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