2. Physics 1710—Warm-up Quiz I estimate my grade on exam 2 to be: A.A B.B C.C D. D E.N/A F.F

3. Center of Gravity (comparison) Physics 1710—Chapter 13 Apps: Gravity Ball F g = - mg Moon F g = - GmM/r 2 CM CG

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

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

6. 1′ Lecture The moduli of elasticity (Y, E, B) characterizes the stress- strain relation:The moduli of elasticity (Y, E, B) characterizes the stress- strain relation: stress= modulus strain; σ = Y ε 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.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 = - řG Mm/d 2 F 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 Physics 1710—Chapter 13 Apps: Gravity

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

8. Elasticity Stress σ – Strain ε “Curve”Stress σ – Strain ε “Curve” Strain ε = ΔL/L (%) Stress σ (N/m 2 ) σ = Y ε Elastic limit Failure Physics 1710—Chapter 13 Apps: Gravity

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

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

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

12. 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 A.k larger for shorter spring. B.k larger for longer spring. C.k the same for both springs. D. None of the above. Physics 1710—Chapter 13 Apps: Gravity

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

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

15. 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

16. 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

17. 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

18. 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

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

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

21. 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

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. 10 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 Answer Now ! 1234567891011121314151617181920 2122232425262728293031323334353637383940 0% 0 of 1 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. Summary: 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 of a planet’s orbit is proportional to the square of the period. is proportional to the square of the period. F = - G M m/ d 2 F = - G M m/ d 2 Physics 1710—Chapter 13 Apps: Gravity