1. Gravitation Newton’s law of universal gravitation (1687): r Sir Henry Cavendish (1731-1810) (a torsion balance) Principle of superposition of forces from different sources: Concept of fields for forces that act at a distance Weight depends on r. On earth’s surface w = mg Cavendish’s mass of the earth (since R E = 6380 km) Magnitude F~1 μN for m 1 =m 2 = = 100 kg at r = 1 m is very small.

2. Gravitational Potential Energy: W grav = U 1 - U 2 Let us choose U to be zero at infinity r = ∞: Gravitational force is conservative ! Derivation of the gravitational potential energy U = mgy near the earth’s surface ( r = R E +y )

3. Journey to the center of the earth For a point mass m inside spherical shell (M E – M) The force exerted by the shell is zero! Only the “inner” mass M contributes to the total gravitational force

4. Exam Example 27: Motion in the gravitational field of two bodies (problem 13.62) 0 X1X1 X2X2 X0X0 X ΔX=X-X 0 M1M1 M2M2 m Data: masses M 1, M 2, m; positions x 1, x 2, x 0, x; v(t=0)=0 Find: (a) change of the gravitational potential of the test particle m; (b) the final speed of the test particle at the final position x; (c) the acceleration of the test particle at the final position x. Solution : (a) (b) Energy conservation: (c) Newton’s 2 nd law:

5. The Motion of Satellites Closed orbits Open orbits V 1 (r=R E ) V 2 (r→∞) Orbits are closed or open. Open orbits are parabolic or hyperbolic. Closed orbits are elliptical or circular. There is only one speed that a satellite can have in a given circular orbit r : Period: “Synchronous satellites” for digital satellite system TV: T=1 sidereal day= 24 h(365-1)/365= 23 h 56 m, r =40000 km Global Positioning System (GPS) : 24 satellites Space speeds: 1 st - to stay in orbit r = R E V 1 =(GM E /R E ) 1/2 =(gR E ) 1/2 =8 km/s 2 nd - to leave the earth K = - U, mV 2 2 /2 = GM E m/R E, V 2 =(2GM E /R E ) 1/2 =(2) 1/2 V 1 =11.2 km/s (parabolic or escape speed) 3 rd - to leave the sun V 3 = 16.7 km/s > V 2 (sun) – V earth-sun = =[(2) 1/2 -1](GM sun /R earth’s-orbit ) 1/2 = 0.41 (29.8 km/s)= 12.3 km/s since both the sun’s and earth’s grav. attractions act. V3V3 V e-s =29.8km/s R e =150 Mkm Earth Sun Remark: Conservation of energy yields another relation between r and v,, that does not contradict to r×v=const along a trajectory. Pioneer-10: Earth (1972)→ Jupiter→ Pluto (1983)→ Solar system’s border (2002)→ star in Mly

6. On December 18, 2004, Voyager 1 passed the termination shock. This marks the point where the solar wind slows to subsonic speeds. This is the unofficial date of departure from the Solar System. While the spacecraft still remains under the sun's influence, at the termination shock particles from the interstellar medium interact with solar particles, signaling that the hypothetical heliopause is not far from this point. Six years later in 2010 Voyager 1 entered an area of the heliosheath where the solar wind outward speed is 0, or flowing sideways relative to the sun. This signals that Voyager 1 is getting very close to entering the interstellar medium.termination shockheliopauseinterstellar medium On December 5, 2011, it was announced that Voyager 1 had entered a new region referred to as a "cosmic purgatory" by NASA. Within this stagnation region, charged particles streaming from the sun slow and turn inward, and the solar system's magnetic field has doubled in strength as interstellar space appears to be applying pressure. Energetic particles originating in the solar system have declined by nearly half, while the detection of high-energy electrons from outside has increased by 100 fold. The inner edge of the stagnation region is located approximately 113 astronomical units from the sun, while the outer edge is unknown. At a distance of 120 astronomical units (1.8×10 10 km) as of February 2012, it is the farthest man-made object from Earth. Voyager 1 is now in the heliosheath, which is the outermost layer of the heliosphere.Earthheliosheathheliosphere It will most likely be the first probe to leave the Solar System.

7. Exam Example 28: Satellite in a Circular Orbit R E =6380 km r MEME m Data: r = 2R E, R E = 6380 km Find: (a) derive formula for speed v and find its value; (b) derive formula for the period T and find its value; (c) satellite’s acceleration. Solution: use the value g = GM E /R E 2 = 9.8 m/s 2 (a) The only centripetal force is the gravitational force: (b) The period T is a time required for one orbital revolution, that is (c) Newton’s second law with the central gravitational force yields a tan = 0 and a rad = a c = F g /m = GM E /r 2 = (GM E /R E 2 ) (R E /r) 2 = g/4 = 2.45 m/s 2

8. Kepler’s Laws of Planetary Motion (1609, 1619) Heliocentric world system (Copernicus, 1543) vs. Geocentric world system (Ptolemei) 1. Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse Eccentricity e = (Center O to Focus S) / (Semi-major axis a) The earth’s orbit has e = 0.017. Pluto (e =0.248) is not 9 th planet, it is a dwarf planet! 2. A line from the sun to a given planet sweeps out equal areas in equal times. Proof is based on the angular momentum conservation that follows from the fact that the gravitational force is a central force 3. The periods of the planets are proportional to the 3/2 powers of the major axis lengths of their orbits: Recent discovery: planets in orbit around thousands of other stars via detection of the apparent “wobble” of a star near the center of mass. Note: Kepler (1571-1630) discovered laws of planetary motion ~100 years before Newton (1642-1727) formulated laws of mechanics (1687) ! 1600-Giordano Bruno burned at the stake by Church for heresy: Copernican system, sun=star, eternal plurality of worlds 1633- Galileo Galilei sentenced by Church to imprisonment

9. Exam Example 29: Satellite in an Elliptical Orbit (problem 13.77) (perigee) (apogee) hphp haha 2R E Data: h p, h a, R E = 6380 km, M E = 6·10 24 kg Find: (a) eccentricity of the orbit e; (b) period T; (c) a rad; (d) ratio of speed at perigee to speed at apogee v p /v a ; (e) speed at perigee v p and speed at apogee v a ; (f) escape speeds at perigee v 2p and at apogee v 2a. Solution: (a) r p =h p +R E, r a = h a +R E, a =(r p +r a )/2, ea = a – r p, e = 1 – r p /a = 1- 2r p /(r p +r a ) = = (r a - r p )/(r a + r p ) = (h a -h p )/(h a +h p +2R E ) (d) Conservation of angular momentum (L a = L p ) or Kepler’s second law: r a v a = r p v p, v p /v a = r a /r p (b) Period of the elliptical orbit is the same as the period of the circular orbit with a radius equal to a semi-major axis R = a, i.e., (e) Conservation of mechanical energy K + U = const : (f) Conservation of mechanical energy for an escape from a distance r (the second space speed) : (c) Newton’s 2 nd law and law of gravitation: a rad = F grav / m = GM E /r 2.

10. Apparent Weight and the Earth’s Rotation At the equator: Small additional variations: (i) imperfect spherical symmetry, (ii) local variations in density, (iii) differences in elevation.

11. Black Holes Event horizon at Schwarzschild radius Supermassive black hole at the center of our Milky Way galaxy Sagittarius A*: R S = 8·10 6 km, M = 4·10 6 M sun = 8·10 36 kg “Black holes have no hair” and can be entirely characterized by energy, momentum, angular momentum, charge, and location. Primordial Black Holes and γ-Ray Bursts Black-hole potential well Objects fall into the black hole in a finite proper time Δt ~ R S /c, but infinitely long for a distant observer, and are affected by gravitational red shift, time dilation, and tidal forces effects ! All information is lost inside black hole for outside observers. E 0 =K 0 +U 0 <0 (trapped state) E=E 0 +W nc (dissipative fall into black hole) Energy r 0 E 0 =K 0 +U 0 >0 (free escape) R Ssun = 3 km, R Searth = 1 cm At r = 1.5 R its velocity is v = c. Circular motion can occur only at r > 1.5 R S. 4R S 2R S 1.5R S RSRS Cone of gravitational capture ( for v 2 /2 = GM/r ) Black hole is a bright source due to an accretion disk !