Transcript Slide 1
Gravity Physics 6B Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB GRAVITY Any pair of objects, anywhere in the universe, feel a mutual attraction due to gravity. There are no exceptions – if you have mass, every other mass is attracted to you, and you are attracted to every other mass. Look around the room – everybody here is attracted to you! Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB GRAVITY Any pair of objects, anywhere in the universe, feel a mutual attraction due to gravity. There are no exceptions – if you have mass, every other mass is attracted to you, and you are attracted to every other mass. Look around the room – everybody here is attracted to you! Newton’s Law of Universal Gravitation gives us a formula to calculate the attractive force between 2 objects: m m Fgrav G 1 2 r2 m1 and m2 are the masses, and r is the center-to-center distance between them G is the gravitational constant – it’s tiny: G≈6.674*10-11 Nm2/kg2 Use this formula to find the magnitude of the gravity force. Use a diagram or common sense to find the direction. The force will always be toward the other mass. m1 r m2 F2 on 1 F1 on 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB GRAVITY Any pair of objects, anywhere in the universe, feel a mutual attraction due to gravity. There are no exceptions – if you have mass, every other mass is attracted to you, and you are attracted to every other mass. Look around the room – everybody here is attracted to you! Newton’s Law of Universal Gravitation gives us a formula to calculate the attractive force between 2 objects: m m Fgrav G 1 2 r2 m1 and m2 are the masses, and r is the center-to-center distance between them G is the gravitational constant – it’s tiny: G≈6.674*10-11 Nm2/kg2 Use this formula to find the magnitude of the gravity force. Use a diagram or common sense to find the direction. The force will always be toward the other mass. m1 r *Note: If you are dealing with spherical objects with uniform density (our typical assumption) then you can pretend all the mass is concentrated at the center. m2 F2 on 1 F1 on 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Three planets are aligned as shown. The masses and distances are given in the diagram. Find the net gravitational force on planet H (the middle one). 1012 m 3 x 1012 m Planet of the Apes: Planet Hollywood: Daily Planet: mass=6 x 1024 kg mass=6 x 1020 kg mass=3 x 1025 kg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Three planets are aligned as shown. The masses and distances are given in the diagram. Find the net gravitational force on planet H (the middle one). 1012 m 3 x 1012 m FApes on H FDP on H Planet of the Apes: Planet Hollywood: Daily Planet: mass=6 x 1024 kg mass=6 x 1020 kg mass=3 x 1025 kg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Three planets are aligned as shown. The masses and distances are given in the diagram. Find the net gravitational force on planet H (the middle one). 1012 m 3 x 1012 m FApes on H FDP on H Planet of the Apes: Planet Hollywood: Daily Planet: mass=6 x 1024 kg mass=6 x 1020 kg mass=3 x 1025 kg Our formula will find the forces (we supply the direction from looking at the diagram): m m Fgrav G 1 2 r2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Three planets are aligned as shown. The masses and distances are given in the diagram. Find the net gravitational force on planet H (the middle one). 1012 m 3 x 1012 m FApes on H FDP on H Planet of the Apes: Planet Hollywood: Daily Planet: mass=6 x 1024 kg mass=6 x 1020 kg mass=3 x 1025 kg m m Fgrav G 1 2 r2 24 20 1 1 N m2 6 10 kg 6 10 kg FA peson H 6.67 10 2 kg2 101 2m Our formula will find the forces (we supply the direction from looking at the diagram): This is negative because the force points to the left Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Three planets are aligned as shown. The masses and distances are given in the diagram. Find the net gravitational force on planet H (the middle one). 1012 m 3 x 1012 m FApes on H FDP on H Planet of the Apes: Planet Hollywood: Daily Planet: mass=6 x 1024 kg mass=6 x 1020 kg mass=3 x 1025 kg Our formula will find the forces (we supply the direction from looking at the diagram): m m Fgrav G 1 2 r2 24 20 11 Nm 2 6 10 kg 6 10 kg FApes on H 6.67 10 2.4 1011N 2 2 kg 1012 m This is negative because the force points to the left Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Three planets are aligned as shown. The masses and distances are given in the diagram. Find the net gravitational force on planet H (the middle one). 1012 m 3 x 1012 m FApes on H FDP on H Planet of the Apes: Planet Hollywood: Daily Planet: mass=6 x 1024 kg mass=6 x 1020 kg mass=3 x 1025 kg Our formula will find the forces (we supply the direction from looking at the diagram): m m Fgrav G 1 2 r2 24 20 11 Nm 2 6 10 kg 6 10 kg FApes on H 6.67 10 2.4 1011N 2 2 kg 1012 m 25 20 11 Nm 2 3 10 kg 6 10 kg FDP on H 6.67 10 1.3 1011N 2 2 kg 3 1012 m This is negative because the force points to the left This is positive because the force points to the right Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: Three planets are aligned as shown. The masses and distances are given in the diagram. Find the net gravitational force on planet H (the middle one). 1012 m 3 x 1012 m FApes on H FDP on H Planet of the Apes: Planet Hollywood: Daily Planet: mass=6 x 1024 kg mass=6 x 1020 kg mass=3 x 1025 kg Our formula will find the forces (we supply the direction from looking at the diagram): m m Fgrav G 1 2 r2 24 20 11 Nm 2 6 10 kg 6 10 kg FApes on H 6.67 10 2.4 1011N 2 2 kg 1012 m 25 20 11 Nm 2 3 10 kg 6 10 kg FDP on H 6.67 10 1.3 1011N 2 2 kg 3 1012 m Add the forces to get the net force on H: 11 Fnet 1.1 10 N This is negative because the force points to the left This is positive because the force points to the right Net force is Prepared by Vince Zaccone to the left For Campus Learning Assistance Services at UCSB GRAVITY One more useful detail about gravity: The acceleration due to gravity on the surface of a planet is right there in the formula. Here is the gravity formula, modified for the case where m is the mass of an object on the surface of a planet. Fgrav G mplanet m Rplanet 2 m Rplanet Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB GRAVITY One more useful detail about gravity: The acceleration due to gravity on the surface of a planet is right there in the formula. Here is the gravity formula, modified for the case where m is the mass of an object on the surface of a planet. Fgrav G mplanet m Rplanet 2 We already know that Fgrav is the weight of the object, and that should just be mg (if the planet is the Earth) mg G mplanet m Rplanet 2 m Rplanet Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB GRAVITY One more useful detail about gravity: The acceleration due to gravity on the surface of a planet is right there in the formula. Here is the gravity formula, modified for the case where m is the mass of an object on the surface of a planet. Fgrav G mplanet m Rplanet 2 We already know that Fgrav is the weight of the object, and that should just be mg (if the planet is the Earth) mg G mplanet m Rplanet 2 m Rplanet This part is g Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 12-3 Kepler’s Laws of Orbital Motion Johannes Kepler made detailed studies of the apparent motions of the planets over many years, and was able to formulate three empirical laws: 1. Planets follow elliptical orbits, with the Sun at one focus of the ellipse. 12-3 Kepler’s Laws of Orbital Motion 2. As a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount of time. 12-3 Kepler’s Laws of Orbital Motion 3. The period, T, of a planet increases as its mean distance from the Sun, r, raised to the 3/2 power. This can be shown to be a consequence of the inverse square form of the gravitational force. 12-3 Kepler’s Laws of Orbital Motion What should you use for the constant? 12-3 Kepler’s Laws of Orbital Motion What should you use for the constant? Let’s derive the formula to find out. 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. m m Fgrav G 1 2 r2 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. m m Fgrav G 1 2 r2 Notice that the gravity force points toward the center: i.e. it is the centripetal force. 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. m m Fgrav G 1 2 r2 Notice that the gravity force points toward the center: i.e. it is the centripetal force. We have a formula for centripetal force: m v2 Fc ent r 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. Set these equal to get our formula: M m m v2 G 2 r r M is the mass of the central object. 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. Set these equal to get our formula: The mass of the orbiting object will cancel out: M v2 G 2 r r How do we get the period (T) involved in our formula? 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. Set these equal to get our formula: The mass of the orbiting object will cancel out: M v2 G 2 r r Each orbit is one trip around the circle, so the distance traveled is 2∏r. Speed is distance/time, so we get: 2r v T 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. Set these equal to get our formula: Plug in our expression for v: G M r 2 (2Tr )2 r Now we can rearrange this to solve for T. Do this now. 12-3 Kepler’s Laws of Orbital Motion Start with Newton’s Law of Gravity applied to a circular orbit. Set these equal to get our formula: Plug in our expression for v: G M r 2 (2Tr )2 r Now we can rearrange this to solve for T. Do this now. 3 2 T r2 GM Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. a) Is Deimos closer to or farther from Mars than Phobos? Explain. Answer: Deimos is farther from Mars. Inspection of the formula in Kepler’s 3rd Law should make this clear. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant. Previously we derived an expression for the constant. Our formula is: 3 2 T r2 GM What do we use for M in this formula? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant. Previously we derived an expression for the constant. Our formula is: 3 2 T r2 GM Here M is the mass of the object being orbited, in this case Mars. We can look this up in a table: MMars = 6.45x1023 kg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant. Previously we derived an expression for the constant. Our formula is: 3 2 T r2 GM Here M is the mass of the object being orbited, in this case Mars. We can look this up in a table: MMars = 6.45x1023 kg Rearrange this formula to solve for radius, then plug in the numbers: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant. Previously we derived an expression for the constant. Our formula is: 3 2 T r2 GM Here M is the mass of the object being orbited, in this case Mars. We can look this up in a table: MMars = 6.45x1023 kg Rearrange this formula to solve for radius, then plug in the numbers: r 3 2 GM 3 T 42 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Example: The Martian moon Deimos has an orbital period that is greater than the other Martian moon, Phobos. Both moons have approximately circular orbits. a) Is Deimos closer to or farther from Mars than Phobos? Explain. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s. We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant. Previously we derived an expression for the constant. Our formula is: 3 2 T r2 GM Here M is the mass of the object being orbited, in this case Mars. We can look this up in a table: MMars = 6.45x1023 kg Rearrange this formula to solve for radius, then plug in the numbers: 2 r 3 2 GM 3 3 T 42 (6.67 101 1 N m2 )(6.45 102 3kg) kg 42 5 2 3 (1.10 10 s) 2.36 107m Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB 12-4 Gravitational Potential Energy Gravitational potential energy of an object of mass m a distance r from the Earth’s center: 12-5 Energy Conservation Total mechanical energy of an object of mass m a distance r from the center of the Earth: This confirms what we already know – as an object approaches the Earth, it moves faster and faster.