Physics 121, Sections 9, 10, 11, and 12 Lecture 15 Today's Topics Homework 6 Chap6:#6,12,20,24,29,38,52,57,78,and83 Midterm 1 Average: 65% Chapter 6 Escape velocity Chapter 7 Linear momentum Collision Physics 121: Lecture 15, Pg 1
Physics 121: Lecture 15, Pg 1 Physics 121, Sections 9, 10, 11, and 12 Lecture 15 Today’s Topics: Homework 6: Chap. 6: # 6, 12, 20, 24, 29, 38, 52, 57, 78, and 83. Midterm 1: Average: 65% Chapter 6: Escape velocity Chapter 7: Linear momentum Collision
Problem: How High? a projectile of mass m is launched straight up from the surface of the earth with initial speed vo. What is the maximum distance from the center of the earth RMAx it reaches before falling back down MAX .m Physics 121: Lecture 15, Pg 2
Physics 121: Lecture 15, Pg 2 Problem: How High? A projectile of mass m is launched straight up from the surface of the earth with initial speed v0 . What is the maximum distance from the center of the earth RMAX it reaches before falling back down. RMAX RE v0 m M
Problem: How High No non-conservative and we know forces W NC 0 △K=-△AU MAX .m MAX mvs=GMm MAX Physics 121: Lecture 15, Pg 3
Physics 121: Lecture 15, Pg 3 Problem: How High... No non-conservative forces: WNC = 0 K = -U RMAX v0 m hMAX And we know: RE M = − E MAX 2 0 R 1 R 1 mv GMm 2 1
Problem: How High mv=GMm MAX 尺 m 6=2GM MAX MAX GM 尺 R=1 MAX M 29R RR MAX 尺 MAX E MAX 2gR 2aR Physics 121: Lecture 15, Pg 4
Physics 121: Lecture 15, Pg 4 E 2 0 MAX E MAX E E MAX E 2 E E E MAX 2 0 E MAX 2 0 2gR v R R 1 R R 2gR 1 R R R 1 R GM 2 R 1 R 1 v 2GM R 1 R 1 mv GMm 2 1 − = = − − = = − = − RMAX RE v0 m hMAX M R R v gR MAX E E = 1 − 2 0 2 Problem: How High
Escape Speed (velocity) E MAX 29R If we want the projectile to escape to infinity we need to make the denominator in the above equation zero 0 97 0=√29RE 2gRE 2 We call this value of vo the escape velocity v Physics 121: Lecture 15, Pg 5
Physics 121: Lecture 15, Pg 5 Escape Speed (Velocity) If we want the projectile to escape to infinity we need to make the denominator in the above equation zero: R R v gR MAX E E = 1 − 2 0 2 1 2 0 0 2 − = v gRE v gRE 0 2 2 =1 v0 = 2gRE We call this value of v0 the escape velocity vesc