Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Worksheet 9: Gravitation 1. The only force that keeps planets (or other objects, like comets) in orbit around the Sun is Newton's gravitational force. a. Explain why the torque produced by this force about the Sun is zero. The gravitational force is directed along the line that goes from the Sun to the planet, so the lever arm r⟂ is zero and 𝜏 = F r⟂. b. As a consequence of this, the angular momentum of a given planet with respect to the Sun: - increases with r, the distance to the Sun. decreases with r, the distance to the Sun. is constant along the planet’s orbit (conserved). Explain. If the net torque is zero, angular momentum is conserved. All planets and closed-orbit comets move along elliptical orbits, the Sun being at one focus of the ellipse. planet Sun The point where the distance of the planet to the Sun is maximum/minimum is called aphelion/perihelion, respectively. Let vA/ vP and rA/ rP be the speed of the planet and its distance to the Sun at the aphelion/perihelion. (Worksheet 9) c. Use your result from part b and the formula L = I ω, I = mr 2 , and ω = v / r to prove that vA rA = vP rP At the aphelion and perihelion, where the velocity of the planet is perpendicular to the line between planet and the Sun, the magnitude of angular momentum is L = Iω , with I = mr 2 and ω = v / r where I = moment of inertia of the planet around the Sun m = mass of the planet r = distance to the Sun ω = angular speed of the planet at that point v = tangential speed of the planet at that point Using conservation of angular momentum, ( I Aω A = I Pω P → mrA2 ) vr = ( mr ) vr A A 2 P P → rAv A = rP v P P d. From this we conclude, the speed of the planet relative to the Sun: - increases with r, the distance to the Sun. decreases with r, the distance to the Sun. is constant along the orbit (conserved). Explain. Angular momentum is proportional to the distance and the speed of the planet. If the distance increases, speed must decrease, and vice versa. (Worksheet 9) 2. Uranus has a radius of 25,560 km and a surface acceleration due to gravity of 8.86 m/s2. Its moon Miranda is in a circular orbit about Uranus at an altitude of 104,000 km above the planet’s surface. Miranda has a mass of 6.6×1019 kg and a radius of 235 km. Calculate: a. The mass of Uranus The acceleration of gravity on the surface of any planet is g = the mass and radius of the planet. Thus GM where M and R are R2 g U RU2 (8.86 m/s2 )(25.56 ×106 m) 2 MU = = = 8.68 ×1025 kg −11 2 2 G 6.67 ×10 Nm /kg b. The escape velocity on Uranus? The escape velocity from the surface of Uranus is vU = 2G ( )( MU (8.68 × 1025 kg) = 2 6.67 × 10−11 Nm 2 /kg 2 = 21.3 km/s RU 25.6 × 106 m ) c. The acceleration of Miranda for its motion about Uranus. The distance between Miranda and the center of Uranus is r = 104 ×106 m +25.56 ×106 m =129.56 ×106 m Since the only force acting on Miranda is the gravitational pull from Uranus, the acceleration of Miranda is the gravitational acceleration by Uranus at that distance mM M U = mM a r2 M 8.68 ×1025 kg a = g U (r ) = G 2U = 6.67 ×10−11 Nm 2 /kg 2 = 0.345 m/s 2 6 2 r (129.56 ×10 m) G d. The acceleration due to Miranda’s gravity at the surface of Miranda. 19 GM M kg −11 2 2 6.6 ×10 gM = = 6.67 ×10 Nm /kg = 0.080 m/s 2 2 2 RM (235000 m) (Worksheet 9) e. Describe the motion of an object released 1 m above Miranda’s surface, on the side of Miranda that faces Uranus. gU gM Miranda Uranus This question can be tricky. If we draw a figure, it is pretty clear that the net acceleration of the object will be toward Uranus. But that does not mean that the object will really “fall” toward Uranus (i.e., that it will get closer to Uranus). The object will fall toward Miranda, just as one would initially expect. The object is orbiting Uranus along with Miranda, and gU is exactly the right acceleration to account for that part of the motion, just like it accounts for the orbiting motion of Miranda. The part of the acceleration that comes from Miranda (gM) will pull the object back to Miranda’s surface. Note that this is the same situation that we have when an object is dropped near the surface of the Earth, which is orbiting the Sun: the gravitational pull of the Sun also acts on the object, and it is very important! It is what keeps the object moving with the Earth. We usually do not worry about this acceleration because we look at the motion of the object relative to the Earth (which we assume to be at rest, or at least an inertial reference frame). This is an excellent approximation for any object whose motion is so short that the motion of the Earth along its orbit can be taken to be rectilinear.