−10 seconds slower. This amounts to
the distance error of

cΔt = (2.997295 × 108) × (5.56 × 10−10)

≈ 0.167 meters .

• In one day, the error is around

0.167 × 24 × 60 × 60 = 14,428.8 meters = 14.4 kilometers.
Mathematics on the Battle elds: GPS 18 2023-09-28 13:22 Mark Lau
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 Example
Consider a spaceship traveling from Earth to the nearest star system: a distance of 4 light
years away, at a speed v = 0.8c (i.e., 80 percent of the speed of light). To make the numbers
easy, amounts of time needed to accelerate, decelerate, and turn around are negligible. Crew
members on Earth have waited for 10 years before the spaceship returns to Earth. When it
returns to Earth, how much younger will the crew on the spaceship be than the crew on Earth?

• Let te be time recorded on Earth. Then, time recorded on spaceship is

v2
t = te 1 − 2 = te 1 − 0.82 = 0.6te.
c

• A round trip has the distance 2d. Hence, in Earth time, the trip takes

2d 2d 2 × 4 yrs × c
= = = 10 years.
v 0.8c 0.8c

• Let te = 10 years. Then, ts = 6 years. Hence, the crew will be 4 years younger.

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 • To understand gravitational time dilation,
let’s consider the following thought
experiment.

• Here, we have the same mirror clock

considered before.

• The observer is now stationary on a

mountain-top on Earth.

• A copy of the same mirror clock is falling

from the sky.

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 Move slower
• The gravitational acceleration is greater when an
object is closer to the core of the Earth.

• Due to a greater gravitational acceleration, the

clock is moving faster as it is closer to the ground.

• When the clock is moving faster, the path of the

light beam is longer and therefore the clock is
ticking more slowly according to the observer.

Move faster

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 • https://youtu.be/yuD34tEpRFw (starting from 6:04, the whole video is worth
watching though.)

• Let t0 be the time measured by an observer having a distance of r units from a

mass M.

• Let tf be the time measured by another observer extremely far away from any
mass (i.e. zero gravitational pull).

• Let G and c be the gravitational constant (6.67 × 10-11 m3kg-1s-2) and speed of
light. (G is a universal constant, having the same value anywhere in the universe.)

• Then, the time measured at the distance r from the mass M is given by

2GM
Clock ticks slower t0 = tf 1− ≤ tf Clock ticks faster
rc 2

Observer t0
tf
r
Far from mass M.
Zero gravitation
M
potential
Mathematics on the Battle elds: GPS 22 2023-09-28 13:22 Mark Lau
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 Example
• A spaceship is initially extremely far away from a blackhole such that its gravitational
pull on the spaceship can be ignored. Now, the spaceship is approaching the
blackhole. How close to this blackhole must the spaceship get in order to
experience time twice as slowly? That means, what is the distance r between the
spaceship and the blockhole? This blackhole weighs 1.3127 × 1040 kilograms.

• Let t0 (tf ) be the time recorded on the spaceship when it is close to (far from) the
black hole. Then, we set
r
t0 2GM
= 0.5 = 1−
tf rc 2
2GM Blackhole
r=
0.75c 2
2 6.673 × 10−11 × 1.3127 × 1040 13
= × ≈ 2.599 × 10 meters .
0.75 299,792,4582
Mathematics on the Battle elds: GPS 23 2023-09-28 13:22 Mark Lau
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 Time Dilation & GPS
• Here are some data:

• Velocity of a typical GPS satellite: v = 3,874 m/s

• Speed of light: c = 2.998 × 108 m/s

• Mass of the Earth: M = 5.974 × 1024 kg

• Radius of a typical GPS satellite’s orbit: rs = 26,541,000 m

• Radius of the Earth: re = 6,357,000 m.

• Gravitational constant: G = 6.674 × 10−11m3kg−1s−2.

Mathematics on the Battle elds: GPS 24 2023-09-28 13:22 Mark Lau
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 • Let ts be the time recorded by a
clock on a GPS satellite moving in
orbit.

• Let tss be the time recorded by a

stationary GPS satellite in the orbit.

• Time dilation of the moving satellite Stationary

due to velocity only:
v
Clock ticking
2
slower v
ts = tss 1 − 2 ≤ tss. Clock ticking ts tss
c faster

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 Let tr be the time recorded by a receiver on the surface of the Far away. Zero gravity.
•
Earth.

• Let tf be a GPS satellite extremely far away from Earth.

tf
• Let tss be the time measured on a stationary GPS satellite in
Earth’s orbit.

• Let re and rs be the radii of the Earth and the satellite’s circular
orbit. Then, use the gravity time dilation equation:
Stationary
2GM 2GM
tr = tf 1− , tss = tf 1−
rec 2 rsc 2
Receiver

• Hence, time dilation between a stationary satellite and a receiver tr tss

on Earth, assuming re < rs: rs
re
2GM
1− Earth
rsc 2
tss = tr ≥ tr. G, M
2GM
1−
rec 2

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 • Recall equations for velocity and gravity time dilation:

2GM
1−
v2 rsc 2
ts = tss 1 − 2 , tss = tr .
c 2GM
1−
rec 2

• We put them together to obtain:

2GM 2GM
1− 1−
v2 rsc 2 v2 rsc 2
ts = tr 1− 2 × ⇒ ts = tr 1 + 1− 2 × − 1 ⇒ ts = tr(1 + α) .
c 2GM c 2GM
1− 1−
rec 2 rec 2

• Now, we calculate: Faster clock ts = (1 + 4.4718 × 10−10)tr ≥ tr Slower clock.

• That mean, for every second recorded by the receiver on Earth, the satellite is
4.4718 × 10−10 = 0.44718 × 10−9 = 0.44718 nanoseconds faster.

• In one day, the satellite is 24 × 60 × 60 × 0.4718 ≈ 40,763 nano seconds faster.

• In one year, it is about 0.01467 seconds faster.

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 Civilian Use of GPS
• On Sept 1, 1983, Korean Airlines Flight 007 (KAL007) bound for
Seoul from New York City was shot down by a Soviet Union ghter
jet by an air-to-air missile.

• The plane deviated from its original route and ew through Soviet
prohibited airspace.

• The Soviet Union once claimed that they believed it was a spy
plane.

• All 269 passengers and crew members died.

• To prevent similar incidents in the future, in 1983, President Ronald

Reagan signed an executive order allowing civilian use of the GPS.

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 • However, selective availability was enabled by U.S.
government at that time, adding 50 meters of error
horizontally and 100 meters of error vertically to GPS
signals.

• Authorized groups like the U.S. military and allies could

access a second GPS signal for better accuracy.

• Selective availability was abandoned in the year 2000.

Nowadays, horizontal position accuracy is usually well under
10 meters and vertical accuracy of 10 to 20 meters is
achievable.

• More details: https://youtu.be/W3t_RDAFSsM (Length: 3:57)

— End —

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