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SPACE
IN MATHS
TECH…
All about space
TIME
EVENT
0
Big Bang
100 seconds
1st atoms
3 minutes
1st helium atoms
1 million years
1st hydrogen atoms
1 billion years
Galaxies start to form
4.5 billion years
Stars start to form
Astronomers use a scale of ‘-1.5 to 2’ for measuring the brightness of stars.
Below is a graph showing some stars near to Earth:
Brightness of stars
Brightness (scale: -1.5 to 2)
Be
te
lg
eu
se
1
0.5
-1.5
-2
Moons
Si
riu
s
ar
l
ig
e
R
Ve
ga
Ac
he
rn
C
ap
e
lla
ri
C
C
en
ta
u
pu
s
us
ct
ur
Ar
oc
an
o
Al
ph
a
-1
Pr
-0.5
yr
on
0
Series1
One of the first people to make a good measurement of the distance to a planet
was the great astronomer Gian Domenico Cassini. In 1672, Cassini used a
technique called parallax to measure the distance to Mars.
You can understand parallax by holding your thumb up at arm's length and
looking at it first with one eye, and then your other. Notice how your thumb
seems to shift back and forth against the objects that are farther away. Because
your two eyes are separated by a few inches, each views your thumb from a
different position. The amount that your thumb appears to move is its parallax.
When astronomers measure the parallax of an object and know the separation
between the two positions from which it is observed, they can calculate the
distance to the object. Using observations on Earth separated by thousands of
miles -- like looking through two eyes that are very far apart -- parallax
measurements can reveal the great distances to planets.
Although he didn't get quite the right answers, Cassini's results were very close
to the correct values. The Sun is about 93 million miles from Earth. As Earth
and Mars move in their separate orbits, they never come closer than 35 million
miles to each other. Saturn, the most distant planet known when Cassini was
alive, is around 900 million miles away. Imagine how exciting it must have been
for him to discover that the solar system is so fantastically big!
Key:
MASS (Earth = 1)
0.11
14.54
0.82
318
17.15
0.06
1
95.16
GALAXIES
THERE ARE 1256 BILLION GALAXIES RECORDED BY
HUBBLE TELESCOPE. THERE IS ALSO THE SLOAN
GREAT WALL WITH 2,000,000. TODAY 90% OF STARS
ARE DWARFS. THERE ARE 200-500 BILLION STARS.
LOCAL GROUP
OF GALAXIES
THERE ARE 40 GALAXIES IN THIS AND ITS
STRETCH IS 10 MILLIONLY. IT INCLUDES THE MILKY
WAY GALAXY TOO.
MILKY WAY IS 100,000LY ACROSS, AND 4,000LY
DEEP. IT INCLUDES 200-400 BILLION STARS
SPIRAL GALAXIES
NAME
DIAMETER
LUMINOSITY
MILKY WAY
10,000
14,000
ANDROMEDA
150,000
40,000
TRIANGULAR
40,000
4,000
DIAMETER OF ALL SPIRAL
GALAXIES – LIGHT YEARS
160000
140000
120000
100000
DIAMETER
80000
60000
40000
20000
0
MILKY WAY
ANDROMEDA
TRIANGULAR
LUMINOSTY OF ALL SPIRAL GALAXIES
IN TERMS OF MILLIONS OF SUNS
LUMINOSITY
40,000
35,000
30,000
25,000
LUMINOSITY
20,000
15,000
10,000
5,000
0
MILKY WAY
ANDROMEDA
TRIANGULAR
IRREGULAR
GALAXIES
NAME
DIAMETER
LUMINIOSITY
LMC
30,000
2,000
SMC
20,000
250
PEGASUS
7,000
50
DIAMETER OF ALL IRREGULAR
GALAXIES IN TERMS OF LIGHT YEARS
LMC
30,000
25,000
20,000
15,000
10,000
5,000
DIAMETER
0
PEGASUS
SMC
LUMINOSITY OF ALL IRREGULAR
GALAXIES IN TERMS OF MILLIONS OF
SUNS
LMC
2000
1800
1600
1400
1200
1000
800
600
400
200
0
PEGASUS
LUMINIOSITY
SMC
ELLIPTICAL GALIXIES
NAME
DIAMETER
LUMINIOSITY
SAGITTARIUS
15,000
30
URSA MINOR
1,000
0.3
DRACO
500
0.3
SCULPTOR
1,000
1.5
CARINA
500
0.4
FORNAX
3,000
20
LEO II
500
1
LEO I
1,000
10
SEXTANTS
1,000
0.8
DIAMETER OF ALL ELLIPTICAL
GALAXIES IN TERMS OF LIGHT YEARS
16,000
14,000
12,000
10,000
8,000
6,000
4,000
2,000
0
DIAMETER
LUMINOSITY OF ALL ELLIPTICAL
GALAXIES IN TERMS OF MILLIONS OF
SUNS
35
30
30
25
20
20
15
10
10
5
0.3
0.3
0
1.5
0.4
0.8
1
LUMINOSITY
DIAMETER OF ALL THREE TYPES
(MEAN)
ELLIPTICAL
DIAMETER
SPIRAL
IRREGULAR
0
10,000
20,000
30,000
40,000
50,000
60,000
70,000
LUMINIOSITY OF ALL THREE
TYPES (MEAN)
LUMINIOSITY
19,333
20000
18000
16000
14000
12000
10000
8000
6000
4000
2000
0
LUMINIOSITY
767
IRREGULAR
7.14
SPIRAL
ELLIPTICAL
CONCLUSIONS
AFTER SEEING THE GRAPHS WE HAVE 2
CONCLUSIONS•SPIRAL GALAXIES HAVE MORE
LUMINIOSITY
•SPIRAL GALAXIES HAVE A BIGGER
DIAMETER
BRIGHTEST STARS
NAME
APPARENT
ABSOLUTE
DISTANCE FROM SUN (LIGHT YEARS)
MAGNITUDE MAGNITUDE
SIRIUS
-1.46
+1.14
8.65
CANOPUS
-0.73
-4.6
1,200
VEGA
+0.04
+0.5
26
RIGEL
+0.10
-7.0
900
PROCYON
+0.35
+2.6
11.4
ARCTURUS
-0.06
-0.3
36
HADAR
+0.63
-4.6
490
MEASUREMENTS
1 light-year = 9460730472580800 meters (exactly)
The parsec (symbol: pc) ,equal to about
30.9 trillion kilometers
The siriometer equal to one million astronomical units, i.e., one
million times the average distance between the Sun and Earth.
Peta- is a prefix in the metric system denoting
1015 or 1000000000000000
Distance
 The
furthest planet from the
sun is Neptune and it is
4,503,000,000 km away.
 The closest planet from the
sun is Mercury and it is
57,910,000 km away.
Standard index form






Example : Write 15 000 000 in standard index
form.
Solution
15 000 000 = 1.5 × 10 000 000
This can be rewritten as:
1.5 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 1. 5 × 10 7
How long does it take for the
planets to orbit the sun?








Mercury takes 88 days.
Venus takes only 224.7 days.
Earth take 365 days.
Mars takes 1.88 days.
Jupiter takes 11.86 days.
Saturn takes 10,759 days.
Uranus takes 84.3 years .
Neptune takes 164.79 years .
SATELLITES
The motion of objects is governed by Newton's laws
Consider a satellite with mass Msat orbiting a central
body with a mass of mass MCentral. The central body
could be a planet, the sun or some other large mass
capable of causing sufficient acceleration on a less
massive nearby object
Fnet = ( Msat • v2 ) / R
This net centripetal force is the result of the gravitational
force that attracts the satellite towards the central body
and can be represented as
Fgrav = ( G • Msat • MCentral ) / R2
Since Fgrav = Fnet, the above expressions for centripetal
force and gravitational force can be set equal to each
other. Thus,
(Msat • v2) / R = (G • Msat • MCentral ) / R2
Thus, the acceleration of a satellite in circular motion
about some central body is given by the following
equation
PROBLEMS
Q1. A satellite wishes to orbit the earth at a height of 100 km
(approximately 60 miles) above the surface of the earth.
Determine the speed, acceleration and orbital period of the
satellite. (Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)
v = SQRT [ (G•MCentral ) / R ]
The substitution and solution are as follows:
v = SQRT [ (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x
106 m) ]
v = 7.85 x 103 m/s
The acceleration can be found from either one of the following
equations:
Equation (1) was derived above. Equation (2) is a general equation
for circular motion. Either equation can be used to calculate the
acceleration. The use of equation (1) will be demonstrated here.
a = (G •Mcentral)/R2
a = (6.673 x 10-11 N m2/kg2) • (5.98 x 1024 kg) / (6.47 x 106 m)2
a = 9.53 m/s2
Q2. The period of the moon is approximately 27.2 days (2.35 x 106 s).
Determine the radius of the moon's orbit and the orbital speed of the moon.
(Given: Mearth = 5.98 x 1024 kg, Rearth = 6.37 x 106 m)
The radius of orbit can be calculated using the following equation:
The equation can be rearranged to the following form
R3 = [ (T2 • G • Mcentral) / (4 • pi2) ]
The substitution and solution are as follows:
R3 = [ ((2.35x106 s)2 • (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) )
/ (4 • (3.1415)2) ]
R3 = 5.58 x 1025 (1)
m3v = SQRT [ (G •
MCentral ) / R ]
(2) v = (2 • pi • R)/T
By taking the cube root of 5.58 x 1025 m3, the radius can be
determined as follows:
R = 3.82 x 108 m
The orbital speed of the satellite can be computed from either of the
following equations:
v = SQRT [ (6.673 x 10-11 N m2/kg2)*(5.98x1024 kg) / (3.82 x 108 m)
]
v = 1.02 x 103 m/s
Q3. Q3.
A geosynchronous
A geosynchronous
satellite
satellite
is a is
satellite
a satellite
thatthat
orbits
orbits
the earth
the earth
withwith
an orbital
an orbital
period
period
of o
24 hours,
24 hours,
thusthus
matching
matching
the period
the period
of the
of earth's
the earth's
rotational
rotational
motion.
motion.
A geostationary
A geostationary
satellite
satellite
orbits
orbits
the earth
the earth
in 24inhours
24 hours
along
along
an orbital
an orbital
pathpath
thatthat
is parallel
is parallel
to an
toimaginary
an imaginary
plane
plane
drawn
drawn
through
through
the Earth's
the Earth's
equator.
equator.
Such
Such
a satellite
a satellite
appears
appears
permanently
permanently
fixedfixed
above
above
the same
the same
location
location
on the
on Earth.
the Earth.
If a If
geostationary
a geostationary
satellite
satellite
wishes
wishes
to orbit
to orbit
the the
earth
earth
in 24
inhours
24 hours
(86400
(86400
s), then
s), then
howhow
highhigh
above
above
the earth's
the earth's
surface
surface
mustmust
it beitlocated?
be located
24 kg,
24 R
(Given:
(Given:
Mearth
Mearth
= 5.98x10
= 5.98x10
kg,
Rearth
= 6.37
= 6.37
x 10x6 10
m)6 m)
earth
The radius of orbit can be found using the following equation:
The equation can be rearranged to the following form
R3 = [ (T2 * G * Mcentral) / (4*pi2) ]
The substitution and solution are as follows:
R3 = [ ((86400 s)2 • (6.673 x 10-11 N m2/kg2) • (5.98x1024 kg) ) / (4 •
(3.1415)2) ]
R3 = 7.54 x 1022 m3
By taking the cube root of 7.54 x 1022 m3, the radius can be determined to
be R = 4.23 x 107 m
The radius of orbit indicates the distance that the satellite is from the center
of the earth. Since the earth's surface is 6.37 x 106 m from its center (that's
the radius of the earth), the satellite must be a height of
4.23 x 107 m - 6.37 x 106 m = 3.59 x 107 m
above the surface of the earth. So the height of the satellite is 3.59 x107 m.
Degrees longitude orbital ground track shifts eastward
with each orbit.
Number of orbits so that Mir flies over Moscow. Mir Moscow = Longitude Distance
Time for Mir's orbit to cross Moscow.
Distance (circumference) Mir travels during one orbit. (The
altitude is the distance from Earth's center to Mir.)
Mir's orbital speed.
Shuttle speed change needed to raise orbit 7 kilometers. (It
is stated in the video that a change in velocity of 0.4 meters
per second raises the Shuttle 1 kilometer.)
Earth's orbit around the Sun is elliptical, but in many cases it is sufficiently accurate to
approximate the orbit with a circle of radius equal to the mean Earth-Sun distance of
1.49598 x 10^8 km. This distance is called the "Astronomical Unit" (AU). Listed in the
chart that follows are actual Earth-Sun distances, given to five significant digits, on the
first day of each month of a representative year. (The "American Ephemeris" lists daily
distances and the actual times for these distances to seven significant digits.)
a. To how many significant digits is it reasonable to approximate the Earth-Sun distance
as though the orbit were circular?
Solution: To two significant digits, each of the distances in the table can be given as 1.5
x 10^8 km.
b. What are the largest possible absolute and relative errors in using the Astronomical
Unit as the Earth-Sun distance in a computation instead of one of the distances from the
table?
Solution: (1.49598 - 1.4710) x 10^8 = 0.0240 x 10^8 km (smallest table
value)
(1.49598 - 1.5208) x 10^8 = -0.0248 x 10^8 km (largest table value)
absolute error less than/equal to 0.0248 x 10^8 km
relative error less than/equal to 0.0248/1.49598 = 0.0166, or 1.7 percent.
Since 100 cm = 1 m, in order to change 623 cm to m, we perform the
multiplication
(623 cm/1) X (1m/100 cm),
"canceling" the cm in numerator and denominator to get
623/100 m, or 6.23 m.
More complex conversions can be done using multiplication by several
factor units and those readers wishing to convert between British and
metric units can also use this method. For example, the speed of light,
3.00 x 10^5 km/sec, can be found in miles per hour:
(3.00 x 10^5/1) X (km/sec) X (1 mile/1.61 km) X (60 sec/1min) X (60
min/1 hour)
= 6.71 x 10^8 miles per hour.
SPACE SUIT
A space suit should allow its user natural unencumbered
movement. Nearly all designs try to maintain a constant volume
no matter what movements the wearer makes. This is
because mechanical work is needed to change the volume of a
constant pressure system. If flexing a joint reduces the volume
of the spacesuit, then the astronaut must do extra work every
time he bends that joint, and he has to maintain a force to keep
the joint bent. Even if this force is very small, it can be seriously
fatiguing to constantly fight against one's suit. It also makes
delicate movements very difficult. The work required to bend a
joint is dictated by the formula -
THANK YOU
A PPT BY – PULKIT GERA (DIGITAL ENGINEER)
&
Debangshu Mukherjee
GROUP L – LOADED MATH
GROUP MEMBERSDEBANGSHU MUKHERJEE
PULKIT GERA
EDWARD FAGAN
KASEY BEARDALL EDWARDS
SPECIAL THANKSMAAM KAMALIKA BOSE