1. PHY 231 1 PHYSICS 231 Lecture 17: We have lift-off! Remco Zegers Walk-in hour: Thursday 11:30-13:30 am Helproom Comet Kohoutek

2. PHY 231 2 Previously… v r acac Centripetal acceleration: a c =v 2 /r=  2 r Caused by force like: Gravity Tension Friction  F=ma c for rotating object

3. PHY 231 3 The gravitational force, revisited G=6.673·10 -11 Nm 2 /kg 2 Newton: The gravitational force works between every two massive particles in the universe, yet is the least well understood force known.

4. PHY 231 4 Gravitation between two objects A B The gravitational force exerted by the spherical object A on B can be calculated by assuming that all of A’s mass would be concentrated in its center and likewise for object B. Conditions: B must be outside of A A and B must be ‘homogeneous’

5. PHY 231 5 Gravitational acceleration F=mg g=Gm EARTH /r 2 On earth surface: g=9.81 m/s 2 r=6366 km On top of mount Everest: r=6366+8.850km g=9.78 m/s 2 Low-orbit satellite: r=6366+1600km g=6.27 m/s 2 Geo-stationary satellite: r=6366+36000km g=0.22 m/s 2

6. PHY 231 6 Losing weight easily? You are standing on a scale in a stationary space ship in low-orbit (g=6.5 m/s 2 ). If your mass is 70 kg, what is your weight? F=mg=70*6.5=455 N And what is your weight if the space ship would be orbiting the earth? Weightless!

7. PHY 231 7 Gravitational potential energy So far, we used: PE gravity =mgh Only valid for h near earth’s surface. More general: PE gravity =-GM Earth m/r PE=0 at infinity distance from the center of the earth See example 7.12 for consistency between these two. Example: escape speed: what should the minimum initial velocity of a rocket be if we want to make sure it will not fall back to earth? KE i +PE i =0.5mv 2 -GM Earth m/R Earth KE f +PE f =0 v=  (2GM earth /R Earth )=11.2 km/s

8. PHY 231 8 Kepler’s laws Johannes Kepler (1571-1630)

9. PHY 231 9 Kepler’s First law Ellipticity e(0-1) An object A bound to another object B by a force that goes with 1/r 2 moves in an elliptical orbit around B, with B being in one of the focus point of the ellipse; planets around the sun. p+q=constant

10. PHY 231 10 Kepler’s second law A line drawn from the sun to the elliptical orbit of a planet sweeps out equal areas in equal time intervals. Area(D-C-SUN)=Area(B-A-SUN)

11. PHY 231 11 Kepler’s third law Consider a planet in circular motion around the sun: T2T2 r3r3 r 3 =T 2 /K s r 3 =constant*T 2 T: period-time it takes to make one revolution

12. PHY 231 12 Chapter 8. Torque It is much easier to swing the door if the force F is applied as far away as possible (d) from the rotation axis (O). Torque: The capability of a force to rotate an object about an axis. Torque  =F·d (Nm) Torque is positive if the motion is counterclockwise Torque is negative if the motion is clockwise Top view

13. PHY 231 13 Decompositions What is the torque applied to the door? F // FLFL Force parallel to the rotating door: F // =Fcos60 0 =150 N Force perpendicular to rotating door: F L =Fsin60 0 =260 N Only F L is effective for opening the door:  =F L ·d=260*2.0=520 Nm F= Top view

14. PHY 231 14 Multiple force causing torque. 0.6 m 0.3 m 100 N 50 N Two persons try to go through a rotating door at the same time, one on the l.h.s. of the rotator and one the r.h.s. of the rotator. If the forces are applied as shown in the drawing, what will happen? Top view  1 =F 1 ·d 1 =-100*0.3=-30 Nm  2 =F 2 ·d 2 =50*0.6 =30 Nm Nothing will happen! The 2 torques are balanced. + 0 Nm 

15. PHY 231 15 Center of gravity. F pull d pull Vertical direction (I.e. side view) F gravity d gravity ?   =F pull d pull +F gravity d gravity We can assume that for the calculation of torque due to gravity, all mass is concentrated in one point: The center of gravity: the average position of the mass d cg =(m 1 d 1 +m 2 d 2 +…+m n d n ) (m 1 +m 2 +…+m n ) 1 2 3………………………n

16. PHY 231 16 Center of Gravity; more general The center of gravity

17. PHY 231 17 Object in equilibrium CG FpFp -F p d-d Top view Newton’s 2nd law:  F=ma F p +(-F p )=ma=0 No acceleration, no movement… But the block starts to rotate!   =F p d+(-F p )(-d)=2F p d There is movement! Translational equilibrium:  F=ma=0 The center of gravity does not move! Rotational equilibrium:   =0 The object does not rotate Mechanical equilibrium:  F=ma=0 &   =0 No movement!