SPACE TECH… IN MATHS BY GROUP

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

Brightness of stars

-2

-1.5

-1

-0.5

0

0.5

1

Bet

elge

use

Pro

cyro

n

Arc

turu

s

Can

opus

Alpha

Cen

taur

i

Cap

ella

Veg

a

Rigel

Ach

erna

r

Sirius

Moons

Bri

gh

tne

ss

(s

ca

le:

-1.5

to

2)

Series1

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:

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)

0.060.8210.1131895.1614.5417.15

GALAXIESTHERE 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 GALAXIESTHERE 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

NAME DIAMETER LUMINOSITY

MILKY WAY 10,000 14,000

ANDROMEDA 150,000 40,000

TRIANGULAR 40,000 4,000

SPIRAL GALAXIES

0

20000

40000

60000

80000

100000

120000

140000

160000

MILKY WAY ANDROMEDA TRIANGULAR

DIAMETER

DIAMETER OF ALL SPIRAL

GALAXIES – LIGHT YEARS

0

5,000

10,000

15,000

20,000

25,000

30,000

35,000

40,000

MILKY WAY ANDROMEDA TRIANGULAR

LUMINOSITY

LUMINOSITY

LUMINOSTY OF ALL SPIRAL GALAXIES

IN TERMS OF MILLIONS OF SUNS

NAME DIAMETER LUMINIOSITY

LMC 30,000 2,000

SMC 20,000 250

PEGASUS 7,000 50

IRREGULAR

GALAXIES

DIAMETER OF ALL IRREGULAR

GALAXIES IN TERMS OF LIGHT YEARS

0

5,000

10,000

15,000

20,000

25,000

30,000

LMC

SMCPEGASUS

DIAMETER

LUMINOSITY OF ALL IRREGULAR

GALAXIES IN TERMS OF MILLIONS OF

SUNS

0

200

400

600

800

1000

1200

1400

1600

1800

2000

LMC

SMCPEGASUS

LUMINIOSITY

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

ELLIPTICAL GALIXIES

DIAMETER OF ALL ELLIPTICAL

GALAXIES IN TERMS OF LIGHT YEARS

0

2,000

4,000

6,000

8,000

10,000

12,000

14,000

16,000

DIAMETER

0 0 0 0 0 0 0 0 0

30

0.3

0.3 1.5 0.4 0.8

20

1

10

0

5

10

15

20

25

30

35

LUMINOSITY

LUMINOSITY OF ALL ELLIPTICAL

GALAXIES IN TERMS OF MILLIONS OF

SUNS

DIAMETER OF ALL THREE TYPES

(MEAN)

0 10,000 20,000 30,000 40,000 50,000 60,000 70,000

IRREGULAR

SPIRAL

ELLIPTICAL

DIAMETER

LUMINIOSITY OF ALL THREE

TYPES (MEAN)

0

2000

4000

6000

8000

10000

12000

14000

16000

18000

20000

IRREGULAR SPIRAL ELLIPTICAL

767

19,333

7.14

LUMINIOSITY

LUMINIOSITY

CONCLUSIONS

AFTER SEEING THE GRAPHS WE HAVE 2

CONCLUSIONS-

•SPIRAL GALAXIES HAVE MORE

LUMINIOSITY

•SPIRAL GALAXIES HAVE A BIGGER

DIAMETER

BRIGHTEST STARS

NAME APPARENT

MAGNITUDE

ABSOLUTE

MAGNITUDE

DISTANCE FROM SUN (LIGHT YEARS)

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

MEASUREMENTS1 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

DistanceThe 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 .

SATELLITESThe 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 ) / RThis net centripetal force is the result of the gravitational force that attracts the satellite towards the central body and can be represented asFgrav = ( 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

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

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)

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 formR3 = [ (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 m3

By taking the cube root of 5.58 x 1025 m3, the radius can be determined as follows:R = 3.82 x 108 m

(1) v = SQRT [ (G •

MCentral ) / R ](2) v = (2 • pi • R)/T

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. A geosynchronous satellite is a satellite that orbits the earth with an orbital period of

24 hours, thus matching the period of the earth's rotational motion. A geostationary

satellite orbits the earth in 24 hours along an orbital path that is parallel to an imaginary

plane drawn through the Earth's equator. Such a satellite appears permanently fixed

above the same location on the Earth. If a geostationary satellite wishes to orbit the

earth in 24 hours (86400 s), then how high above the earth's surface must it be located?

(Given: Mearth = 5.98x1024 kg, Rearth = 6.37 x 106 m)

The radius of orbit can be found using the following equation:The equation can be rearranged to the following formR3 = [ (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 mThe 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 of4.23 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.

Q3. A geosynchronous satellite is a satellite that orbits the earth with an orbital period of

24 hours, thus matching the period of the earth's rotational motion. A geostationary

satellite orbits the earth in 24 hours along an orbital path that is parallel to an imaginary

plane drawn through the Earth's equator. Such a satellite appears permanently fixed

above the same location on the Earth. If a geostationary satellite wishes to orbit the

earth in 24 hours (86400 s), then how high above the earth's surface must it be located?

(Given: Mearth = 5.98x1024 kg, Rearth = 6.37 x 106 m)

The radius of orbit can be found using the following equation:The equation can be rearranged to the following formR3 = [ (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 mThe 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 of4.23 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.

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 -

SPACE SUIT

THANK YOUA PPT BY – PULKIT GERA (DIGITAL ENGINEER)

&

Debangshu Mukherjee

GROUP L – LOADED MATH

GROUP MEMBERS-

DEBANGSHU MUKHERJEE

PULKIT GERA

EDWARD FAGAN

KASEY BEARDALL EDWARDS

SPECIAL THANKS-

MAAM KAMALIKA BOSE

SIR RICHARD DAVIES