PHY 151: Lecture Newton’s Law of Universal Gravitation 11.2 Structural Models 11.3 Kepler’s Laws 11.4 Energy Considerations in Planetary and Satellite Motion 11.5 Atomic Spectra and the Bohr Theory of Hydrogen 11.6 Context Connection: Changing from a Circular to an Elliptical Orbit Context 2: A Successful Mission Plan

PHY 151: Lecture 11 Gravity, Planetary Orbits, and the Hydrogen Atom 11.1 Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation - 1 Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them: G is the universal gravitational constant:

Newton’s Law of Universal Gravitation - 2 This is an example of an inverse square law The magnitude of the force varies as the inverse square of the separation of the particles The law can also be expressed in vector form:

Newton’s Law of Universal Gravitation - 3

Newton’s Law of Universal Gravitation - 4 The gravitational force exerted by a finite- sized, spherically symmetric mass distribution on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at its center

Newton’s Law of Universal Gravitation - 5 For example, the force on a particle of mass m at the Earth’s surface has the magnitude where M E is the Earth’s mass and R E is the Earth’s radius This force is directed toward the center of the Earth

Newton’s Law of Universal Gravitation - 6 G was first measured by Henry Cavendish in 1798 The apparatus shown allowed the attractive force between two spheres to cause the rod to rotate The mirror amplifies the motion It was repeated for various masses

Newton’s Law of Universal Gravitation - 7 Gravitation is a field force that always exists between two particles, regardless of the medium between them The force decreases rapidly as distance increases A consequence of the inverse square law

Newton’s Law of Universal Gravitation - 8 Always distinguish between G and g G is the universal gravitational constant It is the same everywhere g is the acceleration due to gravity g = 9.80 m/s 2 at the surface of the Earth g will vary by location

Newton’s Law of Universal Gravitation - 9 Use the mental representation of a field A source mass creates a gravitational field throughout the space around it A test mass located in the field experiences a gravitational force The gravitational field is defined as

Newton’s Law of Universal Gravitation - 10 Consider an object of mass m near the Earth’s surface The gravitational field at some point has the value of the free fall acceleration: –At the surface of the Earth, r = R E and g = 9.80 m/s 2

Newton’s Law of Universal Gravitation - 11 Figure a: The gravitational field vectors in the vicinity of a uniform spherical mass –the vectors vary in magnitude and direction Figure b: The gravitational field vectors in a small region near the Earth’s surface –the vectors are uniform

Example 11.1 Using the known radius of the Earth and that g = 9.80 m/s 2 at the Earth’s surface, find the average density of the Earth. –Solve for the mass of the Earth: –Substitute into the definition of density:

PHY 151: Lecture 11 Gravity, Planetary Orbits, and the Hydrogen Atom 11.2 Structural Models Skipped

PHY 151: Lecture 11 Gravity, Planetary Orbits, and the Hydrogen Atom 11.3 Kepler’s Laws Skipped

PHY 151: Lecture 11 Gravity, Planetary Orbits, and the Hydrogen Atom 11.4 Energy Considerations in Planetary and Satellite Motion

Energy Considerations in Planetary and Satellite Motion - 1 Consider an object of mass m moving with a speed v in the vicinity of a massive object M M >> m We can assume M is at rest The total energy of the two object system: The gravitational potential energy is

Energy Considerations in Planetary and Satellite Motion - 2 Since U g goes to zero as r goes to infinity, the total energy becomes

Energy Considerations in Planetary and Satellite Motion - 3 For a bound system, E < 0 Total energy becomes Multiplying by r and dividing by 2: Substituting:

Energy Considerations in Planetary and Satellite Motion - 4 This shows the total energy must be negative for circular orbits The kinetic energy of an object in a circular orbit is one-half the magnitude of the potential energy of the system

Energy Considerations in Planetary and Satellite Motion - 5 The total mechanical energy is also negative in the case of elliptical orbits The total energy is r is replaced with a, the semimajor axis

Example 11.3 A space transportation vehicle releases a 470-kg communications satellite while in an orbit 280 km above the surface of the Earth. A rocket engine on the satellite boosts it into a geosynchronous orbit. How much energy does the engine have to provide?

Example 11.3 –Find the initial radius of the satellite’s orbit when it is still in the vehicle’s cargo bay: –Find the difference in energies for the satellite– Earth system with the satellite at the initial and final radii:

Example 11.3 Substitute numerical values:

Energy Considerations in Planetary and Satellite Motion - 6 An object of mass m is projected upward from the Earth’s surface with an initial speed, v i Use energy considerations to find the minimum value of the initial speed needed to allow the object to move infinitely far away from the Earth

Energy Considerations in Planetary and Satellite Motion - 7 Total energy of the isolated object-Earth system is conserved, so

Energy Considerations in Planetary and Satellite Motion - 8 This minimum speed is called the escape speed Note, v esc is independent of the mass of the object The result is independent of the direction of the velocity and ignores air resistance

Energy Considerations in Planetary and Satellite Motion - 9 The Earth’s result can be extended to any planet: The table at right gives some escape speeds from various objects

Energy Considerations in Planetary and Satellite Motion - 10 This explains why some planets have atmospheres and others do not Lighter molecules have higher average speeds and are more likely to reach escape speeds This also explains the composition of the atmosphere

Example 11.4 Calculate the escape speed from the Earth for a 5000-kg spacecraft and determine the kinetic energy it must have at the Earth’s surface to move infinitely far away from the Earth.

Example 11.4 Find the escape speed: Evaluate the kinetic energy of the spacecraft:

Energy Considerations in Planetary and Satellite Motion - 11 A black hole is the remains of a star that has collapsed under its own gravitational force The escape speed for a black hole is very large due to the concentration of a large mass into a sphere of very small radius If the escape speed exceeds the speed of light, radiation cannot escape and it appears black

Energy Considerations in Planetary and Satellite Motion - 12 The critical radius at which the escape speed equals c is called the Schwarzschild radius, R S The imaginary surface of a sphere with this radius is called the event horizon This is the limit of how close you can approach the black hole and still escape

Energy Considerations in Planetary and Satellite Motion - 13 Although light from a black hole cannot escape, light from events taking place near the black hole should be visible If a binary star system has a black hole and a normal star, the material from the normal star can be pulled into the black hole

Energy Considerations in Planetary and Satellite Motion - 14 This material forms an accretion disk around the black hole Friction among the particles in the disk transforms mechanical energy into internal energy

Energy Considerations in Planetary and Satellite Motion - 15 The orbital height of the material above the event horizon decreases and the temperature rises The high-temperature material emits radiation, extending well into the x-ray region These x-rays are characteristics of black holes

Energy Considerations in Planetary and Satellite Motion - 17 There is evidence that supermassive black holes exist at the centers of galaxies Theory predicts jets of materials should be evident along the rotational axis of the black hole

Energy Considerations in Planetary and Satellite Motion -18 Gravity waves are ripples in space-time caused by changes in a gravitational system The ripples may be caused by a black hole forming from a collapsing star or other black holes The Laser Interferometer Gravitational Wave Observatory (LIGO) is being built to try to detect gravitational waves

PHY 151: Lecture 11 Gravity, Planetary Orbits, and the Hydrogen Atom 11.5 Atomic Spectra and the Bohr Theory of Hydrogen Skipped

PHY 151: Lecture 11 Gravity, Planetary Orbits, and the Hydrogen Atom 11.6 Context Connection: Changing from a Circular to an Elliptical Orbit Skipped

PHY 151: Lecture 11 Gravity, Planetary Orbits, and the Hydrogen Atom Context 2 A Successful Mission Plan Skipped

Centripetal Force Satellites in Orbit - 1 Satellite of mass m is in orbit around earth Gravitational pull of earth provides centripetal force  F c = GM e m/r 2 = mv 2 /r  GM e /r = v 2  v = sqrt(GM e /r) Orbital velocity for a given radius is independent of mass of satellite

Centripetal Force Satellites in Orbit - 2 v = sqrt(GM e /r)(from prior slide) v = 2  r/T(definition of period) 2  r/T = sqrt(GM e /r) T 2 = 4  2 r 3 /Gm e r 3 = T 2 Gm e /4  2 Relationships between period, T, and orbital radius r

Low Earth Orbit - 1 Velocity of a satellite in low earth orbit M e = 5.98 x kg r e = 6.37 x 10 6 m Height of satellite = 120 miles = 0.19 x 10 6 m r orbit = 6.56 x 10 6 m G = 6.67 x nm 2 /kg 2 v = sqrt(GM e /r) v = sqrt(6.67x x 5.98 x / 6.56 x 10 6 ) v = 7797 m/s = 17,446 mi/hour

Low Earth Orbit - 2 Period of satellite in low earth orbit T 2 = 4  2 r 3 /Gm e r e = 6.38 x 10 6 m Height of satellite = 120 miles = 0.19 x 10 6 m r orbit = 6.56 x 10 6 m m e = 5.98 x kg T 2 = 4  2 (6.38 x 10 6 ) 3 /[6.673 x x 5.98 x ] T 2 = s 2 T = 5286 s = 88 minutes

Geosynchronous Orbit - 1 A satellite in geosynchronous orbit stays above a specific place on the earth For this to occur, the period of the orbit must be 24 hours = s r 3 = T 2 Gm e /4  2 r 3 = (86400) 2 x [6.673 x x 5.98 x ] /4  2 r 3 = 7.55 x m 3 r = 4.23 x 10 7 m = 26,268 miles This is distance from center of the earth

Geosynchronous Orbit - 2 r = 4.23 x 10 7 m = 26,268 miles This is distance from center of the earth Radius of earth is 6.38 x 10 6 m Distance from surface is 4.23 x 10 7 – 0.64 x 10 7 = 3.59 x 10 7 m 22,312 miles

Satellites in Orbit – Example 3 Earth orbits the sun as a satellite. What is sun’s mass? Distance from earth to sun, r se = 1.5 x m Period of earth around sun  T = days x 24 x 60 x 60 = x 10 7 s d = vt v = d/t = 2  r se /T GM sun /r se = v 2 = 4  2 r se 2 /T 2 M sun = 4  2 r se 3 /GT 2 M sun = 4  2 (1.5 x ) 3 /6.67x /(3.156 x 10 7 ) 2 M sun = 2 x kg M sun /M earth = 2 x / 6 x = 330,000

Satellites in Orbit – Example 4 Moon is satellite of Earth Io is satellite of Jupiter What is ratio of Jupiter’s mass to earth’s mass Period, T moon, of the moon is 27 days Period, T io, of Io is 1.5 days Radii of the moon and Io are approximately the same M earth = 4  2 r me 3 /GT moon 2 M jupiter = 4  2 r ij 3 /GT io 2 M jupiter /M earth = T moon 2 /T io 2 = (27/1.5) 2 = 324

Satellites in Orbit – Example 5 Earth and Jupiter are satellites of the sun What is ratio of Jupiter’s period to earth’s period 1 AU = 1.5 x m, distance from earth to sun Earth’s distance from sun is 1 AU (astronomical unit) Jupiter’s distance from sun is 5.2 AU M sun = 4  2 r se 3 /GT earth 2 T earth 2 = 4  2 r se 3 /GM sun T jupiter 2 = 4  2 r sj 3 /GM sun T jupiter /T earth = sqrt(r sj 3 /r se 3) = sqrt(5.2 3 /1 3 ) = 11.86

Stars Orbiting Edge of Galaxy - 1 Mass of galaxy is 2 x kg This is the mass of visible stars and dust A star orbits the galaxy at 2.7 x m from the center of the galaxy This is close to the edge of the galaxy Assume 90% or more of the galactic mass is inside the star’s orbit Find a formula for the orbital velocity of star’s at the edge of the galaxy

Stars Orbiting Edge of Galaxy - 2

Graph

Actual Data Rotation curve of a typical spiral galaxy: predicted (A) and observed (B). The discrepancy between the curves is attributed to dark matter.