2. Physics 1710—Warm-up Quiz If two (spherical) asteroids are in contact and are attracted with a force of 1 Newton, how much more or less force would they experience if they both had been twice as large (assuming the same density)? A.The same B.Twice as much (2x) C.Half as much (1/2x) D. 4 x E.8 x F.16 x

3. Solution: F = G M 1 M 2 /r 2 M 1 = M 2 = 4π/3 ρR 3 R = r F = G (4π/3 ρR 3 ) 2 /R 2 F∝R4F∝R4 F 2 /F 1 = (R 2 /R 1 ) 4 = 2 4 = 16 Physics 1710—Chapter 12 Apps: Gravity

4. 1′ Lecture F g = - řG Mm/d 2F g = - řG Mm/d 2 G = 6.673 x 10 –11 N m 2 /kg 2 ~ 2/3 x 10 –10 N m 2 /kg 2G = 6.673 x 10 –11 N m 2 /kg 2 ~ 2/3 x 10 –10 N m 2 /kg 2 The gravitational force constant g is equal to g = G M/(R+h) 2, M and R are the mass and radius of the planet.The gravitational force constant g is equal to g = G M/(R+h) 2, M and R are the mass and radius of the planet. The gravitation field is the force divided by the mass.The gravitation field is the force divided by the mass. The gravitation potential energy for a point mass is proportional to the product of the masses and inversely proportional to the distance between their centers.The gravitation potential energy for a point mass is proportional to the product of the masses and inversely proportional to the distance between their centers. Physics 1710—Chapter 13 Apps: Gravity

5. 1′ Lecture (cont’d.) Kepler’s LawsKepler’s Laws –The orbits of the planets are ellipses. –The areal velocity of a planet is constant. –The cube of the radius r 3 of a planet’s orbit is proportional to the square of the period T 2. is proportional to the square of the period T 2. Physics 1710—Chapter 13 Apps: Gravity

6. Which best corresponds to the actual shape of a planet’s orbit? A. A. B. B. C. C. D.None of the above. Physics 1710—Chapter 13 Apps: Gravity

7. Solution: Kepler’s First Law: The orbits of the planets are ellipses. r = r o /[1+ε cos θ] ε = eccentricity Physics 1710—Chapter 12 Apps: Gravity r o r o /(1+ε) r o /(1-ε) r o Planet ε ☿ 0.2056 ♀ 0.0068 ♂ 0.0934 ♃ 0.0483 ♄ 0.0560 ♅ 0.0461 ♆ 0.0097 ♇ 0.2482

8. History: Kepler’s First Law: The orbits of the planets are ellipses. What is the significance? Physics 1710—Chapter 12 Apps: Gravity Repudiation of Aristotle F is inverse square law

9. Isaac Newton’s Universal Law of Gravitation F = - G M m/ d 2 Physics 1710—Chapter 13 Apps: Gravity d moon d apple

10. Kepler’s Laws: The orbits are ellipses.The orbits are ellipses. The areal velocity is a constant. The areal velocity is a constant. T 2 ∝ r 3 implies F ∝ 1/ r 2, only. T 2 ∝ r 3 implies F ∝ 1/ r 2, only. Physics 1710—Chapter 13 Apps: Gravity ⊙ T 2 r 3 = r 3

11. Comment: Kepler’s Second Law: The areal velocity is constant. ½ r 2 ω= constant. Physics 1710—Chapter 12 Apps: Gravity Why? mr 2 ω = L = constant. L is conserved if no torque, i.e. F is “central force.”

12. Kepler’s Laws: The orbits are ellipses. (Contrary to Aristotle and Ptolemy.)The orbits are ellipses. (Contrary to Aristotle and Ptolemy.) A central force: F ∝ 1/ r 2 or r 2 The areal velocity is a constant. The areal velocity is a constant. Angular momentum is conserved: ½ v r ∆t = constant implies that rmv = L = constant. T 2 ∝ r 3 implies F ∝ 1/ r 2, only. T 2 ∝ r 3 implies F ∝ 1/ r 2, only. Physics 1710—Chapter 13 Apps: Gravity ⊙ T 2 r 3 = r 3

13. How did Newton figure out UL of G? Fact: a moon circling a planet has an acceleration of a = v 2 /r Fact: a moon circling a planet has an acceleration of a = v 2 /r Fact: a = F/m. Fact: a = F/m. Fact: Kepler had found that the square of the period T was proportional to the cube of the radius of the orbit r : Fact: Kepler had found that the square of the period T was proportional to the cube of the radius of the orbit r : T 2 = k r 3. Thus: v = 2π r / T Physics 1710—Chapter 12 Apps: Gravity

14. And T 2 = (2 π r) 2 /(F r /m) = k r 3 Thus: F = (2 π ) 2 m/(k r 2 ) An “inverse square law,” with k = 1/ [(2 π ) 2 G M] F = G Mm/ r 2, But what value is G? Physics 1710—Chapter 12 Apps: Gravity

15. Isaac Newton’s Universal Law of Gravitation F = - G M m/ d 2 Physics 1710—Chapter 13 Apps: Gravity d moon d apple F = – g m G M ♁ / R ♁ 2 g = G M ♁ / R ♁ 2

16. Gravitation ♁ G M ♁ / R ♁ 2 g ♁ = G M ♁ / R ♁ 2 G M ♁ R ♁ 2 = (9.80 N/kg)(6.37x10 6 m) G M ♁ = g R ♁ 2 = (9.80 N/kg)(6.37x10 6 m) = 3.99x10 14 N m 2 /kg Physics 1710—Chapter 12 Apps: Gravity Need to know G or M.

17. Henry Cavendish And the Cavendish Experiment Physics 1710—Chapter 13 Apps: Gravity m M d

18. G = 6.673 x 10 -11 N ‧m 2 /kg 2 G ≈ 2/3 x 10 -10 N ‧m 2 /kg 2 (to an accuracy of 0.1%) Physics 1710—Chapter 13 Apps: Gravity So, ♁ 3.99x10 14 N m 2 /kg)/(6.673 x 10 -11 N ‧m 2 /kg 2 ) M ♁ = ( 3.99x10 14 N m 2 /kg)/(6.673 x 10 -11 N ‧m 2 /kg 2 ) = 5.98 x10 24 kg

19. No Talking! Think! Confer! Peer Instruction Time What is g on Mars? M = 0.107 M ♁, R = 0.532 R ♁ What is g ♂ on Mars? M ♂ = 0.107 M ♁, R ♂ = 0.532 R ♁ Physics 1710—Chapter 13 Apps: Gravity

20. What is g on Mars? What is g ♂ on Mars? A.25.9 m/sec 2 B.9.80 m/sec 2 C.3.70 m/sec 2 D.1.97 m/sec 2 E.None of the above. Physics 1710—Chapter 13 Apps: Gravity

21. Peer Instruction Time What is g on Mars? What is g ♂ on Mars? Physics 1710—Chapter 13 Apps: Gravity ♁ M/M ♁ )(R ♁ /R 2 = (0.107)(1/0.532) 2 g ♂ /g ♁ =( M ♂ /M ♁ )(R ♁ /R ♂ ) 2 = (0.107)(1/0.532) 2 = 0.378; = 0.378; g ♂ = 3.70 m/sec 2 M♁M♁ M♂M♂

22. Gravitational Force: Physics 1710—Chapter 13 Apps: Gravity m M d F = - G M m/ d 2 What is the order of magnitude of the attraction between two people (m~ 100 kg) separated by a distance of ~ 1m?

23. A.9.8 N. B.980. N C.6.7 X 10 - 7 N D.6.7 X 10 - 9 N E.None of the above Physics 1710—Chapter 13 Apps: Gravity

24. Gravitational Force: Physics 1710—Chapter 13 Apps: Gravity m M d F = - G M m/ d 2 F = - (6.67 x10 –11 N ‧m 2 /kg 2 ) (100 kg)(100kg )/(1 m) 2 F = - 6.67 x10 –7 N Equivalent weight = F/g = 67 ng

25. Gravitational Potential Energy U = -∫ ∞ R Fd r U = -∫ ∞ R G Mm/r 2 d r U = GMm/R U = 0 as r →∞ Physics 1710—Chapter 13 Apps: Gravity

26. Total Energy for Gravitationally Bound Mass E = K + U E = ½ m v 2 – GMm/r E = L 2 /2mr 2 – GMm/r Bound orbit if E ≤ 0 and - dE/dr | r o = 2 (L 2 /2mr o ) - GMm = 0 E = - GMm/2r o Physics 1710—Chapter 13 Apps: Gravity

27. Total Energy for Gravitationally Bound Mass E = K + U E = - GMm/2r o Physics 1710—Chapter 13 Apps: Gravity L 2 /2mr 2 < – GMm/r E - L 2 /2mr 2 < – GMm/r r o = - 2E/(GMm)

28. Escape Velocity If K is such that E > 0, then K = ½ m v 2 = GMm/ r Thus at r= R: v = √[ 2GM/R] Physics 1710—Chapter 13 Apps: Gravity

29. Summary: The force of attraction between two bodies with mass M and m respectively is proportional to the product of their masses and inversely proportional to the distance between their centers squared. The force of attraction between two bodies with mass M and m respectively is proportional to the product of their masses and inversely proportional to the distance between their centers squared. F = - G M m/ r 2 The proportionality constant in the Universal Law of Gravitation G is equal to 6.673 x 10 –11 N m 2 /kg 2. The proportionality constant in the Universal Law of Gravitation G is equal to 6.673 x 10 –11 N m 2 /kg 2. Physics 1710—Chapter 13 Apps: Gravity

30. Summary: The gravitational force constant g is equal toThe gravitational force constant g is equal to G M/(R+h) 2, R is the radius of the planet. G M/(R+h) 2, R is the radius of the planet. Kepler’s Laws Kepler’s Laws –The orbits of the planets are ellipses. –The areal velocity of a planet is constant. –The cube of the radius of a planet’s orbit is proportional to the square of the period. is proportional to the square of the period. The gravitation field is the force divided by the mass. The gravitation field is the force divided by the mass. g = F g / m Physics 1710—Chapter 13 Apps: Gravity

31. Summary: The gravitation potential energy for a point mass is proportional to the product of the masses and inversely proportional to the distance between their centers: The gravitation potential energy for a point mass is proportional to the product of the masses and inversely proportional to the distance between their centers: U = GMm / r The escape velocity is the minimum speed a projectile must have at the surface of a planet to escape the gravitational field. The escape velocity is the minimum speed a projectile must have at the surface of a planet to escape the gravitational field. v = √[ 2GM/R] Physics 1710—Chapter 13 Apps: Gravity