Newtons Law of Gravitation Applications of Circular Motion Bellwork 011911 A 150kg bucket of water is tied by a rope and whirled in a circle with a radius of 100 m At the top of the circular loop the speed of the bucket is 400 ms Determine the acceleration the net force and the ind ID: 771643
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Newton’s Law of Gravitation Applications of Circular Motion
Bellwork 01/19/11 A 1.50-kg bucket of water is tied by a rope and whirled in a circle with a radius of 1.00 m. At the top of the circular loop, the speed of the bucket is 4.00 m/s. Determine the acceleration, the net force and the individual force values when the bucket is at the top of the circular loop. m = 1.5 kga = _____m/s/sFnet = ______ N
Bellwork Solution F grav = m • g = 14.7 N (g is 9.8 m/s/s)a = v2 / R = (4 • Fgrav)/1 = 16 m/s/sFnet = m • a = 1.5 kg •16 m/s/s = 24 N, downFnet = Fgrav + Ftens, soFtens = Fnet - FgravFtens = 24 N - 14.7 N = 9.3 N **Challenge: What if the bucket was at the bottom of the circle?** m = 1.5 kg a = ____m/s/s F net = _____ N
Review Centripetal Acceleration Work with your best partner using the rally coach method to answer the questions on your worksheet in the section called “Centripetal Acceleration”
linear angular a = (v f - vo)/t α = (ω f - ω o )/t v f = v o + at ω f = ω o + αt s = ½( v f + v o )t θ = ½(ω f + ω o )t s = v o t + ½at 2 θ = ω o t + ½αt 2 v f 2 = v o 2 + 2as ω f 2 = ω o 2 + 2αθ
Einstein’s Ideas About Gravity Masses cause a distortion in space The larger the mass, the larger it’s distortion The Laws (Newton’s and Kepler’s) describe WHAT happens.The Theory of Gravity is the attempt to explain WHY the laws are true.Demo: Write three observations for each of the masses in “space”Is the gravitational force influenced by the radius of the object?Is gravitational force influenced by the mass of the object?Is the gravitational force influenced by the diameter between two objects?
Newton’s Law of Gravitation Newton’s Law of Gravitation: the mutual gravitational attraction between any two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The magnitude of the gravitational force is given byWhere G = the universal gravitation constant, G = 6.67x10-11 Nm2/kg2 .
Applications of Newton’s Second Law a = F/m Thus the acceleration due to gravity by a mass, M acting on another object m is equal to At Earth’s surface, where ME and RE are the mass and radius of the earthAt an altitude, h above the Earth
Gravitational Potential Energy The gravitational potential energy between two masses is given by Note : the denominator is r, not r2 as in Newton’s law of gravitation.The minus sign derives from the choice of zero reference point, where U = 0, which is r = ∞
Let Equations Guide Your Thinking Work with your awesome partner using the rally coach method to answer the remaining questions on your worksheet .
Example One: Bohr’s Model of the Atom The Bohr model of the hydrogen atom consists of a proton of mass 1.67x 10 -27 kg and an orbiting electron of mass 9.11x10-31 kg. In one of its orbits, the electron is 5.3 x10-11 m from the proton and in another orbit it is 10.6 x10-11 m from the proton. What are the mutual attractive forces when the electron is in each of these orbits?If the electron jumps from the larger orbit to the smaller one, what is the change in the potential energy of the atom?
Solution to Example 1 Given: m 1 = 1.67-27 x 10 kg, m2 = 9.11x10-31 kg r1 = 5.3 x10-11 m , r = 10.6-11 x10 m A) r2 = 2r1, so F2 /F 1 = 1/4
B) The change in gravitational potential energy, D U = U 1 – U2 .Therefore, the atom loses potential energy. **BONUS: Predict the color of light emitted by the electron as it transitions from one orbital to the next. E = hn Solution to Example 1, cont.
Example 2 Calculate the acceleration due to gravity at the surface of the Moon, where R M = 1750 km = 1.750 x 106 m MM = 7.4 x 1022 kgHow does the magnitude of ag on the moon compare to the value of ag = g on Earth?
Solution to Example 2 Given: R M = 1750 km = 1.750 x 106 m MM = 7.4 x 1022 kg
Independent Practice P. 288 #17, 19, 20, 33, 25, 26, 29, 34P. 295 #104, 105P. 250-254 #80, 81, 84, 85, 90
Kepler’s Laws and Earth’s Satellites Objective: State and explain Kepler’s laws of planetary motion and describe the orbits and motions of satellites
Warm-Up In The Little Prince , the Prince visits a small asteroid called B612. If the asteroid B612 has a radius of only 20.0 m and a mass of 1.00 x 104 kg, what is the acceleration due to gravity on asteroid B612?What is the escape speed needed to Escape the gravitational pull of Asteroid B612?What would happen if you jumped up on Asteroid B612?
Kepler’s Laws The Law of Orbits: Planets move in elliptical orbits, with the Sun at one of the focal points. A line from the Sun to a planet sweeps out equal areas in equal lengths of time. The square of the orbital period of a planet is directly proportional to the cube of the average distance of the planet from the Sun (see p. 240 for derivation of the equation below).
Example 3: Calculate the mass of Jupiter The planet Jupiter is the larges in the solar system, both in volume and mass. Jupiter has 62 known moons, the four largest being discovered by Galileo in 1610. These moons are Io, Callisto , Ganymede, and Europa. If Io is an average distance of 4.22 x 105 km from Jupiter and has an orbital period of 1.77 days, compute the mass of Jupiter.
Solution to Example 3 Given r = 4.22 x 10 5km = 4.22 x 10 8 m T = 1.77 days (86400 s/day)=1.53 x 10 5 s
Earth’s Satellites KE o +PEo = KE + PE by the conservation of energyIn terms of gravitational potential energy Where vesc is the escape velocity – the initial velocity required to escape from the surface of the Earth.
Staying In Orbit On Earth, the escape velocity is approximately 11 km/s or 7 mi/s A tangential velocity less than the escape velocity is required for a satellite to remain in orbit around the Earth (depending on the altitude of the satellite’s orbit, h).
Guided Practice With your team Calculate the mass of the Sun based on orbital data of the Earth-Sun system (book’s cover or Appendix III) solve problem #107 on page 255. Independent Practice Chapter 7 ConcepTest with clickers P. 254-255; # 91, 92, 93, 95, 97, 104, 106 TEST FRIDAY 01/29/10
EXAMPLE 4: Oliver, whose mass is 65 kg, and Olivia, whose mass is 45 kg, sit 2.0 m apart in their physics classroom. A) What is the force of gravitational attraction between Oliver and Olivia? B) Why don’t they drift toward each other? EXAMPLE 5: Temba is standing in the lunch line 6.38 x 106 m from the center of the earth. Earth’s mass is 5.98 x 1024 kg. A) What is the acceleration due to gravity?B) When Temba eats lunch and increases his mass, does this change the acceleration due to gravity?
EXAMPLE 6: Earth has a mass of 5.98 x 10 24 kg and a radius of 6.38 x 10 6 km. What is the escape speed of a rocket launched on earth in m/s?EXAMPLE 7: Compare this with the escape speed of a rocket launched from the moon if the moon has a mass of 7.35 x 1022 kg and a radius of 1.74 x 106 m.
Bellwork Calculate the total gravitational potential energy for the system in the diagram below.