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Physics 141Mechanics Lecture 18 Kepler's Laws of Planetary Motion Yongli Gao The motion of stars and planets has drawn people's imagination since the.

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Presentation on theme: "Physics 141Mechanics Lecture 18 Kepler's Laws of Planetary Motion Yongli Gao The motion of stars and planets has drawn people's imagination since the."— Presentation transcript:

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2 Physics 141Mechanics Lecture 18 Kepler's Laws of Planetary Motion Yongli Gao The motion of stars and planets has drawn people's imagination since the dawn of civilization. The motion of the planets were particularly puzzling. Kepler derived three laws based on the very accurate measurements of Tycho Brahe. Kepler's first law of planetary motion: All planets move along elliptical orbits, with the sun at one focus. A B F1F1 F2F2 O

3 Basics about an Ellipse An ellipse can be seen as an elongated circle. The distance a=OA is called the semi-major axis, b=OB the semi-minor axis. The eccentricity e=(a-b)/a. For a circle, a=b, and e=0. The sum of the distances from any point on the ellipse to the foci is the same. Mathematically, e is pretty small for planets (e E =0.0167) The closet point to the Sun is the perihelion, and the farthest is the aphelion. A B F1F1 F2F2 O x y r=(x,y)

4 Kepler's Second Law Kepler's second law says that a line connecting a planet to the sun sweeps out equal areas in equal times. The second law is in fact indicating the conservation of anglular momentum. The area swept is The conservation of momentum indicates that the force acting on the planet is along the line connecting to the the sun, and results in no torque. Sun

5 Kepler's Third Law Kepler's third law says that the square of the period T of a planet around the sun is proportional to the cube of a, the semi-major axis of its orbit. C is a constant independent of the planet around the sun. We see that the farther away is the planet the longer its period. Kepler's third law in fact gives a hint to the fundamental interaction in astronomy: gravitational interaction. If we ignore the small eccentricity of a planetary orbit, the motion is then uniform circular, Centripetal force

6 Newton's Law of Gravitation From Kepler's laws one can conclude that the force on the planet is proportional to its mass m and inversely proportional to the square of the distance to the sun R. The force is directly pointing to the sun. The conclusions lead to Newton's law of gravitation: all bodies attract each other. For point particles, every particle attracts any other particle with a gravitational force pointing along the line connecting the particles, and the magnitude is given by where m 1 and m 2 are the masses of the particles, r the distance between them, and G the gravitational constant G=6.67x10 -11 Nm 2 /kg 2.

7 Vector Form of Gravitational Force "-" sign means the force by m 1 on m 2 is attracting m 2 toward m 1. And F 21 =-F 12. x y z r1r1 r2r2 r 12 m1m1 m2m2 F 12 F 21

8 Example: Checking Newton's Gravitational Law How did Newton verify his law? He had formulated his 3 laws of dynamics, he knew the gravitational acceleration g at the surface the earth, the radius of the earth R E. The period of the moon T and its distance to the earth R were also known. Solution: Gravity force at the surface of the earth

9 For the moon, the centripetal force of its orbital motion around the earth is provided by the gravitational attraction, Mr. Newton must have been satisfied to see that his law works both on the earth and on the moon. Kepler's 3rd law can be written as

10 Potential Energy of Gravitational Force Let's imagine that M is fixed somehow and m is moving under the gravitational force on it by M. Potential energy change r F Mm M (fixed) m

11 If we choose the reference point as r i ->∞ and U(r i )=0, we get the gravitational potential energy For a system of n particles, r U

12 The force on the i th particle is Once we get the potential energy, we know everything about the gravitational interaction of the system. Note for the force one has to deal with vector sum.

13 Example: Three Particles If we have three particles m 1, m 2 and m 3 at corners of a triangle of equal sides a. Find the gravitational potential U and force on m 1. Solution: x y m1m1 m2m2 m3m3 a


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