11th Physics Volume

] and the kinetic energy becomes zero as well. Therefore the final total energy of the Therefore, g g cos ( . ) From the expression ( . ), we can infer that at equator, ; g g .

The acceleration due to gravity is minimum. At poles λ = ° ; g g , it is maximum. At the equator, g is minimum. EXAMPLE .

Find out the value of ′ g in your school laboratory? Solution Calculate the latitude of the city or village where the school is located. The information is available in Google search. For example, the latitude of Chennai is approximately degree.

′ = g g ω λ cos Here ω R = (2x3. /86400) x (6400x10 ) = .4x10 − m s − . It is to be noted that the value of λ should be in radian and not in degree. degree is equivalent to .

rad. ′ = ) g cos . g m s − Points to Contemplate Suppose you move towards east-west along the same latitude. Will the value of ′ g change?

Unit Gravitation in which the object is thrown. Irrespective of whether the object is thrown vertically up, radially outwards or tangentially it requires the same initial speed to escape Earth’s gravity. It is shown in Figure . Earth Figure .

Escape speed independent of angle Lighter molecules such as hydrogen and helium have enough speed to escape from the Earth, unlike the heavier ones such as nitrogen and oxygen. (The average speed of hydrogen and helium atoms compaired with the escape speed of the Earth,is presented in the kinetic theory of gases, unit ). . .

Satellites, orbital speed and time period We are living in a modern world with sophisticated technological gadgets and are able to communicate to any place on Earth. This advancement was made possible because of our understanding of solar system. Communication mainly depends on the satellites that orbit the Earth (Figure . ).

Satellites revolve around the Earth just like the planets revolve around the Sun. Kepler’s laws are applicable to man- made satellites also. object becomes zero. This is for minimum energy and for minimum speed to escape.

Otherwise Kinetic energy can be nonzero. E f = According to the law of energy conservation, i ( . ) Substituting ( . ) in ( .

) we get, Mv GMM i Mv GMM i ( . ) Consider the escape speed, the minimum speed required by an object to escape Earth’s gravitational field, hence replace v i with v e . i.e, Mv GMM e GMM M e GM e Using g GM , gR e gR e ( . ) From equation ( .

) the escape speed depends on two factors: acceleration due to gravity and radius of the Earth. It is completely independent of the mass of the object. By substituting the values of g ( . m s − ) and R km e = , the escape speed of the Earth is v kms e - = .

. The escape speed is independent of the direction Unit Gravitation From equation ( . ) GM h h ( . ) GM h / ( .

) Squaring both sides of the equation ( . ), we get GM h π GM c = constant say c R h ( . ) Equation ( . ) implies that a satellite orbiting the Earth has the same relation between time and distance as that of Kepler’s law of planetary motion.

For a satellite orbiting near the surface of the Earth, h is negligible compared to the radius of the Earth R E . Then, GM R GM g R E since GM g = g ( . ) For a satellite of mass M to move in a circular orbit, centripetal force must be acting on the satellite. This centripetal force is provided by the Earth’s gravitational force.

satellite h Figure . Satellite revolving around the Earth. Mv h GMM h ) ( . ) As h increases, the speed of the satellite decreases.

Time period of the satellite: The distance covered by the satellite during one rotation in its orbit is equal to R h and time taken for it is the time period, T. Then Speed v Distance travelled Time taken h Unit Gravitation By substituting these values, the distance to the Moon from the surface of the Earth is calculated to be . × km . .

. Energy of an Orbiting Satellite The total energy of a satellite orbiting the Earth at a distance h from the surface of Earth is calculated as follows; The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is, U GM M h s ( . ) Here M s - mass of the satellite, M E - mass of the Earth, R E - radius of the Earth.

The Kinetic energy of the satellite is K E M v s = ( . ) Here v is the orbital speed of the satellite and is equal to GM h ) ( . ) Substituting the value of v in ( . ), the kinetic energy of the satellite becomes, K E GM M h s By substituting the values of R E = and g = .

m s − , the orbital time period is obtained as T ≅ minutes. EXAMPLE . Moon is the natural satellite of Earth and it takes days to go once around its orbit. Calculate the distance of the Moon from the surface of the Earth assuming the orbit of the Moon as circular.

Solution We can use Kepler’s third law, c R h c h / / c h ; 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.