#### Transcript ch9

Physics 231 Topic 9: Gravitation Alex Brown October 30, MSU Physics 231 Fall 2015 2015 1 What’s up? (Friday Oct 30) 1) The correction exam is now open and is due at 10 pm Tuesday Nov 3th. The exam grades will available on Wednesday Nov 4.. 2) Homework 07 is due Tuesday Nov 10th and covers Chapters 9 and 10. It is a little longer that usual so you may want to start early. MSU Physics 231 Fall 2015 2 MSU Physics 231 Fall 2015 3 MSU Physics 231 Fall 2015 4 Key Concepts: Gravitation Newton’s Law of Gravitation Gravitational Acceleration Planetary Motion Kepler’s Laws Gravitational Potential Energy Conservation of ME Artificial Satellites Covers chapter 9 in Rex & Wolfson MSU Physics 231 Fall 2015 5 The gravitational force Newton: m1m2 F G 2 r G=6.673·10-11 N m2/kg2 N m2/kg2 = m3/(kg s2) The gravitational force works between every two masses in the universe. MSU Physics 231 Fall 2015 6 Gravitation between two objects A B The gravitational force exerted by the spherical object A on B can be calculated as if all of A’s mass would is concentrated in its center and likewise for object B. Conditions: B must be outside of A A and B must be ‘homogeneous’ MSU Physics 231 Fall 2015 7 Gravitation between two objects m1m2 F G 2 r The force of the earth on the moon is equal and opposite to the force of the moon on the earth! MSU Physics 231 Fall 2015 8 Clicker Quiz! Earth and Moon a) one quarter If the distance to the Moon were b) one half doubled, then the force of attraction c) the same between Earth and the Moon would be: d) two times e) four times MSU Physics 231 Fall 2015 9 Clicker Quiz! Earth and Moon a) one quarter If the distance to the Moon were b) one half doubled, then the force of attraction c) the same between Earth and the Moon would be: d) two times e) four times The gravitational force depends inversely on the distance squared. So if you increase the distance by a factor of 2, the force will decrease by a factor of 4. MSU Physics 231 Fall 2015 Mm F G 2 r 10 Gravitational acceleration at the surface of planet with mass M mM F G 2 R F=mg g = GM/R2 MSU Physics 231 Fall 2015 11 Gravitational acceleration at the surface of the earth g=GMe/Re2 G = 6.67x10-11 Me=5.97x1024 kg Radius from Earth’s Center (km) Gravitational Acceleration (m/s2) Earth’s Surface 6366 9.81 Mount Everest 6366 + 8.85 9.78 Mariana Trench 6366 - 11.03 9.85 Polar Orbit Satellite 6366 + 1600 6.27 Geosynchronous Satellite 6366 + 36000 0.22 MSU Physics 231 Fall 2015 12 Gravitational potential energy So far, we used: PEgravity=mgh Only valid for h near earth’s surface. More general: PEgravity=-GMm/r R2=R+h Earlier we noted that we could define the zero of PEgravity anywhere we wanted. So the surface of the earth is as good as anywhere! R Mm R < R2 R Mm PE2 G R2 Thus… PE1 G MSU Physics 231 Fall 2015 PE < PE2 13 Gravitational potential energy So far, we used: PEgravity=mgh Only valid for h near earth’s surface. More general: PEgravity=-GMm/r R2=R+h Earlier we noted that we could define the zero of PEgravity anywhere we wanted. So the surface of the earth is as good as anywhere! Mm Mm PE2 PE1 G (G ) R2 R R G Mm Mm (G ) ( R h) R M 1 1 mGM ( ) m G 2 R R R h MSU Physics 231 Fall 2015 h mgh 14 Gravitational potential energy PEgravity=mgh only valid for h near earth’s surface. More general: PEgravity=-GMm/r PE = 0 at infinite distance from the center of the earth (r = ∞) Application: what should the minimum initial velocity of a rocket be if we want to make sure it will not fall back to earth? KEi + PEi = ½mv2 - GMm/R KEf + PEf = 0 v = (2GM/R) = (2gR) = 11.2 km/s MSU Physics 231 Fall 2015 15 Second cosmic speed Second cosmic speed: speed needed to break free from a planet of mass Mp and radius Rp (gp = GMp/Rp2) v2 =(2GMp/Rp) = (2gpRp) For earth: g = 9.81 m/s2 R = 6.37x106 m MSU Physics 231 Fall 2015 v2 = 11.2 km/s 16 Orbital Velocities What does the word orbit mean? An orbit is the gravitationally curved path of an object around a point in space. To orbit the object, you need to satisfy the kinematic conditions of that type of orbit (more on this shortly…) MSU Physics 231 Fall 2015 17 launch speed 4 km/s 6 km/s MSU Physics 231 Fall 2015 8 km/s 18 First cosmic speed First cosmic speed: speed of a satellite of mass m on a lowlying circular around a planet with orbit of Mp and radius Rp (gp = GMp/Rp2) rsatellite ≈ Rp F = mac mgp = m v2/Rp For earth: g = 9.81 m/s2 so v1 =(gpRp) R = 6.37x106 m MSU Physics 231 Fall 2015 v1=7.91 km/s 19 Period for all orbits Consider an object in circular motion around a larger one F mac 2 v GMm mv 2 2 2 m r m r 2 r r m T 2 3 4 2 r r Kr 3 T GM M 4 2 K GM T 2 Kr 3 MSU Physics 231 Fall 2015 20 4 2 Two common cases K GM Planets and other objects orbiting the sun M (sun) 1.99 1030 kg K (sun) 297 10 21 s 2 / m 3 Moon and satellites orbiting the earth M (earth) 5.97 10 24 kg K (earth) 99 10 15 s 2 / m 3 MSU Physics 231 Fall 2015 21 Clicker Quiz! Averting Disaster a) it’s in Earth’s gravitational field The Moon does b) the net force on it is zero not crash into c) it is beyond the main pull of Earth’s gravity Earth because: d) it’s being pulled by the Sun as well as by Earth e) its velocity is large enough to stay in orbit e) The Moon does not crash into Earth because of its high speed. If it stopped moving, it would fall directly into Earth. With its high speed, the Moon would fly off into space if it weren’t for gravity providing the centripetal force. MSU Physics 231 Fall 2015 22 Clicker Quiz! Averting Disaster a) it’s in Earth’s gravitational field The Moon does b) the net force on it is zero not crash into c) it is beyond the main pull of Earth’s gravity Earth because: d) it’s being pulled by the Sun as well as by Earth e) its velocity is large enough to stay in orbit The Moon does not crash into Earth because of its high speed. If it stopped moving, it would fall directly into Earth. With its high speed, the Moon would fly off into space if it weren’t for gravity providing the centripetal force. MSU Physics 231 Fall 2015 23 Synchronous orbit Synchronous orbit of a satellite: rotation period of satellite of mass m is the same as rotation period of the planet For earth: period T = 24 hours = 86 x 103 s r3 = T2/K = 75 x 1021 r = 42 x 106 m Re = 6.4 x 106 m T 2 Kr 3 4 2 K GM K (earth) 99 10 15 s 2 / m 3 (r/Re) = 6.6 MSU Physics 231 Fall 2015 24 Total mechanical energy for Orbits Consider a planet in circular motion around the sun: F mac v m r M GMm mv 2 2 r r GM 2 v r mv 2 GMm KE 2 2r GMm GMm GMm TE PE KE r 2r 2r GMm TE 2r MSU Physics 231 Fall 2015 25 launch speed = 10 km/s MSU Physics 231 Fall 2015 26 Kepler’s laws Johannes Kepler (1571-1630) MSU Physics 231 Fall 2015 27 Kepler’s First law B A An object A bound to another object B by a force that goes with 1/r2 moves in an elliptical orbit around B, with B being in one of the focus point of the ellipse. MSU Physics 231 Fall 2015 28 area A ab MSU Physics 231 Fall 2015 29 MSU Physics 231 Fall 2015 30 Eccentricity: e c b e 1 a a 2 MSU Physics 231 Fall 2015 31 Eccentricity: e c b e 1 a a 2 circle when e = 0 MSU Physics 231 Fall 2015 32 Kepler’s First law Eccentricity: e c b e 1 a a 2 An object A bound to another object B by a force that goes with 1/r2 moves in an elliptical orbit around B, with B being in one of the focus point of the ellipse. MSU Physics 231 Fall 2015 33 Kepler’s second law A line drawn from the sun to the elliptical orbit of a planet sweeps out equal areas in equal time intervals. Area(D-C-SUN) = Area(B-A-SUN) A L constant t 2m L angular momentum A L for one revolution T 2m 2mA L (A area ab) T MSU Physics 231 Fall 2015 34 PEgravity=-GMEarthm/r Kepler’s second law L m r v constant rmax rmin rmin rmax speed and kinetic energy are smallest speed and kinetic energy are largest MSU Physics 231 Fall 2015 35 PEgravity=-GMEarthm/r Kepler’s third law T 2 K a3 K 297 10 21 s 2 / m 3 a same as for a circular orbit except r is replaced by a = ½(rmin+rmax) the semi-major axis (red line) MSU Physics 231 Fall 2015 36 An Example star A Two planets are orbiting a star. The B orbit of A has a radius of 1x108 km. The distance of closest approach of B to the star is 5x107 km and its maximum distance from the star is 1x109 km. If A has a rotational period of 1 year, what is the rotational period of B? Need to use Kepler’s 3rd Law MSU Physics 231 Fall 2015 37 An Example star B A Rmin = distance of closest approach = 5x10710(perihelion) Rmax = maximum distance = 1x109 (aphelion) RA = 1x108 km RB = ½(Rmin+Rmax) = ½(5x107 + 1x109) = 5.25x108 km R3/T2 = constant RA3/TA2= RB3/TB2 so TB2=(RB3/RA3)TA2 So TB=(5.253 x (1 yr)2) = 12 years MSU Physics 231 Fall 2015 38 Quantities that are constant for a given orbit period T 2 K a 3 A angular momentum L 2m T area A ab ellipse b a 2mA 2m (ab) b L 1/ 2 1/ 2 3 / 2 T (K ) a a GMm total energy TE 2a MSU Physics 231 Fall 2015 39 Quantities that depend on distance r b a r Mm potential energy PE G r L velocity v mr L angular velocity mr 2 kinetic energy KE (1 / 2)mv 2 MSU Physics 231 Fall 2015 40 MSU Physics 231 Fall 2015 41