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I estimate my grade on exam 2 to be:

Physics 1710 —Warm-up Quiz. Answer Now !. 0. 35% 49 of 140. 0. I estimate my grade on exam 2 to be:. A B C D N/A F. Physics 1710 —C hapter 13 Apps: Gravity. 0. CM. CM. Ball F g = - mg. Center of Gravity (comparison). CG. Moon F g = - GmM/r 2.

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I estimate my grade on exam 2 to be:

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  1. Physics 1710—Warm-up Quiz Answer Now ! 0 35% 49 of 140 0 I estimate my grade on exam 2 to be: • A • B • C • D • N/A • F

  2. Physics 1710—Chapter 13 Apps: Gravity 0 CM CM Ball Fg = - mg Center of Gravity (comparison) CG Moon Fg = - GmM/r2

  3. Physics 1710—Chapter 13 Apps: Gravity No Talking! Think! Confer! 0 What is the mass of the earth? How do you know? Peer Instruction Time

  4. Physics 1710—Chapter 13 Apps: Gravity 0 What is the mass of the earth? How do you know? g? Fg = - mg Fg = - mg′ Peer Instruction Time

  5. Physics 1710—Chapter 13 Apps: Gravity 0 1′ Lecture The moduli of elasticity (Y, E, B) characterizes the stress-strain relation: stress= modulus • strain; σ = Y ε 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. Fg = - řG Mm/d 2 G = 6.673 x 10 –11 N m2 /kg2 ~ 2/3 x 10 –10 N m2 /kg2

  6. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Definitions: Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Stress = modulus x strain σ = F/A = Y ε

  7. Physics 1710—Chapter 13 Apps: Gravity Failure Elastic limit 0 Elasticity Stress σ – Strain ε “Curve” Stress σ (N/m2) σ = Y ε Strain ε = ΔL/L (%)

  8. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Tensile/Compressive Stress Young’s ModulusE Stress = modulus x strain σ = F/A = E ε = E ΔL/L ΔL L σ = E ε

  9. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Shear Modulus G Stress = modulus x strain σ = F/A = G ε = G Δx/h L Δx h σ

  10. Physics 1710—Chapter 13 Apps: Gravity 0 Elasticity Stress σ : the deforming force per unit area. Strain ε : the unit deformation. Hydraulic Stress: Bulk Modulus B Stress = modulus x strain σ = F/A = p = B ε = B ΔV/V ΔV p V

  11. Physics 1710—Chapter 13 Apps: Gravity 0 41% 58 of 140 Answer Now ! 0 Which spring should have the larger spring constant, a short spring (0.1 m long) or a longer spring (0.3 m long)? F = -k x • k larger for shorter spring. • k larger for longer spring. • k the same for both springs. • None of the above.

  12. Physics 1710—Chapter 12 Apps: Gravity 0 F = -k x? σ = F/A = E ε = E ΔL/L F = - A E x/ L k = A E/L Solution: What about a thicker (bigger A) wire?

  13. Physics 1710—Chapter 13 Apps: Gravity d apple 0 Isaac Newton’s Universal Law of GravitationF = - G M m/ d 2 d moon

  14. Physics 1710—Chapter 13 Apps: Gravity 0 Kepler’s Laws: 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. Angular momentum is conserved: ½ v r ∆t = constant implies that rmv = L = constant. T 2∝ r 3 implies F ∝ 1/ r 2, only. ⊙ = r3 T 2

  15. Physics 1710—Chapter 12 Apps: Gravity 0 How did Newton figure out UL of G? Fact: a moon circling a planet has an acceleration of a = v 2 /r 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 : T 2 = k r 3 . Thus: v = 2π r / T

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

  17. Physics 1710—Chapter 13 Apps: Gravity d apple 0 Isaac Newton’s Universal Law of GravitationF = - G M m/ d 2 d moon F = – g m g =G M ♁ / R♁2

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

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

  20. Physics 1710—Chapter 13 Apps: Gravity 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%) So, M ⊗ = (3.99x10 14 N m 2/kg)/(6.673 x 10-11 N ‧m 2/kg 2) = 5.98 x10 24 kg

  21. Physics 1710—Chapter 13 Apps: Gravity 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? Gravitational Force: M m d

  22. Physics 1710—Chapter 13 Apps: Gravity 10 0% 0 of 1 Answer Now ! What is the order of magnitude of the attraction between two people (m~ 100 kg) separated by a distance of ~ 1m? • 9.8 N. • 980. N • 6.7 X 10 - 7 N • 6.7 X 10 - 9 N • None of the above

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

  24. Physics 1710—Chapter 13 Apps: Gravity Summary: 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. F = - G M m/ d 2

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