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Gravitation

The document covers the principles of gravitation, including acceleration due to gravity, Newton's law of universal gravitation, and calculations for gravitational force and mass of celestial bodies. It also discusses satellite motion, critical velocity, escape velocity, and provides examples and exercises related to these concepts. Additionally, it explains the differences between natural and artificial satellites, as well as various applications of artificial satellites.

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GRAVITATION Peter Huruma Mammba Department of General Studies DODOMA POLYTECHNIC OF ENERGY AND EARTH RESOURCES MANAGEMENT (MADINI INSTITUTE) –DODOMA peter.huruma2011@gmail.com
GRAVITATION I. Acceleration Due to the Gravity II. Satellite and Planetary Motion III.Gravitational Field, Energy and Potential
Acceleration Due to the Gravity (g) • Is the acceleration that results in an object due to earth's gravity. • If air resistance is neglected, all bodies regardless of size and mass fall with the same acceleration under gravity alone. (i.e. 9.8 m/s2)
The following points may be noted about the g The value of g varies slightly from place to place on the earth's surface. However it is normally sufficiently accurate to use a value of 9.8 The acceleration due to the gravity always acts downward towards the center of the earth.
The following points may be noted about the g Like all accelerations, the acceleration due to the gravity is vector.
Newton’s Law of Universal Gravitation States that every body in the universe attracts every other body with a force which directly proportional to the product of their masses and inversely proportional to the square of distance between their centers.
Gravitation • Every object with mass attracts every other object with mass. • Newton realized that the force of attraction between two massive objects: • Increases as the mass of the objects increases. • Decreases as the distance between the objects increases.
Law of Universal Gravitation FG = G • G = Gravitational Constant= 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 • M1 and M2 = the mass of two bodies • r = the distance between them M1M2 r2
Example 1 Estimate the gravitational force between a 55 Kg woman and a 75 Kg man when they are 1.6 m apart. Solution 𝐹 = 𝐺 𝑚1 𝑚2 𝑟2 = 6.67x10−11 𝑥 55 𝑥 75 1.6 2 = 10−7 𝑁
Example 2 • Two 8 Kg spherical lead balls are placed with their centers 50 cm apart. What is the magnitude of the gravitational force each exerts on the other.
Solution • Here 𝑚1= 𝑚2 = 8 Kg; G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 ; r = 0.5 m 𝐹 = 𝐺 𝑚1 𝑚2 𝑟2 𝐹 = 6.67𝑥10−11 𝑥 8𝑥8 (0.5)2 ∴F= 1.71𝑥10−8
Example 3 • Assuming the orbit of the earth about the sun to be circular (it is actually slightly elliptical) with radius 1.5𝑥1011 𝑚, find the mass of the sun. The earth revolves around the sun in 3.15𝑥107 𝑠𝑒𝑐𝑜𝑛𝑑𝑠.
SOLUTION Centripetal force = Gravitational force Or 𝑚 𝑒 𝑣2 𝑅 = 𝑚 𝑒 𝑚 𝑠 𝑅2 𝑚 𝑠= 𝑅𝑣2 𝐺 v= speed of earth in its orbit around the sun G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2
SOLUTION… 2𝜋𝑅 𝑇 = 2𝜋𝑥 (1.5𝑥1011 ) 3.15𝑥107 = 3𝑥104 𝑚𝑠−1 𝑚 𝑠= (3𝑥104)2 𝑥1.5𝑥1011 6.67𝑥10−11 ∴ 𝒎 𝒔= 𝟐𝒙𝟏𝟎 𝟑𝟎 𝑲𝒈
Expression for the Acceleration Due to Gravity • According to the law of the gravitation, the force of attraction acting on the body due to earth is given by; 𝐹 = 𝐺 𝑚1 𝑚2 𝑟2 …………………….. i The attractive force which the earth exert on the object is simply weight of the object i.e. W= mg……………………..……..ii
From equation (i) and (ii), mg = 𝐺𝑀 𝐸 𝑚 𝑅 𝐸 2 g = 𝑮𝑴 𝑬 𝑹 𝑬 𝟐 ∴The equation for the expression for acceleration due to the gravity on the surface of the earth shows that the value of g does not depend on the mass m of the body.
Differences between G and g G g G is the universal gravitational Constant g is the acceleration due to the gravity G = 6.67 𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 Approximately g = 9.8 𝑚𝑠2, value of g on the earth varies from one place to another Constant through the universe Change every place on the earth. Example on the mon the value of g = 1 6 𝑡ℎ of that of the earth surface
• Using the law of universal gravitation and the measured value of the acceleration due to gravity, we can find, Mass of Earth Density of Earth
Mass of Earth • From equation, g = 𝑮𝑴 𝑬 𝑹 𝑬 𝟐 then; 𝑴 𝑬 = 𝑮g 𝑹 𝑬 𝟐 where ; G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , g = 9.8 𝑚𝑠−2 , 𝑹 𝑬=6.37𝑥106 𝑚 ∴Mass of earth, 𝑴 𝑬= 9.8𝑥 6.37𝑥106 2 6.67𝑥10−11 = 6𝑥1024 𝐾𝑔
Density of Earth • Let 𝜌 be the average density of the earth. Earth is the sphere of radius 𝑅 𝐸. Mass of Earth,𝑀 𝐸 = Volume (𝑉𝐸) x Density (𝜌) 4 3 𝜋𝑅 𝐸 3 𝜌 𝑀 𝐸 = 4 3 𝜋𝑅 𝐸 3 𝜌 ………i But 𝑴 𝑬 = 𝑮𝑹 𝑬 𝟐 g ……….ii
• Substitute the equation (i) into equation (ii); 𝑮𝑹 𝑬 𝟐 g = 4 3 𝜋𝑅 𝐸 3 𝜌 𝜌 = 𝟑g 𝟒𝜋𝑮𝑹 𝑬 G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2, g = 9.8 𝑚𝑠−2, 𝑹 𝑬=6.37𝑥106 𝑚, g = 9.8 𝑚𝑠−2 𝜌 = 3 𝑥 9.8 𝟒𝜋 𝑥 6.37𝑥106 𝑥6.67𝑥10−11 ∴ 𝝆 = 𝟓. 𝟓 𝒙𝟏𝟎 𝟑 𝑲𝒈 𝒎−𝟑
Example 4 What will be the acceleration due to the gravity on the surface of the moon if its radius is 1/5th the radius of the earth and its mass 1/80th of the mass of the earth? SOLUTION The acceleration due to the gravity on the surface of the earth is given by; 𝑔 𝐸 = 𝐺𝑀 𝐸 𝑅 𝐸 2
SOLUTION The acceleration due to the gravity on the surface of the moon is given by; 𝑔 𝑚 = 𝐺𝑀 𝑚 𝑅 𝑚 2 𝑔 𝑚 𝑔 𝐸 = 𝑀 𝑚 𝑀 𝐸 𝑥 𝑅 𝐸 𝑅 𝑚 2 , Now 𝑀 𝑚 𝑀 𝐸 = 1 80 and 𝑅 𝐸 𝑅 𝑚 = 4 𝑔 𝑚 𝑔 𝐸 = 1 80 𝑥 4 1 2 = 1 5 ∴ 𝑔 𝑚 = 𝑔 𝐸 5
Problems for Practice 1 • Two 20 Kg spree are placed with their centers 50 cm apart. What is the magnitude of gravitational force exerts on the other? (1.07 𝑥 107 𝑁) • A spree of mass 40 Kg is attracted by another of mass 15 Kg when their centers are 0.2 m apart, with a force of 9.8 𝑥 10−7 𝑁. Calculate the constant of gravitational. (6.53 𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 )
Problems for Practice … • Assuming the mean density of the earth is 5500 𝐾𝑔𝑚−3, that G is 6.67𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 and that radius of the earth is 6400 Km, Find the value for the acceleration of free fall at the earth’s surface. (9.9 𝑚𝑠−2) • The Acceleration due to the gravity at the moon’s surface is 1.67 𝑚𝑠−2 . If the radius of the moon is 1.74 𝑥 106 𝑚 and 6.67𝑥 10−11 𝑁𝑚2 𝐾𝑔−2, calculate the mass of the moon. (7.58 𝑥 1022 𝐾𝑔)
Satellite and Planetary Motion
Natural and Artificial Satellite • Natural Satellite is a heavenly body revolves around a planet in a close and stable orbit. • For examples, moon is the natural satellite of the earth, • It goes round the earth in about 27.3 days in a nearly circular orbit of radius 3.84 x 10^5 Km.
Earth satellite • Artificial Satellite; is a man-made satellite that orbits around the earth or some other heavenly body. • For example, the communication satellite are used routinely to transmit information around the globe.
Projection of a Satellite • Why it is necessary to have at least a two stages rocket to launch a satellite? • A rocket with at least two stages is required to launch a satellite because; The first stage is used to carry the satellite up to the desired height. In the second stage, rocket is turned horizontally (through 90 degrees) and the satellite is fired with the proper horizontal velocity to perform circular motion around the earth.
Orbital velocity of the satellite • Is the velocity required to put a satellite into a given orbit. • It is also known as Critical velocity (Vc) of the satellite
Critical Velocity of the Satellite • Let M = Mass pf the earth m = Mass of the satellite ( ) h = Height of the satellite above the earth r = R + h (the distance of the satellite form the center of the earth) V - Velocity of projection in a horizontal direction 𝑉𝐶 - Critical velocity 𝑉𝐸- Escape velocity
Critical Velocity of the Satellite • The centripetal force necessary for a circular motion of the satellite around the Earth is proved by the gravitational force of attraction between the Earth and the satellite. Centripetal force = Gravitational force 𝑚𝑉𝑐 2 𝑟 = 𝐺𝑀𝑚 𝑟2 𝑉𝑐 2 = 𝐺𝑀 𝑟 𝑉𝑐 = 𝐺𝑀 𝑟
Critical Velocity of the Satellite • Factors on which Critical velocity of a satellite depends on; 1. Mass of the planet 2. Radius of the planet 3. Height of the planet NB. Critical velocity is not dependent on the mass of the satellite
Critical Velocity of the Satellite • But we know that 𝑔ℎ = 𝐺M (R + h)2 = 𝐺M 𝑟2 GM = 𝑔ℎ(R + h)2 …………………………………….(i) Substitute the equation (i) to the equation of the critical velocity. We get 𝑉𝑐 = 𝑔ℎ(R + h)2 R + h 𝑉𝑐 = 𝑔ℎ(R + h)
Special Case of Critical Velocity of the Satellite • When the satellite orbits very close to the surface of the earth, h ≅ 0. Therefore, equation of the critical velocity will be; • Now g = 9.8 𝑚𝑠−2 , 𝑹 =6.37𝑥106 𝑚 𝑉𝑐 = 9.8 𝑥 6.4 𝑥106 ≈ 8 𝑥 103 𝑚𝑠−1 ∴≈ 8 𝐾𝑚𝑠−1 Thus the orbital speed of the satellite close to the earth’s surface is about 8 𝐾𝑚𝑠−1 𝑉𝑐 = gR
Time Period of the Satellite •Is the time taken by the satellite to complete one revolution around the earth •It is denoted by T T = 𝑪𝒊𝒓𝒄𝒖𝒎𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒐𝒓𝒃𝒊𝒕 𝑶𝒓𝒃𝒊𝒕𝒂𝒍 𝑺𝒑𝒆𝒆𝒅
Time Period of the Satellite 𝑇 = 2𝜋(𝑅+ℎ) 𝑉𝑐 …………………… (i) But; 𝑉𝑐 = 𝐺𝑀 R+h ……………….. (ii) Putting the value of 𝑉𝑐 given by equation (ii) into equation (i), we have, 𝑇 = 2𝜋 𝑥 (𝑅 + ℎ)3 𝐺𝑀
Special Case of Time Period of the Satellite • When the satellite orbits very close to the surface of the earth, h ≅ 0. therefore, equation of the time period will be; 𝑇 = 2𝜋 𝑥 𝑅3 𝐺𝑀 ………………… (i) But; g𝑅2 = 𝐺𝑀 ……………….. (ii) Substitute the value of GM given by equation (ii) into equation (i), we have, • Now • g = 9.8 𝑚𝑠−2, 𝑹 =6.37𝑥106 𝑚 • 𝑇 = 2𝜋 𝑥 6.37𝑥106 9.8 = 5075 s = 84 Minutes ∴ Thus the orbital speed of the satellite revolving very near to the earth’s surface is about 8 𝐾𝑚𝑠−1 and its period of revolution is nearly 84 minutes. 𝑻 = 𝟐𝝅 𝒙 𝑹 g
Example 5 • Calculate the velocity required for a satellite moving in a circular orbit 200 Km above the earth’s surface. Given that radius of the earth is 6380 Km and mass of the earth = 5.98 𝑥 1024 𝐾𝑔 SOLUTION G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , 𝑹 𝑬 = 6.37𝑥106 𝑚 , h =200 𝑥 103 𝑚, M = 5.98 𝑥 1024 𝐾𝑔 𝑉𝑐 = 𝐺𝑀 𝑅+ℎ 𝑉𝑐 = 6.67𝑥10−11 𝑥 5.98 𝑥 1024 6.37𝑥106+200 𝑥 103 ∴ = 7.79 𝑥 103 𝑚𝑠−1
Example 6 •The period of moon around the earth is 27.3 days and the radius of its orbit is 3.9 𝑥 105 𝐾𝑚. If G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , find the mass of the earth SOLUTION T = 27.3 days = 27.3 𝑥 24 𝑥 60 𝑥 60 𝑠, R+h = 3.9 𝑥 108 𝑚, 𝑇 = 2𝜋 𝑥 (𝑅 + ℎ)3 𝐺𝑀 𝑀 = 4𝜋2 𝑅 + ℎ 3 𝐺𝑇2 𝑀 = 4𝜋2 3.9 𝑥 10 3 6.67𝑥10−11 𝑥 27.3 𝑥 24 𝑥 60 𝑥 60 2 ∴ 𝑀 = 6.31 𝑥 1024 𝐾𝑔
Problems for Practice 2 • A satellite is revolving in a circular orbit at a distance of 2620 Km from the surface of the earth. Calculate the orbital velocity and the period of revolution of the satellite. Radius of the earth = 63802 Km, mass of the earth = 6 𝑥 1024 𝐾𝑔 and G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 . (6.67 𝑲𝒎𝒔−𝟏 ; 2.35 Hours) • An earth’s satellite makes a circle around the earth in 90 minutes. Calculate the height of the satellite above the surface of the earth. Given that radius of the earth = 6400 Km and g = 9.8 𝑚𝑠−2. (268 Km)
Uses of Artificial Satellite • They are used to learn about the atmosphere near the earth. • They are used to forecast weather. • They are used to study radiation from the sun and the outer space. • They are used to receive and transmit various radio and television signals. • They are used to know the exact shape and dimensions of the earth. • Space fights are possible due to artificial satellite.
Escape Velocity
Escape velocity • Is the minimum velocity which it is to be projected so that it just overcome the gravitational pull of the earth (or any other planet)
• In order to project a body with escape velocity, we give kinetic energy to it. Let us calculate this Energy. • Suppose a body of mass m is projected upward with escape velocity . When the body is at point P at a distance x from the center of the earth, the gravitational force of attraction exerted by the earth on the body is 𝐹 = 𝐺 𝑀𝑚 𝑥2 In moving a small distance ∆𝑥 against this gravitational force, the small work done at the expense of the KE of the body is given by; ∆𝑊 = 𝐹∆𝑥 = 𝐺 𝑀𝑚 𝑥2 ∆𝑥 P Fig. PH x Q ∆𝑥 R M = Mass of the earth R = Radius of the earth
• Total work done (W) in moving the body from earth’s surface (where x=R) to infinity (where 𝑥 = ∞) is given by; 𝑊 = 𝑅 ∞ 𝐺𝑀𝑚 𝑥2 𝑑𝑥 = 𝐺𝑀𝑚 𝑅 ∞ 𝑥−2 𝑑𝑥 = 𝐺𝑀𝑚 − 1 𝑥 𝑅 ∞ = 𝐺𝑀𝑚 𝑅 ∴ 𝑊 = 𝐺𝑀𝑚 𝑅
• If the body is to be able to do this amount of work (and so escape), it needs to have at least this amount of KE at the moment it is projected. Therefore, escape velocity is given by; 1 2 𝑚𝑉𝑒 2 = 𝐺𝑀𝑚 𝑅 𝑽 𝒆 = 𝟐𝑮𝑴 𝑹
The escape velocity can be written in other equivalent form as shown below; • The acceleration due to the gravity of the surface of the earth is; g = 𝐺𝑀 𝑅2 or 𝐺𝑀 =g𝑅2 Substitute the equation above to the general equation of escape velocity; 𝑉𝑒 = 𝟐g 𝑅2 𝑹 ∴ 𝑉𝑒 = 𝟐g𝑹
Example 7 • What is the escape velocity for a rocket on the surface of the Marks? Mass of Marks = 6.58 𝑥 1023 𝐾𝑔 and radius of Mars = 3.38 𝑥 106 𝑚. Solution 𝑽 𝒆 = 𝟐𝑮𝑴 𝒎 𝑹 𝒎 𝑴 𝒎= 6.58 𝑥 1023 𝐾𝑔, 𝑹𝒎 = 3.38 𝑥 106 𝑚, G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 𝑽 𝒆 = 𝟐 𝑥 6.67𝑥10−11 𝑥 6.58 𝑥 1023 3.38 𝑥 106 ∴ 𝑽 𝒆 = 5.1 𝑥 103 𝑚𝑠−1
Example 8 • Find the velocity of escape at the moon given that its radius is 1.7 𝑥 106 𝑚 and the value of g at its surface is 1.63 𝑚𝑠−2 . SOLUTION 𝑉𝑒 = 𝟐g𝑹 Here; g = 1.63 𝑚𝑠−2, R = 1.7 𝑥 106 𝑚 𝑉𝑒 = 𝟐 𝑥 1.63 𝑥1.7 𝑥 106 𝑉𝑒 = 2.354 𝑥 103 𝑚𝑠−1
Kepler’s First Law • The path of each planet about the sun is an ellipse with sun at the one focus of the ellipse.
What is an ellipse? 2 foci An ellipse is a geometric shape with 2 foci instead of 1 central focus, as in a circle. The sun is at one focus with nothing at the other focus. FIRST LAW OF PLANETARY MOTION
An ellipse also has… …a major axis …and a minor axis Semi-major axis Perihelion Aphelion Perihelion: When Mars or any another planet is closest to the sun. Aphelion: When Mars or any other planet is farthest from the sun.
Kepler’s Second Law • Each planet moves in such a way that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal periods of time.
Kepler also found that Mars changed speed as it orbited around the sun: faster when closer to the sun, slower when farther from the sun… A B But, areas A and B, swept out by a line from the sun to Mars, were equal over the same amount of time. SECOND LAW OF PLANETARY MOTION
Kepler’s Third Law • The square of the period of any planet (time needed for one revolution about the sun) is directly proportional to the cube of the plane’s average distance from the sun. 𝑇2 ∝ 𝑟3 Or 𝑇1 2 𝑇2 2 = 𝑟1 3 𝑟2 3 ∴ 𝑇2 𝑟3 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Derivation of Newton’s Law of Gravitation from Kepler's Third Law • Newton was able to show that Kepler's law could be derived mathematically from his law of universal gravitation and his laws of motion. • He used Kepler’s third law as evidence in favor of his law of universal gravitation.
Derivation continue… • Consider a planet of mass m revolving around the sun of mass M in a circular orbit of radius r. • Suppose 𝑣 is the orbital speed of the planet and T is its time period. • In time, T, the planet travels a distance 2𝜋𝑟. 𝑇 = 2𝜋𝑟 𝑣 or 𝑣 = 2𝜋𝑟 𝑇 • The centripetal force F required to keep the planet in the circular orbit is; 𝐹 = 𝑚𝑣2 𝑟 = This centripetal force is provided by the sun’s gravitational force on the planet.
• According to Kepler’s third law 𝑇2 𝑟3 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (k) or 𝑇2 = 𝑘𝑟3 Putting 𝑇2 = 𝑘𝑟3 in the derive centripetal force, we have ; 𝐹 = 4𝜋2 𝑘 𝑥 𝑚 𝑟2 • From Universal F = G 𝑀𝑚 𝑟2
Gravitational Field due to a Material of a Body • Is the space around the body in which any other mas experiences a force of attraction. E.g. the gravitational field of the earth
Gravitational Field Strength • Is defined as the force per unit mass acting on a test mass placed at that point. 𝐸 = 𝐹 𝑚 Its SI unit is 𝑁𝐾𝑔−1
Gravitational Field Strength Due to the Earth • Since, 𝐸 = 𝐹 𝑚 …………………….. (i) and, 𝐹 = 𝐺𝑀𝑚 𝑅2 ……………………….. (ii) Substitute equation (ii) into equation (i), we have; 𝐸 = 𝐺𝑀 𝑅2 • The field strength at any point in a gravitational field is equal to the gravitational acceleration of any mass placed at that point. • i.e. g = 𝑮𝑴 𝑬 𝑹 𝑬 𝟐 = 𝐸 𝐸 = 𝐺𝑴 𝑬 𝑹 𝑬 𝟐
Gravitational Potential Energy • The gravitational P.E of a body at a point in the gravitational field is defined as the amount of work done in bringing the body from infinity to that point.
Gravitation P.E… • Suppose a body of mass m is situated outside the earth at Point A at a distance r from the center of the earth. It’s Gravitational P.E (𝑼 𝑨) is W.D • Suppose at any instant the body is at point B at a distance x from the center of the earth. The gravitational force exerted by the earth on the body at B is 𝐹 = 𝐺𝑀𝑚 𝑥2 Consider the earth to be a spherical of radius R and Mass M R o c B dx x a r
Gravitation P.E… • Small amount of work done when the body moves from B to C is; 𝑑𝑊 = 𝐹𝑑𝑥 = 𝐺𝑀𝑚 𝑥2 𝑑𝑥 • Total work done by the gravitational force when the body of mass m at a distance r from the center of the earth is; 𝑊 = ∞ 𝑟 𝐺𝑀𝑚 𝑥2 𝑑𝑥 = 𝐺𝑀𝑚 ∞ 𝑟 1 𝑥2 𝑑𝑥 𝑊 = 𝐺𝑀𝑚 𝑥−1 −1 ∞ 𝑟 = −𝐺𝑀𝑚 1 𝑟 − 1 ∞ 𝑊 = − 𝐺𝑀𝑚 𝑟 ∴ Therefore, gravitational potential energy (𝑼 𝑨) of a body of mass m at a distance r from the center of the earth is; 𝑼 𝑨 = − 𝑮𝑴𝒎 𝒓
Kinetic Energy of the Satellite • Suppose a satellite of mas m moves round the earth in a circular orbit at the height h above the surface of the earth. The radius of the orbit of the satellite is (R + h). If v is the speed of the satellite in the orbit, then, KE of the satellite = 1 2 𝑚𝑉𝑐 2 ‘……………………………(i) The orbital velocity of the satellite is 𝑉𝑐 = 𝐺𝑀 𝑟 ………………………………………. (ii)
Kinetic Energy of the Satellite • Suppose a satellite of mas m moves round the earth of mas M in a circular orbit at the height h above the surface of the earth. • The radius of the orbit of the satellite is (R + h). If v is the speed of the satellite in the orbit, then, K.E of the satellite = 1 2 𝑚𝑉𝑐 2 ……(i) The orbital velocity of the satellite is; 𝑉𝑐 = 𝐺𝑀 (R + h) ……………. (ii) • Substitute equation (i) into equation (ii), we get; ∴ 𝐾. 𝐸 = 𝐺𝑀𝑚 2(R + h)
Total Energy of the Satellite • The satellite in orbit around the earth has both K.E and Potential Energy(𝑼 𝑨). Total Energy, E = K.E + 𝑼 𝑨 = 𝐺𝑀𝑚 2(R + h) + − 𝑮𝑴𝒎 R + h ∴ E = - 𝐺𝑀𝑚 2(R + h)
Example 9 • A spaceship is stationed on Mars. How much energy must be expend on the spaceship to rocket it out of the solar system? Mass of the spaceship = 1000Kg, mass of the sun = 2𝑥1030 𝐾𝑔, Mass of mars = 6.4𝑥1023 𝐾𝑔, radius of mars = 3395 Km. radius of orbit of mars = 2.28𝑥108 𝐾𝑚, G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 .
Solution Data given; Mass of the spaceship = 1000Kg, M ass of the sun = 2𝑥1030 𝐾𝑔, Mass of mars = 6.4𝑥1023 𝐾𝑔, radius of mars = 3395 Km, radius of orbit of Mars = 2.28𝑥108 𝐾𝑚,  G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , E = ?. E = - 𝐺𝑀𝑚 2(R + h) E = - 6.67𝑥10−11 𝑥 2𝑥1030 𝑥1000 2 x (2.28𝑥1011+3.395 𝑥 106) E = −2.9 𝑥 1011 𝐽 ∴ The energy extended on the spaceship = 𝟐. 𝟗 𝒙 𝟏𝟎 𝟏𝟏 𝑱
Gravitation

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Gravitation

  • 1.
    GRAVITATION Peter Huruma Mammba Departmentof General Studies DODOMA POLYTECHNIC OF ENERGY AND EARTH RESOURCES MANAGEMENT (MADINI INSTITUTE) –DODOMA peter.huruma2011@gmail.com
  • 2.
    GRAVITATION I. Acceleration Dueto the Gravity II. Satellite and Planetary Motion III.Gravitational Field, Energy and Potential
  • 4.
    Acceleration Due tothe Gravity (g) • Is the acceleration that results in an object due to earth's gravity. • If air resistance is neglected, all bodies regardless of size and mass fall with the same acceleration under gravity alone. (i.e. 9.8 m/s2)
  • 5.
    The following pointsmay be noted about the g The value of g varies slightly from place to place on the earth's surface. However it is normally sufficiently accurate to use a value of 9.8 The acceleration due to the gravity always acts downward towards the center of the earth.
  • 6.
    The following pointsmay be noted about the g Like all accelerations, the acceleration due to the gravity is vector.
  • 7.
    Newton’s Law of UniversalGravitation States that every body in the universe attracts every other body with a force which directly proportional to the product of their masses and inversely proportional to the square of distance between their centers.
  • 8.
    Gravitation • Every objectwith mass attracts every other object with mass. • Newton realized that the force of attraction between two massive objects: • Increases as the mass of the objects increases. • Decreases as the distance between the objects increases.
  • 9.
    Law of UniversalGravitation FG = G • G = Gravitational Constant= 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 • M1 and M2 = the mass of two bodies • r = the distance between them M1M2 r2
  • 10.
    Example 1 Estimate thegravitational force between a 55 Kg woman and a 75 Kg man when they are 1.6 m apart. Solution 𝐹 = 𝐺 𝑚1 𝑚2 𝑟2 = 6.67x10−11 𝑥 55 𝑥 75 1.6 2 = 10−7 𝑁
  • 11.
    Example 2 • Two8 Kg spherical lead balls are placed with their centers 50 cm apart. What is the magnitude of the gravitational force each exerts on the other.
  • 12.
    Solution • Here 𝑚1=𝑚2 = 8 Kg; G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 ; r = 0.5 m 𝐹 = 𝐺 𝑚1 𝑚2 𝑟2 𝐹 = 6.67𝑥10−11 𝑥 8𝑥8 (0.5)2 ∴F= 1.71𝑥10−8
  • 13.
    Example 3 • Assumingthe orbit of the earth about the sun to be circular (it is actually slightly elliptical) with radius 1.5𝑥1011 𝑚, find the mass of the sun. The earth revolves around the sun in 3.15𝑥107 𝑠𝑒𝑐𝑜𝑛𝑑𝑠.
  • 14.
    SOLUTION Centripetal force =Gravitational force Or 𝑚 𝑒 𝑣2 𝑅 = 𝑚 𝑒 𝑚 𝑠 𝑅2 𝑚 𝑠= 𝑅𝑣2 𝐺 v= speed of earth in its orbit around the sun G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2
  • 15.
    SOLUTION… 2𝜋𝑅 𝑇 = 2𝜋𝑥 (1.5𝑥1011 ) 3.15𝑥107 = 3𝑥104 𝑚𝑠−1 𝑚𝑠= (3𝑥104)2 𝑥1.5𝑥1011 6.67𝑥10−11 ∴ 𝒎 𝒔= 𝟐𝒙𝟏𝟎 𝟑𝟎 𝑲𝒈
  • 16.
    Expression for theAcceleration Due to Gravity • According to the law of the gravitation, the force of attraction acting on the body due to earth is given by; 𝐹 = 𝐺 𝑚1 𝑚2 𝑟2 …………………….. i The attractive force which the earth exert on the object is simply weight of the object i.e. W= mg……………………..……..ii
  • 17.
    From equation (i)and (ii), mg = 𝐺𝑀 𝐸 𝑚 𝑅 𝐸 2 g = 𝑮𝑴 𝑬 𝑹 𝑬 𝟐 ∴The equation for the expression for acceleration due to the gravity on the surface of the earth shows that the value of g does not depend on the mass m of the body.
  • 18.
    Differences between Gand g G g G is the universal gravitational Constant g is the acceleration due to the gravity G = 6.67 𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 Approximately g = 9.8 𝑚𝑠2, value of g on the earth varies from one place to another Constant through the universe Change every place on the earth. Example on the mon the value of g = 1 6 𝑡ℎ of that of the earth surface
  • 19.
    • Using thelaw of universal gravitation and the measured value of the acceleration due to gravity, we can find, Mass of Earth Density of Earth
  • 20.
    Mass of Earth •From equation, g = 𝑮𝑴 𝑬 𝑹 𝑬 𝟐 then; 𝑴 𝑬 = 𝑮g 𝑹 𝑬 𝟐 where ; G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , g = 9.8 𝑚𝑠−2 , 𝑹 𝑬=6.37𝑥106 𝑚 ∴Mass of earth, 𝑴 𝑬= 9.8𝑥 6.37𝑥106 2 6.67𝑥10−11 = 6𝑥1024 𝐾𝑔
  • 21.
    Density of Earth •Let 𝜌 be the average density of the earth. Earth is the sphere of radius 𝑅 𝐸. Mass of Earth,𝑀 𝐸 = Volume (𝑉𝐸) x Density (𝜌) 4 3 𝜋𝑅 𝐸 3 𝜌 𝑀 𝐸 = 4 3 𝜋𝑅 𝐸 3 𝜌 ………i But 𝑴 𝑬 = 𝑮𝑹 𝑬 𝟐 g ……….ii
  • 22.
    • Substitute theequation (i) into equation (ii); 𝑮𝑹 𝑬 𝟐 g = 4 3 𝜋𝑅 𝐸 3 𝜌 𝜌 = 𝟑g 𝟒𝜋𝑮𝑹 𝑬 G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2, g = 9.8 𝑚𝑠−2, 𝑹 𝑬=6.37𝑥106 𝑚, g = 9.8 𝑚𝑠−2 𝜌 = 3 𝑥 9.8 𝟒𝜋 𝑥 6.37𝑥106 𝑥6.67𝑥10−11 ∴ 𝝆 = 𝟓. 𝟓 𝒙𝟏𝟎 𝟑 𝑲𝒈 𝒎−𝟑
  • 23.
    Example 4 What willbe the acceleration due to the gravity on the surface of the moon if its radius is 1/5th the radius of the earth and its mass 1/80th of the mass of the earth? SOLUTION The acceleration due to the gravity on the surface of the earth is given by; 𝑔 𝐸 = 𝐺𝑀 𝐸 𝑅 𝐸 2
  • 24.
    SOLUTION The acceleration dueto the gravity on the surface of the moon is given by; 𝑔 𝑚 = 𝐺𝑀 𝑚 𝑅 𝑚 2 𝑔 𝑚 𝑔 𝐸 = 𝑀 𝑚 𝑀 𝐸 𝑥 𝑅 𝐸 𝑅 𝑚 2 , Now 𝑀 𝑚 𝑀 𝐸 = 1 80 and 𝑅 𝐸 𝑅 𝑚 = 4 𝑔 𝑚 𝑔 𝐸 = 1 80 𝑥 4 1 2 = 1 5 ∴ 𝑔 𝑚 = 𝑔 𝐸 5
  • 25.
    Problems for Practice1 • Two 20 Kg spree are placed with their centers 50 cm apart. What is the magnitude of gravitational force exerts on the other? (1.07 𝑥 107 𝑁) • A spree of mass 40 Kg is attracted by another of mass 15 Kg when their centers are 0.2 m apart, with a force of 9.8 𝑥 10−7 𝑁. Calculate the constant of gravitational. (6.53 𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 )
  • 26.
    Problems for Practice… • Assuming the mean density of the earth is 5500 𝐾𝑔𝑚−3, that G is 6.67𝑥 10−11 𝑁𝑚2 𝐾𝑔−2 and that radius of the earth is 6400 Km, Find the value for the acceleration of free fall at the earth’s surface. (9.9 𝑚𝑠−2) • The Acceleration due to the gravity at the moon’s surface is 1.67 𝑚𝑠−2 . If the radius of the moon is 1.74 𝑥 106 𝑚 and 6.67𝑥 10−11 𝑁𝑚2 𝐾𝑔−2, calculate the mass of the moon. (7.58 𝑥 1022 𝐾𝑔)
  • 27.
  • 28.
    Natural and ArtificialSatellite • Natural Satellite is a heavenly body revolves around a planet in a close and stable orbit. • For examples, moon is the natural satellite of the earth, • It goes round the earth in about 27.3 days in a nearly circular orbit of radius 3.84 x 10^5 Km.
  • 29.
    Earth satellite • ArtificialSatellite; is a man-made satellite that orbits around the earth or some other heavenly body. • For example, the communication satellite are used routinely to transmit information around the globe.
  • 30.
    Projection of aSatellite • Why it is necessary to have at least a two stages rocket to launch a satellite? • A rocket with at least two stages is required to launch a satellite because; The first stage is used to carry the satellite up to the desired height. In the second stage, rocket is turned horizontally (through 90 degrees) and the satellite is fired with the proper horizontal velocity to perform circular motion around the earth.
  • 31.
    Orbital velocity ofthe satellite • Is the velocity required to put a satellite into a given orbit. • It is also known as Critical velocity (Vc) of the satellite
  • 32.
    Critical Velocity ofthe Satellite • Let M = Mass pf the earth m = Mass of the satellite ( ) h = Height of the satellite above the earth r = R + h (the distance of the satellite form the center of the earth) V - Velocity of projection in a horizontal direction 𝑉𝐶 - Critical velocity 𝑉𝐸- Escape velocity
  • 33.
    Critical Velocity ofthe Satellite • The centripetal force necessary for a circular motion of the satellite around the Earth is proved by the gravitational force of attraction between the Earth and the satellite. Centripetal force = Gravitational force 𝑚𝑉𝑐 2 𝑟 = 𝐺𝑀𝑚 𝑟2 𝑉𝑐 2 = 𝐺𝑀 𝑟 𝑉𝑐 = 𝐺𝑀 𝑟
  • 34.
    Critical Velocity ofthe Satellite • Factors on which Critical velocity of a satellite depends on; 1. Mass of the planet 2. Radius of the planet 3. Height of the planet NB. Critical velocity is not dependent on the mass of the satellite
  • 35.
    Critical Velocity ofthe Satellite • But we know that 𝑔ℎ = 𝐺M (R + h)2 = 𝐺M 𝑟2 GM = 𝑔ℎ(R + h)2 …………………………………….(i) Substitute the equation (i) to the equation of the critical velocity. We get 𝑉𝑐 = 𝑔ℎ(R + h)2 R + h 𝑉𝑐 = 𝑔ℎ(R + h)
  • 36.
    Special Case ofCritical Velocity of the Satellite • When the satellite orbits very close to the surface of the earth, h ≅ 0. Therefore, equation of the critical velocity will be; • Now g = 9.8 𝑚𝑠−2 , 𝑹 =6.37𝑥106 𝑚 𝑉𝑐 = 9.8 𝑥 6.4 𝑥106 ≈ 8 𝑥 103 𝑚𝑠−1 ∴≈ 8 𝐾𝑚𝑠−1 Thus the orbital speed of the satellite close to the earth’s surface is about 8 𝐾𝑚𝑠−1 𝑉𝑐 = gR
  • 37.
    Time Period ofthe Satellite •Is the time taken by the satellite to complete one revolution around the earth •It is denoted by T T = 𝑪𝒊𝒓𝒄𝒖𝒎𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒐𝒇 𝒕𝒉𝒆 𝒐𝒓𝒃𝒊𝒕 𝑶𝒓𝒃𝒊𝒕𝒂𝒍 𝑺𝒑𝒆𝒆𝒅
  • 38.
    Time Period ofthe Satellite 𝑇 = 2𝜋(𝑅+ℎ) 𝑉𝑐 …………………… (i) But; 𝑉𝑐 = 𝐺𝑀 R+h ……………….. (ii) Putting the value of 𝑉𝑐 given by equation (ii) into equation (i), we have, 𝑇 = 2𝜋 𝑥 (𝑅 + ℎ)3 𝐺𝑀
  • 39.
    Special Case ofTime Period of the Satellite • When the satellite orbits very close to the surface of the earth, h ≅ 0. therefore, equation of the time period will be; 𝑇 = 2𝜋 𝑥 𝑅3 𝐺𝑀 ………………… (i) But; g𝑅2 = 𝐺𝑀 ……………….. (ii) Substitute the value of GM given by equation (ii) into equation (i), we have, • Now • g = 9.8 𝑚𝑠−2, 𝑹 =6.37𝑥106 𝑚 • 𝑇 = 2𝜋 𝑥 6.37𝑥106 9.8 = 5075 s = 84 Minutes ∴ Thus the orbital speed of the satellite revolving very near to the earth’s surface is about 8 𝐾𝑚𝑠−1 and its period of revolution is nearly 84 minutes. 𝑻 = 𝟐𝝅 𝒙 𝑹 g
  • 40.
    Example 5 • Calculatethe velocity required for a satellite moving in a circular orbit 200 Km above the earth’s surface. Given that radius of the earth is 6380 Km and mass of the earth = 5.98 𝑥 1024 𝐾𝑔 SOLUTION G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , 𝑹 𝑬 = 6.37𝑥106 𝑚 , h =200 𝑥 103 𝑚, M = 5.98 𝑥 1024 𝐾𝑔 𝑉𝑐 = 𝐺𝑀 𝑅+ℎ 𝑉𝑐 = 6.67𝑥10−11 𝑥 5.98 𝑥 1024 6.37𝑥106+200 𝑥 103 ∴ = 7.79 𝑥 103 𝑚𝑠−1
  • 41.
    Example 6 •The periodof moon around the earth is 27.3 days and the radius of its orbit is 3.9 𝑥 105 𝐾𝑚. If G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , find the mass of the earth SOLUTION T = 27.3 days = 27.3 𝑥 24 𝑥 60 𝑥 60 𝑠, R+h = 3.9 𝑥 108 𝑚, 𝑇 = 2𝜋 𝑥 (𝑅 + ℎ)3 𝐺𝑀 𝑀 = 4𝜋2 𝑅 + ℎ 3 𝐺𝑇2 𝑀 = 4𝜋2 3.9 𝑥 10 3 6.67𝑥10−11 𝑥 27.3 𝑥 24 𝑥 60 𝑥 60 2 ∴ 𝑀 = 6.31 𝑥 1024 𝐾𝑔
  • 42.
    Problems for Practice2 • A satellite is revolving in a circular orbit at a distance of 2620 Km from the surface of the earth. Calculate the orbital velocity and the period of revolution of the satellite. Radius of the earth = 63802 Km, mass of the earth = 6 𝑥 1024 𝐾𝑔 and G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 . (6.67 𝑲𝒎𝒔−𝟏 ; 2.35 Hours) • An earth’s satellite makes a circle around the earth in 90 minutes. Calculate the height of the satellite above the surface of the earth. Given that radius of the earth = 6400 Km and g = 9.8 𝑚𝑠−2. (268 Km)
  • 43.
    Uses of ArtificialSatellite • They are used to learn about the atmosphere near the earth. • They are used to forecast weather. • They are used to study radiation from the sun and the outer space. • They are used to receive and transmit various radio and television signals. • They are used to know the exact shape and dimensions of the earth. • Space fights are possible due to artificial satellite.
  • 44.
  • 45.
    Escape velocity • Isthe minimum velocity which it is to be projected so that it just overcome the gravitational pull of the earth (or any other planet)
  • 46.
    • In orderto project a body with escape velocity, we give kinetic energy to it. Let us calculate this Energy. • Suppose a body of mass m is projected upward with escape velocity . When the body is at point P at a distance x from the center of the earth, the gravitational force of attraction exerted by the earth on the body is 𝐹 = 𝐺 𝑀𝑚 𝑥2 In moving a small distance ∆𝑥 against this gravitational force, the small work done at the expense of the KE of the body is given by; ∆𝑊 = 𝐹∆𝑥 = 𝐺 𝑀𝑚 𝑥2 ∆𝑥 P Fig. PH x Q ∆𝑥 R M = Mass of the earth R = Radius of the earth
  • 47.
    • Total workdone (W) in moving the body from earth’s surface (where x=R) to infinity (where 𝑥 = ∞) is given by; 𝑊 = 𝑅 ∞ 𝐺𝑀𝑚 𝑥2 𝑑𝑥 = 𝐺𝑀𝑚 𝑅 ∞ 𝑥−2 𝑑𝑥 = 𝐺𝑀𝑚 − 1 𝑥 𝑅 ∞ = 𝐺𝑀𝑚 𝑅 ∴ 𝑊 = 𝐺𝑀𝑚 𝑅
  • 48.
    • If thebody is to be able to do this amount of work (and so escape), it needs to have at least this amount of KE at the moment it is projected. Therefore, escape velocity is given by; 1 2 𝑚𝑉𝑒 2 = 𝐺𝑀𝑚 𝑅 𝑽 𝒆 = 𝟐𝑮𝑴 𝑹
  • 49.
    The escape velocitycan be written in other equivalent form as shown below; • The acceleration due to the gravity of the surface of the earth is; g = 𝐺𝑀 𝑅2 or 𝐺𝑀 =g𝑅2 Substitute the equation above to the general equation of escape velocity; 𝑉𝑒 = 𝟐g 𝑅2 𝑹 ∴ 𝑉𝑒 = 𝟐g𝑹
  • 50.
    Example 7 • Whatis the escape velocity for a rocket on the surface of the Marks? Mass of Marks = 6.58 𝑥 1023 𝐾𝑔 and radius of Mars = 3.38 𝑥 106 𝑚. Solution 𝑽 𝒆 = 𝟐𝑮𝑴 𝒎 𝑹 𝒎 𝑴 𝒎= 6.58 𝑥 1023 𝐾𝑔, 𝑹𝒎 = 3.38 𝑥 106 𝑚, G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 𝑽 𝒆 = 𝟐 𝑥 6.67𝑥10−11 𝑥 6.58 𝑥 1023 3.38 𝑥 106 ∴ 𝑽 𝒆 = 5.1 𝑥 103 𝑚𝑠−1
  • 51.
    Example 8 • Findthe velocity of escape at the moon given that its radius is 1.7 𝑥 106 𝑚 and the value of g at its surface is 1.63 𝑚𝑠−2 . SOLUTION 𝑉𝑒 = 𝟐g𝑹 Here; g = 1.63 𝑚𝑠−2, R = 1.7 𝑥 106 𝑚 𝑉𝑒 = 𝟐 𝑥 1.63 𝑥1.7 𝑥 106 𝑉𝑒 = 2.354 𝑥 103 𝑚𝑠−1
  • 52.
    Kepler’s First Law •The path of each planet about the sun is an ellipse with sun at the one focus of the ellipse.
  • 53.
    What is anellipse? 2 foci An ellipse is a geometric shape with 2 foci instead of 1 central focus, as in a circle. The sun is at one focus with nothing at the other focus. FIRST LAW OF PLANETARY MOTION
  • 54.
    An ellipse alsohas… …a major axis …and a minor axis Semi-major axis Perihelion Aphelion Perihelion: When Mars or any another planet is closest to the sun. Aphelion: When Mars or any other planet is farthest from the sun.
  • 55.
    Kepler’s Second Law •Each planet moves in such a way that an imaginary line drawn from the sun to the planet sweeps out equal areas in equal periods of time.
  • 56.
    Kepler also foundthat Mars changed speed as it orbited around the sun: faster when closer to the sun, slower when farther from the sun… A B But, areas A and B, swept out by a line from the sun to Mars, were equal over the same amount of time. SECOND LAW OF PLANETARY MOTION
  • 57.
    Kepler’s Third Law •The square of the period of any planet (time needed for one revolution about the sun) is directly proportional to the cube of the plane’s average distance from the sun. 𝑇2 ∝ 𝑟3 Or 𝑇1 2 𝑇2 2 = 𝑟1 3 𝑟2 3 ∴ 𝑇2 𝑟3 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡
  • 58.
    Derivation of Newton’sLaw of Gravitation from Kepler's Third Law • Newton was able to show that Kepler's law could be derived mathematically from his law of universal gravitation and his laws of motion. • He used Kepler’s third law as evidence in favor of his law of universal gravitation.
  • 59.
    Derivation continue… • Considera planet of mass m revolving around the sun of mass M in a circular orbit of radius r. • Suppose 𝑣 is the orbital speed of the planet and T is its time period. • In time, T, the planet travels a distance 2𝜋𝑟. 𝑇 = 2𝜋𝑟 𝑣 or 𝑣 = 2𝜋𝑟 𝑇 • The centripetal force F required to keep the planet in the circular orbit is; 𝐹 = 𝑚𝑣2 𝑟 = This centripetal force is provided by the sun’s gravitational force on the planet.
  • 60.
    • According toKepler’s third law 𝑇2 𝑟3 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (k) or 𝑇2 = 𝑘𝑟3 Putting 𝑇2 = 𝑘𝑟3 in the derive centripetal force, we have ; 𝐹 = 4𝜋2 𝑘 𝑥 𝑚 𝑟2 • From Universal F = G 𝑀𝑚 𝑟2
  • 62.
    Gravitational Field dueto a Material of a Body • Is the space around the body in which any other mas experiences a force of attraction. E.g. the gravitational field of the earth
  • 63.
    Gravitational Field Strength •Is defined as the force per unit mass acting on a test mass placed at that point. 𝐸 = 𝐹 𝑚 Its SI unit is 𝑁𝐾𝑔−1
  • 64.
    Gravitational Field StrengthDue to the Earth • Since, 𝐸 = 𝐹 𝑚 …………………….. (i) and, 𝐹 = 𝐺𝑀𝑚 𝑅2 ……………………….. (ii) Substitute equation (ii) into equation (i), we have; 𝐸 = 𝐺𝑀 𝑅2 • The field strength at any point in a gravitational field is equal to the gravitational acceleration of any mass placed at that point. • i.e. g = 𝑮𝑴 𝑬 𝑹 𝑬 𝟐 = 𝐸 𝐸 = 𝐺𝑴 𝑬 𝑹 𝑬 𝟐
  • 65.
    Gravitational Potential Energy •The gravitational P.E of a body at a point in the gravitational field is defined as the amount of work done in bringing the body from infinity to that point.
  • 66.
    Gravitation P.E… • Supposea body of mass m is situated outside the earth at Point A at a distance r from the center of the earth. It’s Gravitational P.E (𝑼 𝑨) is W.D • Suppose at any instant the body is at point B at a distance x from the center of the earth. The gravitational force exerted by the earth on the body at B is 𝐹 = 𝐺𝑀𝑚 𝑥2 Consider the earth to be a spherical of radius R and Mass M R o c B dx x a r
  • 67.
    Gravitation P.E… • Smallamount of work done when the body moves from B to C is; 𝑑𝑊 = 𝐹𝑑𝑥 = 𝐺𝑀𝑚 𝑥2 𝑑𝑥 • Total work done by the gravitational force when the body of mass m at a distance r from the center of the earth is; 𝑊 = ∞ 𝑟 𝐺𝑀𝑚 𝑥2 𝑑𝑥 = 𝐺𝑀𝑚 ∞ 𝑟 1 𝑥2 𝑑𝑥 𝑊 = 𝐺𝑀𝑚 𝑥−1 −1 ∞ 𝑟 = −𝐺𝑀𝑚 1 𝑟 − 1 ∞ 𝑊 = − 𝐺𝑀𝑚 𝑟 ∴ Therefore, gravitational potential energy (𝑼 𝑨) of a body of mass m at a distance r from the center of the earth is; 𝑼 𝑨 = − 𝑮𝑴𝒎 𝒓
  • 68.
    Kinetic Energy ofthe Satellite • Suppose a satellite of mas m moves round the earth in a circular orbit at the height h above the surface of the earth. The radius of the orbit of the satellite is (R + h). If v is the speed of the satellite in the orbit, then, KE of the satellite = 1 2 𝑚𝑉𝑐 2 ‘……………………………(i) The orbital velocity of the satellite is 𝑉𝑐 = 𝐺𝑀 𝑟 ………………………………………. (ii)
  • 69.
    Kinetic Energy ofthe Satellite • Suppose a satellite of mas m moves round the earth of mas M in a circular orbit at the height h above the surface of the earth. • The radius of the orbit of the satellite is (R + h). If v is the speed of the satellite in the orbit, then, K.E of the satellite = 1 2 𝑚𝑉𝑐 2 ……(i) The orbital velocity of the satellite is; 𝑉𝑐 = 𝐺𝑀 (R + h) ……………. (ii) • Substitute equation (i) into equation (ii), we get; ∴ 𝐾. 𝐸 = 𝐺𝑀𝑚 2(R + h)
  • 70.
    Total Energy ofthe Satellite • The satellite in orbit around the earth has both K.E and Potential Energy(𝑼 𝑨). Total Energy, E = K.E + 𝑼 𝑨 = 𝐺𝑀𝑚 2(R + h) + − 𝑮𝑴𝒎 R + h ∴ E = - 𝐺𝑀𝑚 2(R + h)
  • 71.
    Example 9 • Aspaceship is stationed on Mars. How much energy must be expend on the spaceship to rocket it out of the solar system? Mass of the spaceship = 1000Kg, mass of the sun = 2𝑥1030 𝐾𝑔, Mass of mars = 6.4𝑥1023 𝐾𝑔, radius of mars = 3395 Km. radius of orbit of mars = 2.28𝑥108 𝐾𝑚, G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 .
  • 72.
    Solution Data given; Mass ofthe spaceship = 1000Kg, M ass of the sun = 2𝑥1030 𝐾𝑔, Mass of mars = 6.4𝑥1023 𝐾𝑔, radius of mars = 3395 Km, radius of orbit of Mars = 2.28𝑥108 𝐾𝑚,  G = 6.67𝑥10−11 𝑁𝑚2 𝐾𝑔−2 , E = ?. E = - 𝐺𝑀𝑚 2(R + h) E = - 6.67𝑥10−11 𝑥 2𝑥1030 𝑥1000 2 x (2.28𝑥1011+3.395 𝑥 106) E = −2.9 𝑥 1011 𝐽 ∴ The energy extended on the spaceship = 𝟐. 𝟗 𝒙 𝟏𝟎 𝟏𝟏 𝑱