康涅狄格州大学：《普通物理》课程PPT教学课件（英文版）Lecture 15 A projectile of mass m is launched straight

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

Escape Velocity Remembering that g GM we find the escape velocity GM from a planet of mass m and radius r to be: Vesc=2p (where= 6.67X 10-11 m3 kg 1s-2) Rp(m)M(kg)9p(m/s2)Vesc(m/s) Earth 6.378×1065.977×10249.81 11.2x103 Moon1.738×1067.352×10221.62 2.38×103 Jupiter7.150×1071900×10224.8 59.5×103 Sun 6.960×1081.989X1029274 195×103 Physics 121: Lecture 15, Pg 6

Physics 121: Lecture 15, Pg 6 Escape Velocity Remembering that we find the escape velocity from a planet of mass Mp and radius Rp to be: (where G = 6.67 x 10-11 m3 kg-1 s -2 ). v GM R esc p p = 2 g GM RE = 2 Moon Earth Sun Jupiter Rp (m) Mp (kg) gp (m/s2 ) vesc(m/s) 6.378x106 5.977x1024 1.738x106 7.352x1022 7.150x107 1.900x1027 6.960x108 1.989x1029 9.81 1.62 24.8 27.4 11.2x103 2.38x103 59.5x103 195.x103

Lecture 15: Act 1 Escape Velocity Two identical spaceships are awaiting launch on two planets with the same mass. Planet 1 is stationery, while Planet 2 is rotating with an angular velocity o Which spaceship needs more fuel to escape to infinity (a)1 (b) (c)same (2) Physics 121: Lecture 15, Pg 7

Physics 121: Lecture 15, Pg 7 Lecture 15: Act 1 Escape Velocity Two identical spaceships are awaiting launch on two planets with the same mass. Planet 1 is stationery, while Planet 2 is rotating with an angular velocity . Which spaceship needs more fuel to escape to infinity. (a) 1 (b) 2 (c) same (1) (2) 

Lecture 15: Act 1 Solution Both spaceships require the same escape velocity to reach infinity Thus they require the same kinetic energy Both initially have the same potential energy Spaceship 2 already has some kinetic energy due to its rotational motion, so it requires less work (i.e less fuel) Physics 121: Lecture 15, Pg 8

Physics 121: Lecture 15, Pg 8 Lecture 15: Act 1 Solution Both spaceships require the same escape velocity to reach infinity. Thus they require the same kinetic energy. Both initially have the same potential energy. Spaceship 2 already has some kinetic energy due to its rotational motion, so it requires less work (i.e. less fuel)

V=or Aside This is one of the reasons why all of the worlds spaceports are located as close to the equator as possible r2>1 K2=2m(0r2)2 Physics 121: Lecture 15, Pg 9

Physics 121: Lecture 15, Pg 9 Aside This is one of the reasons why all of the world’s spaceports are located as close to the equator as possible. r1 r2 r2 > r1 K2 = m(r2 ) 2 > 2 1 K1 = m(r1 ) 2 2 1 v = r

Algebraic Solution C=△K+ △U=△E For spaceship 1: Wi=(K-Ko+(r-U0 0,0=0(W7=K+U For spaceship 2: W2=(K-Ko+(r-00 Ko= m(or)2, U0=0 E) W2=KF+U,-m(or)2 W,>W2 So spaceship 1 will need more fuel Physics 121: Lecture 15, Pg 10

Physics 121: Lecture 15, Pg 10 Algebraic Solution WNC = K + U = E For spaceship 1: W1 = (Kf - K0 ) + (Uf - U0 ) K0 = 0 , U0 = 0 W1 = Kf + Uf For spaceship 2: W2 = (Kf - K0 ) + (Uf - U0 ) K0 = m(r)2 ,U0 = 0 W2 = Kf + Uf - m(r)2 W1 >W2 So spaceship 1 will need more fuel