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! Gravitational potential energy
m1m2 r Physics for Scientists & Engineers 1
Spring Semester 2006 Lecture 40 U(r) = !G Note: always U < 0 ! Escape speed vesc = 2GM R Earth: 11.2 km/s Sun: 42 km/s BUT, use the velocity of the earth April 4, 2006 Physics for Scientists&Engineers 1 1 April 4, 2006 Physics for Scientists&Engineers 1 2 Kepler's Laws
! Johannes Kepler (15711630) used empirical observations, mainly from the data gathered by Tycho Brahe, and sophisticated calculations, to arrive at his famous laws of planetary motion ! Published in 1609 and 1619. ! This observation occurred decades before Isaac Newton was born in 1643 ! Kepler and other pioneers, like Copernicus and Galileo Galilei, changed the geocentric worldview into a heliocentric worldview ! Main selling point: much greater simplicity in this model, and easy explanation of epicycles Motion of the Planets in the Solar System
! The strange motion of the planets as viewed from Earth is simple to understand if viewed from the sun.
Sun Motion of Sun, Earth, and Mars viewed from the Sun Earth Mars Motion of Sun, Earth, and Mars viewed from the Earth April 4, 2006 Physics for Scientists&Engineers 1 3 April 4, 2006 Physics for Scientists&Engineers 1 4 Motion of the Planets in the Solar System (2)
viewed from the Sun viewed from the Earth Kepler's First Law: Orbits
! All planets move in elliptical orbits with the Sun at one focal point Sun Mercury Venus Earth Mars Jupiter Saturn ! The epicyclic motion is easy to understand when viewed from the sun. ! But Brahe's data was so precise, that even description in terms of circles around the sun was not good enough.
April 4, 2006 Physics for Scientists&Engineers 1 5 April 4, 2006 Physics for Scientists&Engineers 1 6 Ellipses (1)
! An ellipse is a closed curve in a twodimensional plane. It has two focal points (f1,f2), separated by distance 2c. ! Any point on ellipse has the condition that the sum of the distances to the two focal points is: r1 + r2 = 2a where a is the length of the semimajor axis ! Define semiminor axis b: b2 = a2  c2. ! Cartesian coordinates of points on ellipse Ellipses (2)
! If a = b => c = 0, then the ellipse becomes a circle ! Eccentricity of ellipse e= c b2 = 1! 2 a a x 2 y2 + =1 a2 b2
April 4, 2006 Physics for Scientists&Engineers 1 7 ! e = 0 => circle ! Shown here: e = 0.6 ! Eccentricity for Earth's orbit: e = 0.017 ! Pluto: e = 0.244 Venus: e = 0.007 All other planets are between these two extremes
April 4, 2006 Physics for Scientists&Engineers 1 8 Typical orbits Earth's Orbit
! The semiminor axis' length of Earth's orbit is approximately 99.98% of that of the semimajor axis. ! At its closest approach to the Sun, the perihelion, the Earth is 147.1 million km away from the Sun. ! The aphelion, the furthest point from the Sun, is 152.6 million km away from the Sun. ! Change in seasons is not caused by the eccentricity of Earth's orbit. (The point of closest approach to the Sun is reached in early January each year, in the middle of the cold season in the Northern hemisphere.) ! Instead, the seasons are caused by the fact that the Earth's axis of rotation is tilted by an angle of 23.45 degrees relative to the plane of the orbital ellipse, exposing more of the Northern hemisphere to the Sun's rays during our summer months. April 4, 2006 Physics for Scientists&Engineers 1 9 April 4, 2006 Physics for Scientists&Engineers 1 10 Kepler's Second Law: Areas
! A straight line connecting the center of the Sun and the center of any planet sweeps out an equal area in any given time interval Kepler's Third Law: Periods
! The square of the period of a planet is proportional to the cube of the semimajor axis of the orbit dA = const. dt T2 = const. r3
T12 T22 = 3 r13 r2 April 4, 2006 Physics for Scientists&Engineers 1 11 April 4, 2006 Physics for Scientists&Engineers 1 12 Example: Kepler's Third Law
! What is the length of a year on Jupiter? ! Answer: straightforward application of Kepler's Third Law Proof of Kepler's Laws (circular orbits)
! Circular orbit: centripetal force provided by gravity
v2 Mm GM Mass cancels out! =G 2 ! v= r r r v GM ! Angular velocity ! = = (constant, because r=const.) r r3 m !r $ T2 = T1 # 2 & "r%
1 3/2 ! Orbital radius of Jupiter: 778.6 million km ! Period T= 2! r3 r3 = 2! # T 2 = (2! ) 2 " GM GM ! 778.6 $ T2 = (1 year) # " 149.6 & % 3/2 = 11.9 years T 2 4! 2 = r 3 GM
! Third Law
13 April 4, 2006 Physics for Scientists&Engineers 1 14 ! One "year" on Jupiter corresponds to 11.9 years on earth
April 4, 2006 Physics for Scientists&Engineers 1 Proof of Kepler's Laws (2)
! Angular momentum L is constant throughout orbit. Mass of the Sun
! Kepler's third Law: T 2 4! 2 = r 3 GM
4! 2 r 3 GT 2 2 ! Area swept out "A = 1 rs = 1 r "! 2 2 ! Solve for M: M= dA 1 2 d" 1 2 L = 2r = 2r # = = const. dt dt 2m
where ! Insert numbers for Earth's orbit (T=1 y, r=150 million km)
4! 2 (1.496 "1011 m)3 M= = 1.99 "10 30 kg #11 3 1 #2 7 2 (6.67 "10 m kg s )(3.15 "10 s) ! L = mr 2" ! Second Law. ! This holds for all orbits. !
April 4, 2006 Physics for Scientists&Engineers 1 15 ! In the same way we can determine the mass of Earth from the Moon's orbit, or the mass of the galaxy from stars orbits
April 4, 2006 Physics for Scientists&Engineers 1 16 ...
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This note was uploaded on 03/19/2008 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.
 Spring '08
 Wolf
 Physics, Energy, Potential Energy

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