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Gravitation

Gravitation. Magnitude F~1 μ N for m 1 =m 2 = = 100 kg at r = 1 m is very small. Newton’s law of universal gravitation (1687):. r. Sir Henry Cavendish (1731-1810) (a torsion balance). Concept of fields for forces that

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Gravitation

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  1. Gravitation Magnitude F~1μN for m1=m2= = 100 kg at r = 1 m is very small. Newton’s law of universal gravitation (1687): r Sir Henry Cavendish (1731-1810) (a torsion balance) Concept of fields for forces that act at a distance Principle of superposition of forces from different sources: Weight depends on r. On earth’s surface w = mg Cavendish’s mass of the earth (since RE = 6380 km)

  2. Gravitational Potential Energy: Wgrav= U1 - U2 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 = RE+y )

  3. Journey to the center of the earth For a point mass m inside spherical shell (ME – 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) M2 M1 m ΔX=X-X0 0 X X0 X2 X1 Data: masses M1, M2, m; positions x1, x2, x0, 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 2nd law:

  5. The Motion of Satellites Re=150 Mkm Open orbits Orbits are closed or open. Open orbits are parabolic or hyperbolic. Closed orbits are elliptical or circular. V2(r→∞) There is only one speed that a satellite can have in a given circular orbit r : V1(r=RE) Closed orbits 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:1st- to stay in orbit r = RE V1=(GME/RE)1/2=(gRE)1/2=8 km/s 2nd- to leave the earth K = - U, mV22/2 = GMEm/RE , V2 =(2GME/RE)1/2 =(2)1/2 V1 =11.2 km/s (parabolic or escape speed) 3rd- to leave the sun V3= 16.7 km/s > V2(sun) – Vearth-sun= =[(2)1/2-1](GMsun/Rearth’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. V3 Earth Ve-s=29.8km/s 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 1passed 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. 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 (1.8×1010 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. It will most likely be the first probe to leave the Solar System.

  7. Exam Example 28: Satellite in a Circular Orbit m Data: r = 2RE , RE = 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. ME r Solution: use the value g = GME/RE2 = 9.8 m/s2 RE=6380 km (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 atan = 0 and arad = ac = Fg/m = GME/r2 = (GME/RE2) (RE/r)2 = g/4 = 2.45 m/s2

  8. Kepler’s Laws of Planetary Motion (1609, 1619) Heliocentric world system (Copernicus, 1543) vs. Geocentric world system (Ptolemei) 1600-Giordano Bruno burned at the stake by Church for heresy: Copernican system, sun=star, eternal plurality of worlds 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 9th 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: Note: Kepler (1571-1630) discovered laws of planetary motion ~100 years before Newton (1642-1727) formulated laws of mechanics (1687) ! 1633- Galileo Galilei sentenced by Church to imprisonment Recent discovery:planets in orbit around thousands of other stars via detection of the apparent “wobble” of a star near the center of mass.

  9. Exam Example 29: Satellite in an Elliptical Orbit (problem 13.77) Data: hp , ha , RE= 6380 km, ME= 6·1024 kg (apogee) (perigee) Find: (a) eccentricity of the orbit e; (b) period T; (c) arad; (d) ratio of speed at perigee to speed at apogee vp/va; (e) speed at perigee vp and speed at apogee va; (f) escape speeds at perigee v2p and at apogee v2a. 2RE hp ha Solution: (a) rp =hp+RE, ra= ha+RE, a =(rp+ra)/2, ea = a – rp, e = 1 – rp/a = 1- 2rp/(rp+ra) = = (ra- rp)/(ra+ rp) = (ha-hp)/(ha+hp+2RE) (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., (c) Newton’s 2nd law and law of gravitation: arad= Fgrav/ m = GME/r2. (d) Conservation of angular momentum (La= Lp) or Kepler’s second law: rava= rpvp, vp/va= ra/rp (e) Conservation of mechanical energy K + U = const : (f) Conservation of mechanical energy for an escape from a distance r (the second space speed) :

  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 Circular motion can occur only at r > 1.5 RS. Event horizon at Schwarzschild radius At r = 1.5 R its velocity is v = c. RSsun= 3 km, RSearth= 1 cm Black-hole potential well Energy E0=K0+U0>0 (free escape) r 0 E0=K0+U0<0 (trapped state) Black hole is a bright source due to an accretion disk ! E=E0+Wnc (dissipative fall into black hole) Objects fall into the black hole in a finite proper time Δt ~ RS/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. “Black holes have no hair” and can be entirely characterized by energy, momentum, angular momentum, charge, and location. Supermassive black hole at the center of our Milky Way galaxy Sagittarius A*: RS = 8·106 km, M = 4·106 Msun = 8·1036 kg Cone of gravitational capture ( for v2/2 = GM/r ) 2RS 4RS RS Primordial Black Holes and γ-Ray Bursts 1.5RS

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