More on Kepler’s Laws

1. More on Kepler’s Laws

2. Kepler’s 3rd Law • Can be shown that this also applies to an elliptical orbit with replacement of r with a, where a is the semimajor axis. • Ks 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, KSunis replaced with KEarth, where KEarth is obtained by replacing MSun by MEarth in the above equation.

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

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

5. Kepler’s Lawscan be derived fromNewton’s Laws. In particular, Kepler’s 3rd 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 rES = 1.5  1011 m

9. “Weighing” the Sun! • We’ve “weighed” the Earth, now lets “weigh” the Sun!! Assume: Earth & Sun are perfectuniform spheres. & Earth orbit is a perfect circle. • Note: For Earth, Mass ME= 5.99  1024kg Orbit period is T = 1 yr  3 107 s Orbit radius r = 1.5  1011 m So, orbit velocity is v = (2πr/T),v  3 104 m/s • Gravitational Force between Earth & Sun: Fg = G[(MSME)/r2] Circular orbit is circular  centripetal acceleration Newton’s 2nd Law gives: ∑F = Fg = MEa = MEac = ME(v2)/r OR:G[(MSME)/r2] = ME(v2)/r. If the Sun mass is unknown, solve for it: MS = (v2r)/G  2  1030 kg  3.3  105ME

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 mE) and Sun (mass mE), 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 RES, then the satellite’s orbital radius is (RES - 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 Mis 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 Fieldgis defined as • Gravitational Field = Gravitational Forceexperienced 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 fieldg 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