1. More on Kepler’s Laws

2. Can be shown that this also applies to an elliptical orbit with replacement of r with a, where a is the semimajor axis. K s is independent of the planet mass, & is valid for any planet Note: If an object is orbiting another object, the value of K will depend on the mass of the object being orbited. For example, for the Moon’s orbit around the Earth, K Sun is replaced with K Earth, where K Earth is obtained by replacing M Sun by M Earth in the above equation. Kepler’s 3 rd Law

3. The square of a planet’s orbital period is proportional to the cube of its mean distance from the Sun. Kepler’s 3 rd Law

4. Table 13-2, p. 370 Solar System Data

5. Kepler’s Laws can be derived from Newton’s Laws. In particular, Kepler’s 3 rd Law follows directly from the Universal Law Gravitation: Equating the gravitational force with the centripetal force shows that, for any two planets (assuming circular orbits, and that the only gravitational influence is the Sun):

6. Irregularities in planetary motion led to the discovery of Neptune, and irregularities in stellar motion have led to the discovery of many planets outside our solar system.

7. Example 6-8: Where is Mars? Mars’ period (its “year”) was first noted by Kepler to be about 687 days (Earth-days), which is (687 d/365 d) = 1.88 yr (Earth years). Determine the mean distance of Mars from the Sun using the Earth as a reference.

8. Example 6-9: The Sun’s mass determined. Determine the mass of the Sun given the Earth’s distance from the Sun as r ES = 1.5  10 11 m

9. “Weighing” the Sun! We’ve “weighed” the Earth, now lets “weigh” the Sun!! Assume: Earth & Sun are perfect uniform spheres. & Earth orbit is a perfect circle. Note: For Earth, Mass M E = 5.99  10 24 kg Orbit period is T = 1 yr  3  10 7 s Orbit radius r = 1.5  10 11 m So, orbit velocity is v = (2πr/T), v  3  10 4 m/s Gravitational Force between Earth & Sun: F g = G[(M S M E )/r 2 ] Circular orbit is circular  centripetal acceleration Newton’s 2 nd Law gives: ∑F = F g = M E a = M E a c = M E (v 2 )/r OR: G[(M S M E )/r 2 ] = M E (v 2 )/r. If the Sun mass is unknown, solve for it: M S = (v 2 r)/G  2  10 30 kg  3.3  10 5 M E

10. Example 6-10: Lagrange point. The mathematician Joseph-Louis Lagrange discovered five special points in the vicinity of the Earth’s orbit about the Sun where a small satellite (mass m) can orbit the Sun with the same period T as Earth’s (= 1 year). One of these “Lagrange Points,” called L1, lies between the Earth (mass m E ) and Sun (mass m E ), on the line connecting them. That is, the Earth and the satellite are always separated by a distance d. If the Earth’s orbital radius is R ES, then the satellite’s orbital radius is (R ES - d). Determine d.

11. Sect. 6-6: Gravitational Field

12. The Gravitational Field is defined as the gravitational force per unit mass: The Gravitational Field due to a single mass M is given by:

13. A Gravitational Field  g, exists at all points in space. If a particle of mass m is placed at a point where the gravitational field is g, it experiences a force: The field exerts a force F on the particle. The Gravitational Field g is defined as Gravitational Field = Gravitational Force experienced by a “test” particle placed at that point divided by the mass of the test particle. –The presence of the test particle is not necessary for the field to exist A source particle creates the field

14. The gravitational field g vectors point in the direction of the acceleration a particle would experience if it were placed in that field. Figure The field magnitude is that of the freefall acceleration, g, at that location. The gravitational field g describes the “effect” that any object has on the empty space around itself in terms of the force that would be present if a second object were somewhere in that space