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SPACE TECH… . BY GROUP . IN MATHS . All about space. 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:.
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SPACE TECH… BY GROUP IN MATHS
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) 0.11 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
LUMINOSTY OF ALL SPIRAL GALAXIES IN TERMS OF MILLIONS OF SUNS
LUMINOSITY OF ALL IRREGULAR GALAXIES IN TERMS OF MILLIONS OF SUNS
LUMINOSITY OF ALL ELLIPTICAL GALAXIES IN TERMS OF MILLIONS OF SUNS
CONCLUSIONS • AFTER SEEING THE GRAPHS WE HAVE 2 CONCLUSIONS- • SPIRAL GALAXIES HAVE MORE LUMINIOSITY • SPIRAL GALAXIES HAVE A BIGGER DIAMETER
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
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 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 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) 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 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. 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 MEMBERS- DEBANGSHU MUKHERJEE PULKIT GERA EDWARD FAGAN KASEY BEARDALL EDWARDS SPECIAL THANKS- MAAM KAMALIKA BOSE SIR RICHARD DAVIES
