2. Physics 1710—Warm-up Quiz In the movie “Armaggedon” a comet “the size of Texas’” (800 miles or 1250 km in diameter) was heading for Earth. If the density were 800 kg/m 3, the mass would be about 8.x10 20 kg. How far would a ball on the surface fall in 1.0 second? A.About 5 m B.About 7 cm C.About 2 cm D.None of the above

3. Gravitation G M/ R 2 g = G M/ R 2 g=(6.67x10 -11 N m 2 /kg 2 )(8x10 20 kg)/(625x10 3 m) 2 g = 0.14 N/kg d = ½ g t 2 = 0.5 (0.14 m/s 2 )(1 sec) 2 = 0.0 7 m = 7 cm = 0.0 7 m = 7 cm Physics 1710—Chapter 12 Apps: Gravity

4. 1′ Lecture 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. The escape velocity is the initial velocity at the surface required to leave a body. v escape = √[ 2GM/R]The escape velocity is the initial velocity at the surface required to leave a body. v escape = √[ 2GM/R] Physics 1710—Chapter 13 Apps: Gravity

5. No Talking! Think! Confer! Peer Instruction Time Does the position of the moon affects us as astrology claims? Physics 1710—Chapter 13 Apps: Gravity

6. Does the position of the moon affects us as astrology claims? A.Yes, I strongly agree B.Yes, I Agree C.No, I disagree D.No, I strongly disagree Physics 1710—Chapter 13 Apps: Gravity

7. Astrology: Does the moon affect us? F = - G M m/ d 2 Physics 1710—Chapter 13 Apps: Gravity d moon =3.84x10 8 m M = 7.36x10 23 kg; d = 3.84x10 8 m |F/m|= (7x10 -11 )(7x10 23 )/(4x10 8 ) 2 = 3x10 -4 N/kg ~ 30 μ g M =7.36x10 23 kg

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

9. Escape Velocity If K is such that E = 0 as r →∞, then K R = ½ m v 2 = U R = GMm/ R and v escape = √[ 2GM/R] = √[2gR] Physics 1710—Chapter 13 Apps: Gravity K = ½ m v 2 U ∞ = 0 U R = GMm/R K ∞ = 0

10. Were the astronauts in danger of jumping off the “Armaggedon” comet? (Hint: What is the escape velocity from our “Armaggedon” comet? M = 8x10 20 kg, R =6.25x10 5 m, g = 0.14 N/kg) A.Yes. Definitely. B.Maybe. C.No. Definitely not. D.Not enough information. Physics 1710—Chapter 13 Apps: Gravity

11. Escape Velocity v escape = √[ 2GM/R] = √[2gR] = √[2(0.14 N/kg)(625 km)] ~ 400 m/s ~ 900 mph. Safe! Physics 1710—Chapter 13 Apps: Gravity v escape = √[ 2GM/R] = √[2gR]??

12. Escape Velocity of Earth? v escape = √[ 2GM/R] = √[2gR] = √[2(9.8 N/kg)(6.3 x10 6 m)] ~ 11,000 m/s 25,000 mph. Stuck! Physics 1710—Chapter 13 Apps: Gravity v escape = √[ 2GM/R] = √[2gR]??

13. 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

14. Total Energy for Gravitationally Bound Mass E = K + U E = - GMm/2r o Physics 1710—Chapter 13 Apps: Gravity < 0 E < 0 r o = - 2E/(GMm) L 2 /2mr 2 – GMm/r r

15. 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

16. 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

17. 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 escape = √[ 2GM/R] Total Energy E is conserved for two body geavitational problem; bodies are bound for E ≤ 0 Total Energy E is conserved for two body geavitational problem; bodies are bound for E ≤ 0 E = L 2 /2mr 2 – GMm/r Physics 1710—Chapter 13 Apps: Gravity