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64 CHAPTER Key Ideas3 Stage 2 Physics 3.3 Application Weather & Communication Satellites The U.S.S.R. launched the first satellite, Sputnik 1, on October 4, 1957. Since then man has launched many thousands of satellites into orbit around the Earth. These satellites have a wide variety of purposes and objectives. Among these are: Weather satellites which observe, photograph and analyse weather patterns. Communication satellites which provide communication links between widely separated areas and to remote communities. Surveillance satellites which observe areas of the Earth’s surface for environmental, defence, military and agricultural purposes. Global positioning satellites which enable aircraft, ships and any person with the appropriate equipment to precisely locate their position on the Earth by comparing the signals from several satellites. 3.3.1 Possible Orbits A satellite in a circular orbit is moving with uniform circular motion. Therefore, there is a centripetal acceleration acting towards the centre of the circle, and so there must be a force that causes this centripetal acceleration. The direction of this force must also be towards the centre of the circle. But this force is the force of gravitational attraction. The gravitational force acts along the line joining the centres of the two bodies, and so the force on the satellite is directed towards the centre of the Earth. The implication of the above two points is that the only circular orbits possible are those such that their centre coincides with the centre of the Earth. Figure 3.4(a) illustrates some possible orbits, while Figure 3.4(b) illustrates two satellite orbits that are not possible. Fig. 3.4 (a) 64 Essentials Text Book Fig. 3.4 (b) Gravitation GRAVITATION and Satellites AND SATELLITES 65 3.3.2 Geostationary Satellites A geostationary satellite is one that is always above the same point on the Earth’s surface. You will find them sometimes referred to as geosynchronous orbits. These satellites are extremely useful for communication purposes, for the satellite is always in the same position relative to the Earth, and this simplifies communication in the following ways. Communication relay stations can keep their dishes and antennae constantly pointed in the same direction. They do not need complicated tracking mechanisms to follow a satellite that is moving relative to the Earth. Communication facilities are not lost when the satellite moves out of the “line of sight”. Geostationary satellites are also extremely useful for maintaining continuous surveillance on the same part of the Earth’s surface for military reasons, to monitor weather patterns and for global positioning reasons. Currently there are more than 200 satellites in geostationary orbits above the Earth. Geostationary Orbits The Earth rotates from west to east with a period of 24 hours. To maintain its position above a given point on the Earth’s surface the geostationary satellite must also rotate in the same direction as the Earth (i.e. from west to east with a period of 24 hours). Because the centre of the orbit must coincide with the centre of the Earth, and the satellite rotates from west to east, geostationary satellites must be in an equatorial orbit (i.e. they rotate above the equator of the Earth). Thus, geostationary satellites can only be stationary above some point on the equator. Notes on Geostationary Orbit The period of a geostationary orbit is fixed at 24 hours. Therefore, this determines the radius of the geostationary orbit, as period depends only on radius as we showed in Section 3.2.1. The height of a geostationary orbit is about 35,900km above the surface of the Earth. We derive this result in Example 6 below. In terms of general satellite orbits this is very high. The vast majority of the satellites above the Fig. 3.5 Earth are in orbits much lower than this. Figure 3.5 portrays a geostationary satellite orbiting the Earth, drawn approximately to scale. The height of a geostationary satellite above the Earth’s surface causes its own problems. In the case of communication satellites. Signal strength must be high. Receiving equipment must be quite sophisticated, in order to achieve a decipherable signal from this distance. There is a problem with time delays because of the distance that the signal has to travel. Consider communication between Adelaide and London. The signal leaves Adelaide, is beamed up to a satellite and then may be relayed via several other geostationary satellites to London. The large distance travelled means that there is a noticeable time delay before the signal reaches London. This is why, when watching a TV interview with someone overseas, we detect a hesitation before the interviewee responds to a question. Essentials Text Book 65 66 CHAPTER Key Ideas3 Stage 2 Physics In the case of surveillance and weather satellites. The height of the orbit limits the amount of close up detail that can be observed. It is not possible to gain a good view of the surface of the Earth at high latitudes (i.e. close to the poles). Because the satellite is positioned above the equator, when viewing regions near the equator from a geostationary satellite, we are viewing the surface of the Earth at an angle of 90 to the surface. Because of the curvature of the Earth’s surface, when looking at high latitudes from a geostationary satellite, the viewing angle is not at 90 to the surface but gets closer and closer to 0 as we view nearer to one of the poles. Thus, for these regions of the Earth we get a very unclear and distorted image. Example 6 Calculate the height of a satellite that is in geostationary orbit above the Earth. The mass of the Earth ME = 5º9771024kg, the equatorial radius of the Earth rE = 6º378106m. Period T 24 hrs 24 3600 s 8 º 64 10 4 s v v2 i.e. 4S 2 r 2 T2 r3 r GM E r GM E r GM E 2Sr as v r T GM E T 2 4S 2 6 º 673 10 11 5 º 977 10 24 8 º 64 2 10 8 4S 2 3 75 º 4175 10 21 4 º 225 10 7 m This is the radius of the satellite’s orbit. The height of the orbit above the surface is given by h r rE 42 º 250 10 6 6 º 378 10 6 35 º 872 10 6 m 35,872 km 3.3.3 Direction of Launch of Equatorial Orbit Satellites Consider a satellite in an equatorial orbit at a height of 1000km above the Earth’s surface. It can be shown that the speed of this satellite in its orbit is approximately 7,350ms–1. To maintain a stable orbit at this height the satellite must have this speed. This speed is fixed and it does not depend on whether the Earth is stationary, or whether it is spinning on its polar axis below the satellite. The Earth is rotating on its polar axis with a period of 24 hours. It can be shown that a point on the equator has a linear speed of 465ms–1 from West to East. 66 Essentials Text Book Gravitation GRAVITATION and Satellites AND SATELLITES 67 Thus if we wish to launch this satellite in an equatorial orbit it already has a speed of 465ms–1 from West to East before it leaves the launch pad. Therefore if we launch the satellite in a West to East direction we only need to give it an additional speed of 6885ms–1 relative to the Earth. If the satellite were to be launched in an East to West direction we would have to give it an additional velocity of 7815ms–1. Therefore, it is advantageous to launch low-altitude equatorial-orbit satellites in a West to East direction, as about 930ms–1 smaller velocity change must be imparted to the satellite, with consequential savings in fuel etc. 3.3.4 Satellites in Polar Orbits Geostationary satellites have several disadvantages for surveillance and meteorological (weather monitoring) applications. Because of the great height of the orbit above the Earth’s surface the amount of fine detail that can be observed is limited. Because these satellites are located above the equator geostationary satellite images of polar regions are distorted because of the low angle at which the satellite sees the region. Low altitude polar orbit satellites are used to overcome many of these disadvantages. These satellites rotate around the Earth in North-South orbits, passing over the Earth’s North and South Poles, at altitudes of approximately 850km above the Earth’s surface. They orbit the Earth approximately 14º2 times per day. As the satellite in the polar orbit depicted in Figure 3.6 rotates about the Earth it surveys and photographs a northsouth strip of the Earth’s surface. Meanwhile the Earth is rotating from West to East under the satellite. Therefore, when the satellite has completed one revolution (about 1hr and 40mins later) it will now pass over a strip approximately 2800km to the west of the previous one. The photographic images of these strips can then be pieced together to produce an image of a much larger area. Rotation of the Earth The width of the field of view of these satellites on the Fig. 3.6 Earth’s surface is about 2900km. Therefore successive orbits give slightly overlapping images, thus making the job of piecing them together easier. The United States has two fully functional polar orbiting weather satellites. They are called NOAA 12 and NOAA 14. They orbit the Earth with periods of approximately 102 minutes at altitudes between 820km and 860km. Thus, they complete about 14º2 orbits per day. Their orbits are not truly polar. Instead of being inclined at 90 to the equator they are inclined at 98 to the equator. This means they are in retrograde orbits – they orbit in the opposite direction to which the Earth turns. The advantage of this particular orbit is that the satellites cover (i.e can “see”) the same part of the Earth’s surface at the same time each day, and they do this day after day for the life of the satellite. This kind of orbit is called a sun synchronous orbit. Low altitude polar orbit satellites have the advantage that they are photographing areas of the Earth virtually directly beneath them (therefore with minimal distortion) from a low altitude. This enables analysts to see great detail. In addition one of these satellites will cover and observe every point the Earth’s surface once each day. Essentials Text Book 67 68 CHAPTER Key Ideas3 Stage 2 Physics 3.4 Exercises Conceptual Questions 1 The value of gravitational acceleration is the same throughout the universe. Is this true or false? Explain your answer. 2 The mass of an apple in the greengrocer’s shop is 0º2kg. With what force does the apple attract the Earth? 3 An observer, stationed on the Moon, has a very highpower telescope and, with it, he observes an apple fall off a tree down to the ground on Earth. If the apple is attracting the Earth, why did he not see the Earth fall towards the apple? 4 The orbit of the Earth about the Sun is not truly circular. It is slightly elliptical. The closest distance (at perihelion) is 1º4961011m and the furthest distance (at aphelion) is 1º5211011m. Is the Earth travelling faster at perihelion or at aphelion? Explain your answer. 5 Explain why it takes more fuel for a spacecraft to travel from the Earth to the Moon than for the return journey. 6 It was proposed to put military satellites carrying bombs in orbit around the Earth. By releasing it at the correct time and location, one could drop a bomb on any point on the Earth’s surface. Is this feasible? 7 The polar diameter of the Earth is less than the equatorial diameter. Will the value of g be greater at the North Pole or at the Equator? Explain your answer. 8 At high altitude the value of g is less than at the Earth’s surface. As a body falls towards the Earth’s surface the value of g that it experiences increases. Consider a body continuing to fall down a very deep vertical mineshaft below the Earth’s surface. Does the value of g continue to increase as the body approaches the centre of the Earth? Explain your answer. 9 My mass is 100kg. What is my weight on the Earth’s surface? What would be my weight at the centre of the Earth (if I could survive there)? 10 The force of gravity acting on a body on the Earth’s surface is directly proportional to the mass of the body. Therefore, explain why heavier bodies do not fall faster than light ones. 11 A satellite is orbiting the Earth. A piece breaks off an antenna on the satellite. Describe the subsequent motion of this piece. 12 When a satellite is moving in a circular orbit there is a gravitational force F acting on the satellite. In a given time the satellite moves a certain distance s. Therefore work is done on the satellite (as W = Fs). Therefore the satellite must gain energy. Therefore the satellite’s speed must increase (as kinetic energy K= ½mv2). Thus it is impossible for a satellite to move in a circular orbit with constant speed. Find and explain the flaw in the above argument. 68 Essentials Text Book 13 “ACME Fine Foods” advertises that their prices are the same at all their stores throughout the world. Therefore explain why it is better to buy 1kg of freshly sliced leg ham at their Darwin store than at their South Pole store. 14 At night you are on the side of the Earth that is facing away from the Sun, while during the day you are on the side of the Earth that is facing the Sun. If you had a very sensitive bathroom scale, would you weigh more at midday or at midnight? Explain. 15 Consider four bodies of equal mass. Body A is a large solid sphere, body B is a small solid sphere, body C is a large hollow sphere and body D is a small hollow sphere. A particle P is placed, in turn, an equal distance from the centres of each of the above spheres. Rank the four spheres in the order of the magnitude of the gravitational force that they would exert on P. 16 If the mass of the Earth were to suddenly double, what effect would this have on the orbit of the Moon? 17 Four students are explaining satellite motion. (A) Ianthe says: “The satellite is no longer in the Earth’s gravitational field and is no longer attracted to the Earth. The net force on the satellite is zero and it is kept in orbit by its tangential velocity v”. (B) Jim says: “The satellite is no longer in the Earth’s gravitational field and the force of gravity on it is negligible. There is, however, an inward centripetal force mv2/r which maintains the satellite in its orbit”. (C) Kate says: “There is only the force of gravity acting on the satellite and this causes an inward acceleration of v2/r”. (D) Leo says: “There is a gravitational force acting on the satellite, but it is kept in its orbit by an outward force mv2/r which cancels the effect of the gravitational force”. Which of the above explanations is correct? 18 A gravitational force is acting between two bodies. What is its direction? 19 Is it possible to have a geostationary satellite rotating from East to West. 20 Why is there not a satellite permanently stationed above Adelaide? 21 A mistake was made in positioning a geostationary satellite and it was placed in an orbit above the equator that was 1km too high. Is the satellite moving West to East relative to the Earth or East to West? Gravitation GRAVITATION and Satellites AND SATELLITES 69 Descriptive / Explanatory Questions 1 Given the mass and the radius of some planet, explain how you could calculate the value of g, the gravitational acceleration, at the surface of the planet. 2 Explain how the gravitational force between two bodies is consistent with Newton’s third law. 3 Show that the speed v of a satellite orbiting a larger body of mass M in a stable circular orbit of radius r is given by v GM . r 4 Prove that the period of a satellite in stable circular orbit is independent of the mass of the satellite and dependent only on the mass of the central body and the radius of the orbit. 5 Explain why a geostationary satellite must move in an orbit that coincides with the equatorial plane of the Earth. 6 Is it possible for a satellite to move in a stable orbit around the Earth, with the centre of the orbit not coinciding with the centre of the Earth? Explain. 7 Explain why it is advantageous to launch satellites, which are to circle the Earth in an equatorial orbit, in a West to East direction in preference to an East to West direction. 8 Explain why low-altitude polar orbits are preferred to geostationary orbits for meteorological and spying surveillance. In your answer include three disadvantages of geostationary orbits for these purposes and three advantages of low-altitude polar orbits. 9 An astronaut of mass 90kg flies in a direct line from the Earth to the Moon. On Earth g = 9º80ms–2 while on the Moon g = 1º62ms–2. Describe and explain how his weight changes during the flight. 14 A satellite is moving in a circular orbit of radius r and period T. Show that the radius and period of the 4S2 r where g is the orbit are related by T 2 g gravitational acceleration at the satellite’s orbit height. 15 At midnight on December 31, 2999 the Masters of the Universe plan to decrease the universal gravitational constant G from its current value of 6º6710–11Nm2kg–2 to 4º010–11Nm2kg–2. Explain what effect you expect this to have on each of the following. (1) The value of gravitational acceleration g on the surface of the Earth. (2) The radius of the Earth’s orbit about the Sun. (3) The length of the year. (4) The diameter of the Earth. 10 The planet Jupiter has 12 moons. Explain what measurements an astronomer could take to determine the mass of Jupiter. 11 For satellites moving in low polar orbits there are still some residual traces of atmosphere at the height of the satellite. Describe and explain the effects of air resistance on the motion of the satellite. 12 Photographs of astronauts in the space shuttle or on space station Mir show them in a state of weightlessness, i.e floating around the inside of the vehicle. However at a height of 1000km above the surface of the Earth the magnitude of gravitational acceleration is g = 7º3ms–2. Explain how it is possible for them to attain this weightless state at this altitude. 13 The diagram shows a M mass M and five smaller equal masses m. The smaller masses are evenly spaced along the line AB. What is the A B direction of the net m m m m m gravitational force on M due to the five smaller masses? Explain your answer. Essentials Text Book 69 70 CHAPTER Key Ideas3 Stage 2 Physics Analytical / Computational Questions 1 (1) Determine the gravitational force between an 80kg man and a 60kg woman if they are standing 10m apart. (2) What will be the gravitational force if they move towards each other so that the distance between their centres of mass is reduced to 1m? (3) What will be gravitational force between the 80kg man and a 30kg child, if they are 10m apart? (4) What will be gravitational force between the 80kg man and a 30kg child, if they are 5m apart? 10 How high above the surface of the Earth must we go before the value of gravitational acceleration is reduced to 4º9ms–2? 2 Find the gravitational force of attraction between the Earth and the Sun given that the mass of the Earth is 5º9771024kg, the mass of the Sun is 1º9911030kg and their mean distance apart is 1º4961011m. 3 Two bodies, of mass 5º0kg and 8º0kg respectively, experience a gravitational force of 4º61010N. Calculate the distance between their centres of mass. 12 The Russian space station MIR is in circular orbit above the Earth at a height of 365º4km. Find (1) its orbital speed; (2) its period of revolution about the Earth; (3) the number of times it circles the Earth in one day. 4 The gravitational force of attraction between two bodies, 3º0m apart, is 3º335109N. The mass of one body is double the mass of the other. Find the masses of the two bodies. 5 The mass of the Sun is 1º9911030kg and the mass of Jupiter is 1º9011027kg. The mean radius of Jupiter’s orbit is 7º7841011m. Find (1) the gravitational force of attraction between Jupiter and the Sun; (2) Jupiter’s orbital speed; (3) the length of a Jovian year (i.e. the time it takes for Jupiter to complete one revolution). Give your answer in Earth years. 6 The mass of Jupiter is 1º90110 kg and its radius is 7º144107m. Calculate the value of gravitational acceleration, g, on the surface of Jupiter. 7 How far from the Earth must a spacecraft be along a line towards the Moon so that the Moon’s gravitational pull on the body balances that of the Earth. The Moon is 3º844108m away from the Earth and the Earth’s mass is 81º33 times the mass of the Moon. 8 9 70 27 Fred’s planet has three moons named Alpha, Beta and Gamma. Their relative masses are mE = 2mD and mJ = 3mD. The relative radii of their orbits are given by rE = 2rD and rJ = 3rD. The gravitational force of attraction between Fred’s planet and Alpha is 2º501020N. Find (1) the gravitational attraction between Fred’s planet and Beta; (2) the gravitational attraction between Fred’s planet and Gamma. On its journeys “where no man has gone before”, the starship Enterprise discovers a new planet. Its mass is double the mass of the Earth and its radius is ¾ that of the Earth. Determine the value of gravitational acceleration on the surface of this planet. Essentials Text Book 11 A satellite is in circular orbit above the Earth. Its orbital radius is 6º77106m. Find (1) its orbital speed; (2) its period of revolution about the Earth; (3) the number of times it circles the Earth in one day; (4) the value of gravitational acceleration, g, at this satellite’s orbit. 13 A satellite, in circular orbit above the Earth, circles the Earth 10 times in one day. Find (1) its period; (2) the radius of its orbit; (3) its orbital speed; (4) its height above the Earth. 14 A satellite is in geostationary orbit above Earth. (1) Explain what this means and state the period of the satellite. (2) Calculate the height of the satellite above the surface of the Earth. (3) Calculate the orbital speed of the satellite. 15 A satellite is in circular orbit about the Earth. It has an orbital speed of 5595ms–1. Find (1) its height above the Earth’s surface; (2) its period of rotation. 16 The following table gives the value of gravitational acceleration g at various distances r from the centre of the Earth. r (106)m g (ms–2) (1) (2) (3) (4) (5) 6º4 9º80 7º0 8º19 8º0 6º27 9º0 4º96 10 4º01 1 Draw a graph of g against 2 . r What does your graph tell about the proportionality relationship between g and r? Use your graph to find the value of g at a distance of 8º5106m from the centre of the Earth. Use your graph to find the distance from the centre of the Earth at which the value g is 3º0ms–2. Use your graph to find the mass of the Earth. The mass of Venus is about 0º8 times the mass of the Earth and its radius is slightly less than that of the Earth. On your graph draw a line which would approximately represent similar data for the planet Venus. Gravitation GRAVITATION and Satellites AND SATELLITES 71 Analytical / Computational Questions 17 The following table gives the orbital speeds v and the altitudes h of satellites in circular orbits about the Earth. Take the radius of the Earth to be 6º4106m. h (km) 500 1000 1500 2000 2500 v (10 ms ) 7º60 7º34 7º11 6º89 6º69 3 –1 (1) Find the orbital radius, r, of each satellite and 1 draw a graph of v against (2) (3) (4) (5) r . What does your graph tell you about the proportionality relationship between v and r? Use your graph to find the orbital speed of a satellite that is 3000km above the surface of the Earth. Use your graph to find the orbital radius of a satellite with an orbital speed of 5000ms–1. Use your graph to find the mass of the Earth. The mass of Fred’s planet (see question 8) is approximately four times the mass of the Earth and its radius is the same as the radius of the Earth. On your graph draw a line which would represent similar data for Fred’s planet. 18 At the beginning of the seventeenth century, more than half a century before Isaac Newton published his laws of motion and the law of universal gravitation, a German astronomer Johannes Kepler, after examining and analysing detailed astronomical data collected by Tycho Brahe, published three laws of planetary motion. Kepler’s third law states that the square of the period of a planet’s motion about the Sun is directly proportional to the cube of the radius of its orbit. Newton could show that this law, and the other two laws, could be derived mathematically from the law of universal gravitation and the laws of motion, and he used this fact as evidence in favour of his law of universal gravitation. (1) Prove Kepler’s third law mathematically; i.e. ⎛ GM S ⎞ 2 show that r 3 ⎜ ⎟T , where r is the radius 2 ⎝ 4S ⎠ of a planet’s orbit, T is its period and MS is the mass of the Sun. The table below gives some planetary data for our solar system. It gives the mean distance from the sun r of five planets and the length of their orbital period T about the sun (i.e. the length of their year) in Earth years. Planet Mercury 9 r (10 m) 57º9 Venus Earth Mars 108º2 149º6 227º9 Jupiter 778º3 T (years) 0º241 0 615 1º0 1º88 11º86 (2) Using the data for the five planets listed, draw a 18 (3) Use your graph to find the value of r3 for the T2 planets of our solar system. (4) Use this ratio to determine the mass of the Sun. 19 Once a star has burnt out it can collapse under its own gravitational force to become a neutron star or “black hole” (so called because its gravitational pull is so strong that even light cannot escape from it). Consider our Sun collapsing into a neutron star of diameter 20km. Assuming that the mass of the Sun at that time is 1º01030kg (1) Find the values of gravitational acceleration g at distances of 10m and 12m above the surface of this neutron star. (2) Deduce what would happen to your body if you were to “fall” head first into a black hole. 20 An object on the Earth’s equator is subject to three centripetal accelerations. One towards the centre of the Earth due to the rotation of the Earth about its axis. One toward the Sun due to the rotation of the Earth about the Sun. One toward the centre of our galaxy due to the rotation of the Sun about the centre of our galaxy. Determine each of these centripetal accelerations as multiples of g = 9º8ms–2. [Use astronomical data from Appendix C plus the following: Period of Sun’s rotation about the galactic centre = 2º5108yr, distance of Sun from galactic centre = 2º21020m.] 21 A mass M is split into two parts, one of mass m and the other of mass M – m. The two parts are then separated by a distance d. Show that the gravitation force between them is maximum if the two parts are of equal mass. [You need to know how to find the maximum value of a quadratic function of the form y = –ax2 + bx to solve this problem.] 22 Assume that all the stars in our galaxy are uniformly distributed in a sphere about the galactic centre. Also assume that the average mass of the stars is equal to the mass of our Sun. Our solar system is on the outer edge of or galaxy and the Sun rotates about the galactic centre with a period and radius as given in Question 20 above. Estimate the number of stars in our galaxy. 23 An astronaut flies in a direct line from the Earth to the Moon. Using the astronomical data in Appendix C, determine his distance from the Moon when he is truly weightless. 24 A cannonball is fired horizontally from a height of 2m above the surface of the Earth. Determine the speed with which it must be projected so that it travels around the Earth. Assume that the Earth is a perfect sphere and neglect the effect of air resistance. 2 graph of r against T 3 to show this proportionality. Essentials Text Book 71