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CAPSTONE Lecture 4

CAPSTONE Lecture 4. Laws of planet motion. 1. What are the units of G? (Set F=ma=-GmM/r 2 , solve for G, and work out the resulting units.).

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CAPSTONE Lecture 4

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  1. CAPSTONE Lecture 4 Laws of planet motion CAPSTONE.lecture 4.Planet motion

  2. 1. What are the units of G? (Set F=ma=-GmM/r2, solve for G, and work out the resulting units.) CAPSTONE.lecture 4.Planet motion

  3. 2. Calculate the orbital velocity of Earth around the Sun. Use the circumference of the (assumed) circular orbit and the time it takes for one orbit to get the velocity. This should give the same answer as the equation vE=(GMS/r)1/2, where vE is the unknown velocity of Earth in its orbit, G is the Gravitational constant, Ms is the mass of the Sun and r is the Earth-Sun distance. (one astronomical unit, r=1.5 x 1013 cm.) Calculate the orbital velocity of Jupiter around the Sun. Calculate the orbital velocity of the Moon around the Earth CAPSTONE.lecture 4.Planet motion

  4. . 3. a) What is the kinetic energy of a 55 kg runner while running the 100 yard dash in 10 seconds? Use her average velocity. b) What is the kinetic energy of Jupiter in motion around the Sun? 4. a) What is the potential energy of Jupiter in orbit around the Sun? (Ignore the other planets and nearby stars.) b) Verify that K=-U/2 for an object in orbit around another body.(the Virial theorem.) Do this using the numbers above, and analytically. CAPSTONE.lecture 4.Planet motion

  5. Kepler’s Law • P2=R3, P=period, R=average orbit radius. • Consider a body with M1, v1 at distance r from M2, v2. R=r1+r2. r1 and r2 distances from the point about which both objects seem to move (center of mass). CAPSTONE.lecture 4.Planet motion

  6. Center of Mass • (M2v22)/r2=(M1v12/r1). (Huygens) (1) • v22/v12=M1r2/M2r1=(2r2/P2)2/ (r1/P1)2, (2) since, P1=2r1/v1, P2=2r2/v2 (3) But, note, P1=P2 (both complete one round at the same time) v22/v12=M1r2/M2r1=(2r2/P2)2/ (r1/P1)2 • From ( 2), M1r2/M2r1=r22/r12 so M1r1=M2r2 (4) • This is the definition of center of mass. The more massive object is closer to the center of mass. • From eq. 3 and 4, r1/r2=M2/M1=v1/v2 • The more massive object is closer to the center of mass and moves slowest. • For newly discovered, extra-solar planets, it is the very tiny vel. of the massive star that is seen (<10 m/sec) CAPSTONE.lecture 4.Planet motion

  7. Final step • The force acting on any one of the two bodies can be expressed in terms of the overall system, or the center of mass system. The consequences are the same. • Pick the second body, there are two expressions for force that are equal. • F=-GM1M2/(r1+r2)2=-M2v22/r2. (6) • F=…………………..=[-M2v22/r2][42r2/ 42r2] • F=……………………=[M2 42r2][v22/ 42r22] (7) • But, the last term is just [1/P2], where P=2r2/v2 • GM1M2/R2 = M2r24/ P2 (since R=r1+r2) • GM1M2/R2 = M2r24/ P2 (since R=r1+r2) • GM1/R2 = r24/ P2 (8) CAPSTONE.lecture 4.Planet motion

  8. Conclusion • Now, rewrite some terms: • R=r1+r2=r2(1+r1/r2), or r2=R/ (1+r1/r2) • M=M1+M2=M1(1+M2/M1) or M1=M / (1+M2/M1) • Form the ratio M/R M/R =[ M1(1+M2/M1)] / r2(1+r1/r2)] • From the equation 4), r1/r2= M2/M1 • M/R = [ M1(1+M2/M1)] / r2(1+M2/M1)] = M1/r2 • M/R = [ M1(1+M2/M1)] / r2(1+M2/M1)] = M1/r2 • Equation 8 tells us that • GM1/R2 = r24/ P2, so G(M1/r2)/R2 = 4/ P2 • G(M/R)/R2 = 4/ P2 • GM/R3 = 4/ P2 CAPSTONE.lecture 4.Planet motion

  9. If we talk about the Earth/Sun system, we know R=1AU, P=1 year, M = 1 solar mass. Convert G (cm3/g-sec2) to G (AU3/M(solar)-yr2), then 42/G=1 and R(AU)3=P(yr)2. Newton derived from first principles the same empirical law as Kepler for planetary motion. QED CAPSTONE.lecture 4.Planet motion

  10. 42/GM=1 in units of AU, Mo, yr. • P(s)2={4/[GM(Mo)]}R(AU)3, where Mo=1 solar mass • If Kepler finds P(yr)2=R(AU)3, the true form must be • P(yr)2=R(AU)3/M(Mo) • Therefore, 42/GM = 1, M=1, 42/G=1 • or G=4 2 in solar system units ([AU3/Mo-yr2]) • G=6.67 x 10-8 cm3/g-sec2, • 1=1AU/1.5x1013 cm,1=1yr/ x107s, 1=1Mo/ 2x1033. CAPSTONE.lecture 4.Planet motion

  11. G=[6.67 x 10-8 x(1AU/1.5x1013)3]/[(1 Mo/2x1033)(1yr/x107)2)] cm3/g-sec2 G=[6.67x22/(1.5)3]x[10-8x10-39 x1033x1014][AU3/Mo-yr2] G=[(13.34/3.375)] x 2 x 10-47 x 1047 [AU3/Mo-yr2] G=3.95 2 QED CAPSTONE.lecture 4.Planet motion

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