Mass training of trainers General Physics 1 May 20-22, 2017 The Law of Universal Gravitations Prof. MARLON FLORES SACEDON Physics Instructor Visayas State University Baybay City, Leyte

The answer is Gravitations Questions Why doesn’t the moon fall to earth? Why do the planets move across the sky? Why doesn’t the earth fly off into space rather than remaining in orbit around the sun? The answer is Gravitations

What is Gravitations? Gravity or gravitation is a natural phenomenon by which all things with energy are brought toward one another, including stars, planets, galaxies and even light and sub-atomic particles.

Objectives To study the Laws of motion of planets by Johann Kepler. To study and apply the Law of Universal Gravitations.

Laws of motion of planets by Johann Kepler: Law#1. The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)

Laws of motion of planets by Johann Kepler: Law#2. An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas).

Laws of motion of planets by Johann Kepler: Law#3. The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies) 𝑻 𝟐 𝑹 𝟑

Laws of motion of planets by Johann Kepler: NOTE: The average distance value is given in astronomical units where 1 a.u. is equal to the distance from the earth to the sun - 1.4957 x 1011 m. The orbital period is given in units of earth-years where 1 earth year is the time required for the earth to orbit the sun.

There must be a force acted on the apple… It was a force coming from earth which is a gravitational force.

The Law of Universal Gravitations The Law of Universal Gravitations by: Isaac Newton The Law of Universal Gravitations “Any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.” 𝑟 𝐹 𝑔 ∝ 1 𝑟 2 𝑚 𝐵 𝐹 𝑔 ∝ 𝑚 𝐴 𝑚 𝐵 𝐹 𝐴𝐵 𝐹 𝑔 =𝐺 𝑚 𝐴 𝑚 𝐵 𝑟 2 Where: 𝐹 𝐵𝐴 𝐹 𝑔 = gravitational force (N, dyne, poundal) 𝑚 𝐴 𝑚 𝐴 & 𝑚 𝐵 = interacting masses (kg, g, lb) 𝑟 = distance between masses (m, cm, ft) 𝐹 𝐵𝐴 − 𝐹 𝐴𝐵 =0 𝐺 = 6.67𝑥 10 −11 𝑁. 𝑚 2 𝑘𝑔 2 gravitational constant 𝐹 𝐵𝐴 = 𝐹 𝐴𝐵 = 𝐹 𝑔

The Law of Universal Gravitations by: Isaac Newton 𝐹 𝑔 =𝐺 𝑚 𝐴 𝑚 𝐵 𝑟 2 𝐺 = 6.67𝑥 10 −11 𝑁. 𝑚 2 𝑘𝑔 2 gravitational constant Cavendish balance By: Henry Cavendish Torsion Balance

Application Problem 1. Gravitational force of two interacting objects Calculate the force of gravity on two interacting objects whose masses are 8kg and 15kg separated at distance of 2m. 𝑟 𝐹 𝑔 =𝐺 𝑚 𝐴 𝑚 𝐵 𝑟 2 𝐺 = 6.67𝑥 10 −11 𝑁. 𝑚 2 𝑘𝑔 2 𝑚 𝐵 𝐹 𝐴𝐵 𝐹 𝐵𝐴 𝐹 𝑔 =6.67𝑥 10 −11 𝑁. 𝑚 2 𝑘𝑔 2 ∙ 8𝑘𝑔 15𝑘𝑔 2𝑚 2 =2𝑥 10 −9 𝑁 𝑚 𝐴

Application Problem 2. Superposition of gravitational forces Many stars belong to system of two or more stars held together by their mutual gravitational attraction. Figure below shows a three star system at an instant when the stars are at the vertices of a 45 𝑜 right triangle. Find the total gravitational fore exerted on the small star by the two large ones. Calculate 𝑠 12 the distance between 𝑚 1 and 𝑚 2 𝑠 12 𝑠 12 = 2.00𝑥 10 12 2 + 2.00𝑥 10 12 2 =2.83𝑥 10 12 𝑚 +𝑦 +x 2 8.00𝑥 10 30 𝑘𝑔 45 𝑜 2.00𝑥 10 12 𝑚 Calculate the forces 𝐹 21 and 𝐹 31 𝐹 21 =6.67𝑥 10 −11 𝑁. 𝑚 2 𝑘𝑔 2 ∙ 8.00𝑥 10 30 𝑘𝑔 1.00𝑥 10 30 𝑘𝑔 2.83𝑥 10 12 𝑚 2 =6.67𝑥 10 25 𝑁 𝐹 21 𝐹 31 3 8.00𝑥 10 30 𝑘𝑔 1 1.00𝑥 10 30 𝑘𝑔 2.00𝑥 10 12 𝑚 𝐹 31 =6.67𝑥 10 −11 𝑁. 𝑚 2 𝑘𝑔 2 ∙ 8.00𝑥 10 30 𝑘𝑔 1.00𝑥 10 30 𝑘𝑔 2.00𝑥 10 12 𝑚 2 =1.33𝑥 10 26 𝑁

Application Problem 2. Superposition of gravitational forces Many stars belong to system of two or more stars held together by their mutual gravitational attraction. Figure below shows a three star system at an instant when the stars are at the vertices of a 45 𝑜 right triangle. Find the total gravitational fore exerted on the small star by the two large ones. 𝑠 12 +𝑦 +x 2 Σ 𝐹 𝑥 =1.33𝑥 10 26 𝑁+6.67𝑥 10 25 𝑁cos45=1.81𝑥 10 26 𝑁 8.00𝑥 10 30 𝑘𝑔 45 𝑜 Σ 𝐹 𝑦 =6.67𝑥 10 25 𝑁sin45=4.72𝑥 10 25 𝑁 𝐹 𝑔 = (1.81𝑥 10 26 ) 2 + (4.72𝑥 10 25 ) 2 =1.87𝑥 10 26 𝑁 𝜃= 𝑇𝑎𝑛 −1 4.72𝑥 10 25 1.81𝑥 10 26 = 14.62 𝑜 𝐹 21 =6.67𝑥 10 25 𝑁 3 8.00𝑥 10 30 𝑘𝑔 1 1.00𝑥 10 30 𝑘𝑔 𝐹 31 =1.33𝑥 10 26 𝑁

Application Problem 3. Weight and Gravitational acceleration 𝐹 𝑔 =𝐺 𝑚 𝐴 𝑚 𝐵 𝑟 2 𝐹 𝑀𝑚 𝐹 𝐸𝑚 𝐹 𝑚𝑀 𝐹 𝐸𝑀 𝑭 𝒎𝑬 𝐹 𝑀𝐸

Application Problem 3. Weight and Gravitational acceleration 𝐹 𝑔 =𝐺 𝑚 𝐴 𝑚 𝐵 𝑟 2 𝐹 𝐸𝑚 =𝐺 𝑚 𝐸 𝑚 𝑚 𝑅 𝐸 2 Force of gravity exerted by earth to the man 𝑤 𝑚 =𝐺 𝑚 𝐸 𝑚 𝑚 𝑅 𝐸 2 𝐹 𝐸𝑚 Weight of the man 𝑅 𝐸 But weight is equal to mass time gravitational acceleration… 𝑤 𝑚 = 𝑚 𝑚 𝑔 𝑭 𝒎𝑬 𝐹 𝐸𝑚 =𝑤 𝑚 𝑔 𝑀 =𝐺 𝑚 𝑀 𝑅 𝑀 2 𝑚 𝑚 𝑔 𝐸 =𝐺 𝑚 𝐸 𝑚 𝑚 𝑅 𝐸 2 Gravitational acceleration at the surface of the moon 𝑔 𝐸 =𝐺 𝑚 𝐸 𝑅 𝐸 2 Gravitational acceleration at the surface of the earth

Problem Exercise/ Seatwork/ Assignment 1. The moon has a mass of 7.32 𝑥 10 22 𝑘𝑔 and a radius of 1,609.4 𝑘𝑚. Calculate the value of "g" at the surface of the moon. 2. At what distance from the earth will a spacecraft on the way to the moon experience zero net force because the earth and moon pull with equal and opposite forces? 3. Calculate the force of gravity on a spacecraft 12,800 𝑘𝑚 above the earth’s surface if its mass is 850 𝑘𝑔. 4. Calculate the centripetal acceleration of the earth in its orbit around the sun and the net force exerted on the earth. What exerts this force on the earth? Assume the earth’s orbit is a circle of radius 1.49 𝑥 10 11 𝑚. P A B 50𝑐𝑚 30𝑐𝑚 35 𝑜 5. Two uniform spheres, each of mass 0.260 𝑘𝑔, are fixed at point 𝐴 and 𝐵. Find the magnitude and direction of gravitational attraction force on a uniform sphere with mass 0.010 𝑘𝑔 at point 𝑃.

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