11th Physics Volume

; c GM E h T GM / Here h is the distance of the Moon from the surface of the Earth. Here, R E – radius of the Earth = . × m M E – mass of the Earth = . × kg G – Universal gravitational constant = . × Nm kg Unit Gravitation E m E m E m = − Joule The negative energy implies that the Moon is bound to the Earth. Same method can be used to prove that the energy of the Earth is also negative. . . Geo-stationary and polar satellite The satellites orbiting the Earth have different time periods corresponding to different orbital radii. Can we calculate the orbital radius of a satellite if its time period is hours? Kepler’s third law is used to find the radius of the orbit. GM h h GM T h GM T / Substituting for the time period ( hrs = 86400 seconds), mass, and radius of the Earth, h turns out to be , km. Such satellites are called “geo-stationary satellites”, since they appear to be stationary when seen from Earth. India uses the INSAT group of satellites that are basically geo-stationary satellites for the purpose of telecommunication. Another type of satellite which is placed at a distance Therefore the total energy of the satellite is GM M h GM M h s s GM M h s ( . ) The negative sign in the total energy implies that the satellite is bound to the Earth and it cannot escape from the Earth. As h approaches ¥ , the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth’s gravity and is not bound to Earth at large distances. EXAMPLE . Calculate the energy of the (i) Moon orbiting the Earth and (ii) Earth orbiting the Sun. Solution Assuming the orbit of the Moon to be circular, the energy of Moon is given by, GM M where M E is the mass of Earth . × kg; M m is the mass of Moon . × kg; and R m is the distance between the Moon and the center of the Earth . × km G = . N m kg − . Unit Gravitation . . Weightlessness Weight of an object Objects on Earth experience the gravitational force of Earth. The gravitational force acting on an object of mass m is mg. This force always acts downwards towards the center of the Earth. When we stand on the floor, there are two forces acting on us. One is the gravitational force, acting downwards and the other is the normal force exerted by the floor upwards on us to keep us at rest. The weight of an object W is defined as the downward force whose magnitude W is equal to that of upward force that must be applied to the object to hold it at rest or at constant velocity relative to the earth. The direction of weight is in the direction of gravitational force. So the magnitude of of to km from the surface of the Earth orbits the Earth from north to south direction. This type of satellite that orbits Earth from North Pole to South Pole is called a polar satellite. The time period of a polar satellite is nearly minutes and the satellite completes many revolutions in a day. A Polar satellite covers a small strip of area from pole to pole during one revolution. In the next revolution it covers a different strip of area since the Earth would have moved by a small angle. In this way polar satellites cover the entire surface area of the Earth. Strip Figure . Strip of communication region, covered by a polar satellite. Polar orbit Polar orbiting satellite Geostationary satellite Geostationary orbit Figure . Polar orbit and geostationary satellite Unit Gravitation Hence, in this case also the apparent weight of the man is equal to his actual weight. It is shown in Figure . (a) Case (iii) When the elevator is accelerating upwards If an elevator is moving with upward acceleration ( )  a aj with respect to inertial frame (ground), applying Newton’s second law on the man,    F N ma G + Writing the above equation in terms of unit vector in the vertical direction, mgj Nj maj By comparing the components, N = m ( g + a ) ( . ) Therefore, apparent weight of the man is greater than his actual weight. It is shown in Figure . (b) Case (iv) When the elevator is accelerating downwards If the elevator is moving with downward acceleration ( a aj = − ˆ ), by applying Newton’s second law on the man, we can write    F N ma G Writing the above equation in terms of unit vector in the vertical direction, =− mgj Nj maj By comparing the components, N = m (g − a ) ( . ) weight of an object is denoted as, W=N=mg. Note that even though magnitude of weight is equal to mg, it is not same as gravitational force acting on the object. Apparent weight in elevators Everyone who used an elevator would have felt a jerk when the elevator takes off or stops. Why does it happen? Understanding the concept of weight is crucial for explaining this effect. Let us consider a man inside an elevator in the following scenarios. When a man is standing in the elevator, there are two forces acting on him. . Gravitational force which acts downward. If we take the vertical direction as positive y direction, the gravitational force acting on the man is  F mgj G =− . The normal force exerted by floor on the man which acts vertically upward,  N Nj Case (i) When the elevator is at rest The acceleration of the man is zero. Therefore the net force acting on the man is zero. With respect to inertial frame (ground), applying Newton’s second law on the man,   F N mgj Nj G + By comparing the components, we can write N – mg = (or) N = mg ( . ) Since weight, W =N, the apparent weight of the man is equal to his actual weight. Case (ii) When the elevator is moving uniformly in the upward or downward direction In uniform motion (constant velocity), the net force acting on the man is still zero. Unit Gravitation Therefore, apparent weight W = N = m(g-a) of the man is lesser than his actual weight. It is shown in Figure . (c) Weightlessness of freely falling bodies Freely falling objects experience only gravitational force. As they fall freely, they are not in contact with any surface (by neglecting air friction). The normal force acting on the object is zero. The downward acceleration is equal to the acceleration due to the gravity of the Earth. i.e (a = g). From equation ( . ) we get. a = g ∴ N = m ( g – g ) = . This is called the state of weightlessness. When the lift falls (when the lift wire cuts) with downward acceleration a = g, the person inside the elevator is in the state of weightlessness or free fall. It is shown in Figure . (d) When the apple was falling from the tree it was weightless.As soon as it hit Newton’s head, it gained weight! and Newton gained physics! Weightlessness in satellites : There is a wrong notion that the astronauts in satellites experience no gravitational force because they are far away from the Earth. Actually the Earth satellites that orbit very close to Earth experience only gravitational force. The astronauts inside the satellite also experience the same gravitational force. Because of this, they cannot exert any force on the floor of the satellite. Thus, the floor of the satellite also cannot exert any normal force on the astronaut. Therefore, the astronauts inside a satellite are in the state of weightlessness. Not only the astronauts, but all the objects in the satellite will be in the state of weightlessness which is similar to that of a free fall. It is shown in the Figure . . a a g Figure . Apparent weight in the lift (a) (b) (c) (d) N = mg N = m(g–a) N = m(g+a) Apparent Weight < Actual Weight Apparent > Actual Weight Free Fall N = ; g = a Unit Gravitation shortcoming of the geocentric model over heliocentric model. . . Heliocentric system over geocentric system When the motion of the planets are observed in the night sky by naked eyes over a period of a few months, it can be seen that the planets move eastwards and reverse their motion for a while and return to eastward motion again. This is called “retrograde motion” of planets. Figure . shows the retrograde motion of the planet Mars. Careful observation for a period of a year clearly shows that Mars initially moves eastwards (February to June), then reverses its path and moves backwards (July, August, September). It changes its direction of motion once again and continues its forward motion (October onwards). In olden days, astronomers recorded the retrograde motion of all . Figure . The well known scientist Stephen Hawking in the state of weightlessness. ELEMENTARY IDEAS OF ASTRONOMY Astronomy is one of the oldest sciences in the history of mankind. In the olden days, astronomy was an inseparable part of physical science. It contributed a lot to the development of physics in the th century. In fact Kepler’s laws and Newton’s theory of gravitation were formulated and verified using astronomical observations and data accumulated over the centuries by famous astronomers like Hippachrus, Aristachrus, Ptolemy, Copernicus and Tycho Brahe. Without Tycho Brahe’s astronomical observations, Kepler’s laws would not have emerged. Without Kepler’s laws, Newton’s theory of gravitation would not have been formulated. It was mentioned in the beginning of this chapter that Ptolemy’s geocentric model was replaced by Copernicus’ heliocentric model. It is important to analyze and explain the Unit Gravitation Figure . Retrograde motion of planets Retrograde motion of Mars over a year Feb Mar Apr May Oct June Sep Aug Jul Nov Dec East West Figure . “Epicycle” motion of planetary objects around Earth, depicted with respect to months of observation. Earth Planet Epicycle Deferent Center of epicycle Earth visible planets and tried to explain the motion. According to Aristotle, the other planets and the Sun move around the Earth in the circular orbits. If it was really a circular orbit it was not known how the planet could reverse its motion for a brief interval. To explain this retrograde motion, Ptolemy introduced the concept of “epicycle” in his geocentric model. According to this theory, while the planet orbited the Earth, it also underwent another circular motion termed as “epicycle”. A combination of epicycle and circular motion around the Earth gave rise to retrograde motion of the planets with respect to Earth (Figure . ). Essentially Ptolemy retained the Earth centric idea of Aristotle and added the epicycle motion to it. Unit Gravitation on correctness of the Copernicus model over to Ptolemy’s model can be found in astronomy books. Students are encouraged to observe the motion of the planet Mars by naked eye and identify its retrograde motion. As mentioned above, to observe the retrograde motion six to seven months are required. So students may start their observation of Mars from the month of June and continue till April next year. Mars is the little bright planet with reddish color. The position of the planet Mars in the sky can be easily taken from ‘Google’. Activity But Ptolemy’s model became more and more complex as every planet was found to undergo retrograde motion. In the th century, the Polish astronomer Copernicus proposed the heliocentric model to explain this problem in a simpler manner. According to this model, the Sun is at the center of the solar system and all planets orbited the Sun. The retrograde motion of planets with respect to Earth is because of the relative motion of the planet with respect to Earth. The retrograde motion from the heliocentric point of view is shown in Figure . . Figure . shows that the Earth orbits around the Sun faster than Mars. Because of the relative motion between Mars and Earth, Mars appears to move backwards from July to October. In the same way the retrograde motion of all other planets was explained successfully by the Copernicus model. It was because of its simplicity, the heliocentric model slowly replaced the geocentric model. Historically, if any natural phenomenon has one or more explanations, the simplest one is usually accepted. Though this was not the only reason to disqualify the geocentric model, a detailed discussion a a b b e e d d c c g g West East Earth Sun Mars Figure . ‘Retrograde motion’ in heliocentric model Unit Gravitation Venus ° Mercury . ° Figure . Angle of elevation for Venus and Mercury from horizon The trigonometric relation satisfied by this right angled triangle is shown in Figure . . sin r where R = AU. AU sin sin . . Kepler’s Third Law and The Astronomical Distance When Kepler derived his three laws, he strongly relied on Tycho Brahe’s astronomical observation. In his third law, he formulated the relation between the distance of a planet from the Sun to the time period of revolution of the planet. Astronomers cleverly used geometry and trigonometry to calculate the distance of a planet from the Sun in terms of the distance between Earth and Sun. Here we can see how the distance of Mercury and Venus from the Sun were measured. The Venus and Mercury, being inner planets with respect to Earth, the maximum angular distance they can subtend at a point on Earth with respect to the Sun is degree for Venus and . degree for Mercury. It is shown in the Figure . Figure . shows that when Venus is at maximum elongation (i.e., degree) with respect to Earth, Venus makes degree to Sun. This allows us to find the distance between Venus and Sun. The distance between Earth and Sun is taken as one Astronomical unit ( AU). . Earth Sun Venus Venus Evening sky Morning sky Earth Sun Sun Su n Mercury Earth Sun Sun Venus Figure . Angle of elevation for Mercury from horizon Unit Gravitation Venus can be observed with the naked eye. We can see Venus during sunrise or sunset. Students are encouraged to observe the motion of Venus and verify that the maximum elevation is at degree and calculate the distance of Venus from the Sun. As pointed out already Google or Stellarium will be helpful in locating the position of Venus in the sky. Activity . . Measurement of radius of the Earth Around B.C a Greek librarian “Eratosthenes” who lived at Alexandria measured the radius of the Earth with a small error when compared with results using modern measurements. The technique he used involves lower school geometry and Here sin ° = . Hence Venus is at a distance of . AU from Sun. Similarly, the distance between Mercury ( θ is . degree) and Sun is calculated as . AU. To find the distance of exterior planets like Mars and Jupiter, a slightly different method is used. The distances of planets from the Sun is given in the table below. Table . a /T for different planets Planet semi major axis of the orbit(a) Period T (days) a /T Mercury . AU . . Venus . AU . . Earth . AU . . Mars . AU . . Jupiter . AU . . Saturn . AU , . . It is to be noted that to verify the Kepler’s law we need only high school level geometry and trigonometry. Figure . Measuring radius of The Earth . . Sun shadow angle Sun rays How to measure the sun shadow angle tan = L/h Shadow length (L) Pole height (h) No shadow Syne S Small shadow O Alexandria Unit Gravitation brilliant insight. He observed that during noon time of summer solstice the Sun’s rays cast no shadow in the city Syne which was located miles away from Alexandria. At the same day and same time he found that in Alexandria the Sun’s rays made . degree with local vertical as shown in the Figure . . He realized that this difference of . degree was due to the curvature of the Earth. The angle . degree is equivalent to radian. So q = rad; If S is the length of the arc between the cities of Syne and Alexandria, and if R is radius of Earth, then S q 500miles , so radius of the Earth R miles R =    miles R = miles mile is equal to . km. So, he measured the radius of the Earth to be equal to R = km, which is amazingly close to the correct value of km. The distance of the Moon from Earth was measured by a famous Greek astronomer Hipparchus in the rd century BC. To measure the radius of the Earth, choose two different places (schools) that are separated by at least km. It is important to note that these two places have to be along the same longitude of the Earth (For example Hosur and Kanyakumari lie along the same longitude of . ° E). Take poles of known length (h) and fix them vertically in the ground (it may be in the school playgrounds) at both the places. At exactly noon in both the places the length of the shadow (L) cast by each pole has to be noted down. Draw the picture like in Figure . . By using the equation tan L h , the angle in radian can be found at each place. The difference in angle ( θ′ ) is due to the curvature of the Earth. Now the distance between the two schools can be obtained from ‘Google maps’. Divide the distance with the angle ( θ′ in radians) which will give the radius of the Earth. Activity . . Interesting Astronomical Facts . Lunar eclipse and measurement of shadow of Earth On January , there was a total lunar eclipse which was observed from various places including Tamil Nadu. It is possible to measure the radius of shadow of the Earth at the point where the Moon crosses. Figure . illustrates this. When the Moon is inside the umbra shadow, it appears red in color. As soon as the Moon exits from the umbra Unit Gravitation Figure . Schematic diagram of umbra disk radius Earth Earth umbra shadow Moon Radius of umbra disc(R s ) The apparent radius of Earth’s umbra shadow = R s = . cm The apparent radius of the Moon = R m = . cm The ratio R s » The radius of the Earth’s umbra shadow is R s The radius of Moon R m = 1737km shadow, it appears in crescent shape. Figure . is the photograph taken by digital camera during Moon’s exit from the umbra shadow. Figure . Image of the Moon when it exits from umbra shadow By finding the apparent radii of the Earth’s umbra shadow and the Moon, the ratio of the these radii can be calculated. This is shown in Figures . and . . Figure . Total lunar eclipse Penumbra (partial shadow) Penumbra (partial shadow) Earth Moon’s orbit Sun Umbra (full shadow) Moon (eclipsed) Lunar Eclipse Unit Gravitation observe solar eclipse. But Moon’s orbit is tilted ° with respect to Earth’s orbit. Due to this ° tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment. This is shown in Figure . . Why do we have seasons on Earth? The common misconception is that ‘Earth revolves around the Sun, so when the Earth is very far away, it is winter and when the Earth is nearer, it is summer’. Actually, the seasons in the Earth arise due to the rotation of Earth around the Sun with . ° tilt. This is shown in Figure . Due to this . ° tilt, when the northern part of Earth is farther to the Sun, the southern part is nearer to the Sun. So when it is summer in the northern hemisphere, the southern hemisphere experience winter. Figure . Calculation of umbra radius Rs Rm The radius of the Earth’s umbra shadow is R s = km ≅ 4446km . The correct radius is km. The percentage of error in the calculation = . % . The error will reduce if the pictures taken using a high quality telescope are used. It is to be noted that this calculation is done using very simple mathematics. Early astronomers proved that Earth is spherical in shape by looking at the shape of the shadow cast by Earth on the Moon during lunar eclipse. . Why there are no lunar eclipse and solar eclipse every month? If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so during new Moon we can Unit Gravitation . Star’s apparent motion and spinning of the Earth The Earth’s spinning motion can be proved by observing star’s position over a night. Due to Earth’s spinning motion, the stars in sky appear to move in circular motion about the pole star as shown in Figure . Pole star is a star located exactly above the Earth’s axis of rotation, hence it appears to be stationary. The Star Polaris is our pole star. Note Point to ponder Using Sun rays and shadows, How will you prove that the Earth’s tilt is . ° ? Figure . Orbital tilt of the Moon Full moon (No eclipse) Full moon (Lunar eclipse) New moon (Solar eclipse) New moon (No eclipse) Orbital plane of earth The sun ° Figure . Seasons on Earth Unit Gravitation Subramanian Chandrasekar formulated the theory of black holes and explained the life of stars. These studies brought him the Nobel prize in the year . Another very notable Indian astrophysicist Meghnad Saha discovered the ionization formula which was useful in classifying stars. This formula is now known as “Saha ionization formula”. In the field of gravitation Amal Kumar Raychaudhuri solved an equation now known as “Raychaudhuri equation” which was a very important contribution. Another notable Indian Astrophysicist Jayant Narlikar made pioneering contribution in the field of astrophysics and has written interesting books on astronomy and astrophysics. IUCAA (Inter University Center for Astronomy and Astrophysics) is one of the important Indian research institutes where active research in astrophysics and gravitation are conducted. The institute was founded by Prof. J.V. Narlikar. Students are encouraged to read more about the recent developments in these fields. . . Recent developments of astronomy and gravitation Till the th century astronomy mainly depended upon either observation with the naked eye or telescopic observation. After the discovery of the electromagnetic spectrum at the end of the th century, our understanding of the universe increased enormously. Because of this development in the late th century it was found that Newton’s law of gravitation could not explain certain phenomena and showed some discrepancies. Albert Einstein formulated his ‘General theory of relativity’ which was one of the most successful theories of th century in the field of gravitation. In the twentieth century both astronomy and gravitation merged together and have grown in manifold. The birth and death of stars were more clearly understood. Many Indian physicists made very important contributions to the field of astrophysics and gravitation. Pole star Q O Rotation of the earth about PQ explains the apparent rotation of stars (in opposite direction) about PQ. If we observe the stars, it slowly changes its position and appears to move in a circle over the night. Pole star Figure . Star’s apparent circular motion due to Earth’s rotation. Unit Gravitation The motion of planets can be explained using Kepler’s laws. Kepler’s first law : All the planets in the solar system orbit the Sun in elliptical orbits with the Sun at one of the foci. Kepler’s second law: The radial vector line joining the Sun to a planet sweeps equal areas in equal intervals of time. Kepler’s third law: The ratio of the square of the time period of planet to the cubic power of semi major axis is constant for all the planets in the solar system. Newton’s law of gravitation states that the gravitational force between two masses is directly proportional to product of masses and inversely proportional to square of the distance between the masses. In vector form it is given by  F Gm m =− Gravitational force is a central force. Kepler’s laws can be derived from Newton’s law of gravitation. The gravitational field due to a mass m at a point which is at a distance r from mass m is given by  Gm =− . It is a vector quantity. The gravitational potential energy of two masses is given by U Gm m . It is a scalar quantity. The gravitational potential at a point which is at a distance r from mass m is given by Gm . It is a scalar quantity. The acceleration due to Earth’s gravity decreases as altitude increases and as depth increases. Due to rotation of the Earth, the acceleration due to gravity is maximum at poles and minimum at Earth’s equator. The (escape) speed of any object required to escape from the Earth’s gravitational field is v gR e e . It is independent of mass of the object. The energy of the satellite is negative. It implies that the satellite is bound to Earth’s gravitational force. Copernicus model explained that retrograde motion is due to relative motion between planets. This explanation is simpler than Ptolemy’s epicycle explanation which is complicated Copernicus and Kepler measured the distance between a planet and the Sun using simple geometry and trigonometry. years ago, Eratosthenes measured the radius of the Earth using simple geometry and trigonometry. SUM MA R Y Unit Gravitation Kepler’s Laws Altitude Variation Depth Variation Latitude Variation Astronomical Measurements Ptolemy Vs Copernicus Newton’s Law of Gravitation Satellites, Energy, Escape Speed Gravitational Field Gravitational Potential energy Gravitational Potential Planetary Motion Elementary Ideas of Astronomy Recent Developments Acceleration due to Gravity g C O N C EP T M AP Unit Gravitation EVALUATION I. Multiple Choice Questions . The linear momentum and position vector of the planet is perpendicular to each other at (a) perihelion and aphelion (b) at all points (c) only at perihelion (d) no point . If the masses of the Earth and Sun suddenly double, the gravitational force between them will (a) remain the same (b) increase times (c) increase times (d) decrease times . A planet moving along an elliptical orbit is closest to the Sun at distance r and farthest away at a distance of r . If v and v are linear speeds at these points respectively. Then the ratio v is (NEET ) (a) r (b) r (c) r (d) r . The time period of a satellite orbiting Earth in a cirular orbit is independent of . (a) Radius of the orbit (b) The mass of the satellite (c) Both the mass and radius of the orbit (d) Neither the mass nor the radius of its orbit . If the distance between the Earth and Sun were to be doubled from its present value, the number of days in a year would be (a) . (b) (c) . (d) . According to Kepler’s second law, the radial vector to a planet from the Sun sweeps out equal areas in equal intervals of time. This law is a consequence of (a) conservation of linear momentum (b) conservation of angular momentum (c) conservation of energy (d) conservation of kinetic energy . The gravitational potential energy of the Moon with respect to Earth is (a) always positive (b) always negative (c) can be positive or negative (d) always zero . The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K A , K B and K C respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then (NEET ) (a) K A > K B >K C (b) K B < K A <K C (c) K A < K B <K C (d) K B > K A >K C S B C Unit Gravitation . The work done by the Sun’s gravitational force on the Earth is (a) always zero (b) always positive (c) can be positive or negative (d) always negative . If the mass and radius of the Earth are both doubled, then the acceleration due to gravity g' (a) remains same (b) g (c) g (d) g . The magnitude of the Sun’s gravitational field as experienced by Earth is (a) same over the year (b) decreases in the month of January and increases in the month of July (c) decreases in the month of July and increases in the month of January (d) increases during day time and decreases during night time. . If a person moves from Chennai to Trichy, his weight (a) increases (b) decreases (c) remains same (d) increases and then decreases . An object of mass kg is hanging on a spring scale which is attached to the roof of a lift. If the lift is in free fall, the reading in the spring scale is (a) N (b) zero (c) N (d) . N . If the acceleration due to gravity becomes times its original value, then escape speed (a) remains same (b) times of original value (c) becomes halved (d) times of original value . The kinetic energy of the satellite orbiting around the Earth is (a) equal to potential energy (b) less than potential energy (c) greater than kinetic energy (d) zero Answers ) a ) c ) a ) b ) b ) b ) b ) a ) c ) b ) c ) b ) b ) b ) b II. Short Answer Questions . State Kepler’s three laws. . State Newton’s Universal law of gravitation. . Will the angular momentum of a planet be conserved? Justify your answer. . Define the gravitational field. Give its unit. . What is meant by superposition of gravitational field? . Define gravitational potential energy. . Is potential energy the property of a single object? Justify. . Define gravitational potential. . What is the difference between gravitational potential and gravitational potential energy? . What is meant by escape speed in the case of the Earth? . Why is the energy of a satellite (or any other planet) negative? . What are geostationary and polar satellites? . Define weight Unit Gravitation . Why is there no lunar eclipse and solar eclipse every month? . How will you prove that Earth itself is spinning? III. Long Answer Questions . Discuss the important features of the law of gravitation. . Explain how Newton arrived at his law of gravitation from Kepler’s third law. . Explain how Newton verified his law of gravitation. . Derive the expression for gravitational potential energy. . Prove that at points near the surface of the Earth, the gravitational potential energy of the object is U mgh . Explain in detail the idea of weightlessness using lift as an example. . Derive an expression for escape speed. . Explain the variation of g with lattitude. . Explain the variation of g with altitude. . Explain the variation of g with depth from the Earth’s surface. . Derive the time period of satellite orbiting the Earth. . Derive an expression for energy of satellite. . Explain in detail the geostationary and polar satellites. . Explain how geocentric theory is replaced by heliocentric theory using the idea of retrograde motion of planets. . Explain in detail the Eratosthenes method of finding the radius of Earth. . Describe the measurement of Earth’s shadow (umbra) radius during total lunar eclipse IV. Exercises . An unknown planet orbits the Sun with distance twice the semi major axis distance of the Earth’s orbit. If the Earth’s time period is T , what is the time period of this unknown planet? Ans: T . Assume that you are in another solar system and provided with the set of data given below consisting of the planets’ semi major axes and time periods. Can you infer the relation connecting semi major axis and time period? Planet (imaginary) Time period(T) (in year) Semi major axis (a) (in AU) Kurinji Mullai Marutham Neithal Paalai Ans: a µ . If the masses and mutual distance between the two objects are doubled, what is the change in the gravitational force between them? Ans: No change . Two bodies of masses m and 4m are placed at a distance r. Calculate the gravitational potential at a point on the line joining them where the gravitional field is zero. Ans: V Gm =− Unit Gravitation . If the ratio of the orbital distance of two planets d d = , what is the ratio of gravitational field experienced by these two planets? Ans: E = E . The Moon Io orbits Jupiter once in . days. The orbital radius of the Moon Io is 421700 km. Calculate the mass of Jupiter? Ans: . × kg . If the angular momentum of a planet is given by  t tj k i What is the torque experienced by the planet? Will the torque be in the same direction as that of the angular momentum? Ans:  τ = ti j . Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their mutual gravitational attraction. Calculate the speed of each