Newton's Law of Universal Gravitation PDF

Newton’s Law of Universal Gravitation STM 003: MODULE 14  Learning Targets 01 02 03 Compare and contrast Solve using Newton's Apply Newton’s law of Newton's Law of Law of Motion in m...

Newton’s Law of Universal Gravitation STM 003: MODULE 14  Learning Targets 01 02 03 Compare and contrast Solve using Newton's Apply Newton’s law of Newton's Law of Law of Motion in motion and kinematics Motion context of its to obtain quantitative practical application. and qualitative conclusions about the velocity and acceleration of one or more bodies, and the contact and noncontact forces acting on one or more bodies.  Force exerted Universal Force by massive objects (G = Universal (more massive, more Gravitational force exerted) Constant) Gravity Weak between Stronger on objects objects that are that have larger distant from each masses. other Newton’s Law Of Universal Gravitation Sir Isaac Newton put forward the universal law of gravitation in 1687 and used it to explain the observed motions of the planets and moons. Newton’s Law of Universal Gravitation states that every particle attracts every other particle in the universe with force directly proportional to the product of the masses and inversely proportional to the square of the distance between them. “Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.”  The Universal Gravitation Equation 𝑚 1𝑚 2 𝑚1𝑚2 𝐹 =𝐺 𝐹∝ 𝑟2 𝑟2 F = force of gravity between two objects (Fgrav) F = force of gravity between two objects (Fgrav) = “proportional to” m1= Mass of object 1 m2= Mass of object 2 m1= Mass of object 1 r = distance between the centers of two bodies m2= Mass of object 2 r = distance between the centers of two bodies Gravitational Constant The constant proportionality (G) in the above equation is known as the universal gravitation constant. Henry Cavendish experimentally determined the precise value of G. The value of G is found to be G = 6.67 x 10-11 N m2/kg2. The Universal Gravitational Law can explain almost anything, right from how an apple falls from a tree to why the moon revolves around the earth.  Some Implications and Derived Formulas from Law of Gravitation Speed of a Satellite in a Circular Orbit 𝑉𝑐= √ 𝐺𝑀 𝑟 Vc= Velocity in circular orbit M= Mass of object r = distance of the object (satellite) from the center of the planet or sun Acceleration due to gravity of a planet: 𝐺𝑚 𝑝 𝑔= 𝑅2𝑝 g= Acceleration due to gravity of the planet mp= Mass of the planet Rp = Radius of the planet Escape velocity: 𝑉𝑒= √ 2 𝐺𝑀 𝑅 Ve= escape velocity M= Mass of object R= Radius Example: Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level, a distance of 6.38 x 106 m from earth's center. 𝑚 1𝑚 2 𝐹 =𝐺 𝑟2 Given: F= (5.98 𝑥 1024)(70) 𝐹 =6.67 𝑥 10 − 11 (6.38 𝑥 10 6)2 m1= 5.98 x 1024 kg 𝐹 =685.94 𝑁 m2= 70 kg r = 6.38 x 106 m Example: Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40000 feet above earth's surface. This would place the student a distance of 6.39 x 106 m from earth's center. 𝑚 1𝑚 2 𝐹 =𝐺 𝑟2 Given: F= (5.98 𝑥 1024)(70) 𝐹 =6.67 𝑥 10 − 11 (6.39 𝑥 10 6)2 m1= 5.98 x 1024 kg 𝐹 =683.79 𝑁 m2= 70 kg r = 6.39 x 106 m Example: Calculate the gravitational force of attraction between the Earth and a 70 kg man standing at a sea level, a distance of 6.38 x 106 m from the earth’s center. 𝑚 1𝑚 2 𝐹 =𝐺 𝑟2 Given: m1 = Mass of the Earth (5.98 x 1024 kg) 𝐹 =6.67 𝑥 10 − 11 (5.98 𝑥 1024)(70) m2 is the mass of the man which is equal to 70 kg (6.38 𝑥 10 6)2 r= Radius of the Earth (6.38 x 106 m) 𝐹 =685.94 𝑁 G = 6.67 x 10-11 N m2/kg2 Example: Suppose a 100 kg astronaut feels a gravitational force of 700 N when placed in the gravitational field of a planet. a. What is the gravitational field strength at the location of the astronaut? b. What is the mass of the planet if the astronaut is 2×10 6 m from its center? 𝐹 =𝑚𝑔 A.) Given: g= ? F= 700 N m= 100 Kg Example: Suppose a 100 kg astronaut feels a gravitational force of 700 N when placed in the gravitational field of a planet. a. What is the gravitational field strength at the location of the astronaut? b. What is the mass of the planet if the astronaut is 2×10 6 m from its center? 𝑚 1𝑚 2 𝐹 =𝐺 Given: 𝑟2 F= 𝐹𝑟 2=𝐺𝑚 1 𝑚 2 𝐺𝑚 1 𝐺 𝑚 1 M1 (Astronaut) = 100 kg M2 (planet)= ? 𝐹𝑟 2 r = 2 x 106 m 𝑀 2= 𝐺𝑚 1 Example: Suppose a 100 kg astronaut feels a gravitational force of 700 N when placed in the gravitational field of a planet. a. What is the gravitational field strength at the location of the astronaut? b. What is the mass of the planet if the astronaut is 2×10 6 m from its center? 𝐹𝑟 2 𝑀 2= Given: 𝐺𝑚 1 F= ( 700 𝑁 ) (2 𝑥 106𝑚)2 𝑀 2= 𝑁𝑚 2 (6.67 𝑥 10 − 11 )(100 𝑘𝑔) 𝑘𝑔 2 M1 (Astronaut) = 100 kg M2 (planet)= ? 𝑀 2=4.2 𝑥 1023𝑘𝑔 r = 2 x 106 m Thank you.