Answers to homework problems: Chapter 7 part 2. 82) a)At a distance x from Earth the gravitational force on object with any mass m due to Earth is F_1 = G * m * M(Earth)/x^2 the gravitational force due to the Moon is F_2 = G * m * M(Moon)/(R-x)^2 where R = 3.8 10^5 km is the distance from center of Earth to center of moon (see the back of the cover), M(Earth) = 5.98 10^24 kg, M(Moon)=7.4 10^22 kg. For equilibrium F_1 = F_2 (value of m and G cancel out and we are left with quadratic equation for x: M(Earth)/x^2 = M(Moon)/(R-x)^2 The solution is x = 0.9 R = 3.42 10^5 km from Earth Note: The quadratic equation has two solutions - the second solution gives x greater than distance from Earth to Moon (on the other side of the Moon). At this point the two gravitational forces from Earth and from the Moon would add up instead of cancelling out and therefore the second solution is unphysical. b) We have considered only Earth and Moon gravitational pull here. There remains pull from the Sun and from all other planets. 97) Total energu is conserved: For M = mass of Earth, m = mass of the package, R = radius of Earth Energy of the package on the Earth surface = -GMm/R +mv^2/2 Energy of the package at a hight h = -GMm/(R+h) The equation becomes independent of m v^2/2-GM/R=-GM/(R+h) ---> v=sqrt(2GMh/(R(R+h))= 3.7 km/s Since the escape velocity v_esc=sqrt(2GM/R)--> v=0.34 v_esc 100) The Kepler's law relates the orbital period (day for stationary satelite) to the radius: r^3 = T^2 GM/(4pi^2) 1 day period T= 24*3600 s , G= 6.67 10^-11 Nm^2/s^2, Earth mass M= 5.98 10^24 kg r =cube root ( T^2 GM/(4pi^2)) = 42250 km Here r is the distance from the center of Earth. To get the altitude we have to subtract the Earth radius. For the equatorial orbit altitude = 42250 km - 6378 km = 35872 km Note: to get a cube root on your calculator take a log(base 10) of the value, divide by 3 and calculate 10 to the power of the result. 101) From appendix III(page A-5): Mass of Venus = 0.815 * Mass of Earth = 4.87 10^24 kg T= 243 days * 24 h/day * 3600 s/h = 20 995 200 s (I assume here Earth definition of a day) r =cube root ( T^2 GM/(4pi^2)) = 1.53 10^9 m Chapter 8 Part 1 20) Torque = distance * force tau = 50 N * 0.9 m = x * 0.6 m --> x = 75 N 21) Here the force is the force of gravity = m*g m_1 * g * r_1 = m_2 * g * r_2 ---> r_2= r_1 * m_1/m_2 = 2.0 m * 35 kg/30 kg = 2.33 m 30) The total torque is the sum of the torque due to the weight of the stick and the weight suspended from it. Here we have to get units straight ! tau_1 = 0.15 kg * 9.8 m/s^2 *0.75 m + 5 kg*m/s^2 * 0.5 m = 3.6 kg*m^2/s^2 For stability the the totla torque must be zero, so the torque due to the ballancing force must be also tau_2 = Force * 1 m --> Force = 3.6 kg*m/s^2 = 3.6 N This force must point up . 32) a)Lets define the horizontal axis at one edge of the bottom book. The center of mass of this book lies at the point half width of the book from the origin i.e., x_1 = 12.5 cm, The center of mass of the second book will be at the center of the second book i.e., shifted by 3 cm to x_2=15.5 cm The center of mass of 2 books is at x_cm(1+2) = m*x_1+m*x_2/2m = 14 cm The center of mass of 3 books is at x_cm(1+2+3) = m*x_1+m*x_2+m*x_3/3 = 15.5 cm x_cm(1+2+3+4) = m*x_1+m*x_2+m*x_3+m*x_4/4 = 17.0 cm x_cm(1+2+3+4+5) = = 18.5 cm x_cm(1+2+3+4+5+6) = = 20.0 cm x_cm(1+2+3+4+5+6+7) = = 21.5 cm x_cm(1+....+8) = = 23.0 cm x_cm(1+....+9) = = 24.5 cm x_cm(1+....+10) = = 26.0 cm The center of mass position of the 10 book stack is outside of the base support while the 9 book stack center of mass is still inside the 25 cm base. So the answer is 9 books. b) The center of mass in vertical dimension is in the middle of the stack y_cm = 1/2 * (9 * 5 cm) = 22.5 cm