Physics 231 Topic 9: Gravitation Alex Brown October 30, 2015

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.

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

The gravitational force Newton: 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.

Gravitation between two objects 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’

Gravitation between two objects The force of the earth on the moon is equal and opposite to the force of the moon on the earth!

Clicker Quiz! Earth and Moon a) one quarter b) one half c) the same d) two times e) four times If the distance to the Moon were doubled, then the force of attraction between Earth and the Moon would be: 9

Clicker Quiz! Earth and Moon a) one quarter b) one half c) the same d) two times e) four times If the distance to the Moon were doubled, then the force of attraction between Earth and the Moon would be: 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. 10

Gravitational acceleration at the surface of planet with mass M F=mg g = GM/R2

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

Gravitational potential energy So far, we used: PEgravity=mgh Only valid for h near earth’s surface. More general: PEgravity=-GMm/r 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! R2=R+h R < R2 Thus… PE < PE2 R

Gravitational potential energy So far, we used: PEgravity=mgh Only valid for h near earth’s surface. More general: PEgravity=-GMm/r 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! R2=R+h R

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

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 v2 = 11.2 km/s

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…)

launch speed 4 km/s 6 km/s 8 km/s

First cosmic speed First cosmic speed: speed of a satellite of mass m on a low-lying circular around a planet with orbit of Mp and radius Rp (gp = GMp/Rp2) rsatellite ≈ Rp F = mac mgp = m v2/Rp so v1 =(gpRp) For earth: g = 9.81 m/s2 R = 6.37x106 m v1=7.91 km/s

Period for all orbits Consider an object in circular motion around a larger one

Two common cases Planets and other objects orbiting the sun Moon and satellites orbiting the earth

Clicker Quiz! Averting Disaster a) it’s in Earth’s gravitational field b) the net force on it is zero c) it is beyond the main pull of Earth’s gravity 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: 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. 22

Clicker Quiz! Averting Disaster a) it’s in Earth’s gravitational field b) the net force on it is zero c) it is beyond the main pull of Earth’s gravity 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: 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. 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 (r/Re) = 6.6

Total mechanical energy for Orbits Consider a planet in circular motion around the sun:

launch speed = 10 km/s

Kepler’s laws Johannes Kepler (1571-1630)

Kepler’s First law A B 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.

Eccentricity: e

Eccentricity: e circle when e = 0

Kepler’s First law Eccentricity: e 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.

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)

Kepler’s second law rmax rmin speed and kinetic energy are largest PEgravity=-GMEarthm/r rmax speed and kinetic energy are smallest rmin rmax rmin speed and kinetic energy are largest

Kepler’s third law PEgravity=-GMEarthm/r a a = ½(rmin+rmax) same as for a circular orbit except r is replaced by the semi-major axis (red line) a = ½(rmin+rmax)

An Example B star A Need to use Kepler’s 3rd Law Two planets are orbiting a star. The 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? B star A Need to use Kepler’s 3rd Law

An Example B star A RA = 1x108 km 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

Quantities that are constant for a given orbit

Quantities that depend on distance r b a r