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Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points Given: G = 6.67259 × 10−11 N m2 /kg2 . A 517 kg geosynchronous satellite orbits a planet similar to Earth at a radius 198000 km from the planet’s center. Its angular speed at this radius is the same as the rotational speed of the Earth, and so they appear stationary in the sky. That is, the period of the satellite is 24 h . What is the force acting on this satellite? Correct answer: 541.362 N. Explanation: 4 π2 (6.67259 × 10−11 N m2 /kg2 ) (86400 s)2 × (1.98 × 108 m)3 = 6.15224 × 1026 kg . = 003 (part 1 of 3) 10 points C The density of a sphere is given by ρ(r) = . r The sphere has a radius of 2.53 m and a mass of 1011 kg. Find the constant C. Correct answer: 25.138 kg/m2 . Explanation: The mass of a differential element of the sphere is Solution: 2πr T v2 r 4 π2 r = Msatellite T2 4 π 2 (1.98 × 108 m) = (517 kg) (86400 s)2 = 541.362 N . 4 π2 3 r GT2 r = 2.53 m and M = 1011 kg . Let : G = 6.67259 × 10−11 N m2 /kg2 , Msatellite = 517 kg , and T = 86400 s . v= Mplanet = 1 dm = ρ dV = C (4 π r2 dr) . r Z Z F = Msatellite (1) 002 (part 2 of 2) 10 points What is the mass of this planet? Correct answer: 6.15224 × 1026 kg. Explanation: Using the general gravitation law and Eq. 1, we have Msatellite Mplanet F =G r2 4 π2 r = Msatellite T2 Msatellite Mplanet 4 π2 r = Msatellite , so G r2 T2 M= r dm = 4 π C R dR , 0 ¶ ¯r µ 1 2 ¯¯ = 4π C R ¯ 2 0 = 2 π C r2 = (12.8018 m2 ) π C M C= (12.8018 m2 ) π 1011 kg = 25.138 kg/m2 . = (12.8018 m2 ) π 004 (part 2 of 3) 10 points Find the gravitational field 3.84 m from the center of the sphere. Correct answer: 4.57492 × 10−9 N/kg. Explanation: Let : r = 3.84 m . Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin g= (6.6726 × 10−11 N · m2 /kg2 ) (1011 kg) (3.84 m)2 = 4.57492 × 10−9 N/kg . 005 (part 3 of 3) 10 points Find the gravitational field 1.43 m from the centre of the sphere. Correct answer: 1.05391 × 10−8 N/kg. Explanation: Let : R = 1.43 m . µ ¶ R 8. ac = g r µ ¶2 R 1 9. ac = g 2 r Explanation: Basic Concepts: The Universal Law of Gravity is m1 m2 F =G . r2 Gravitational potential energy is The gravitational field inside the sphere is GM R2 2 π G C R2 = R2 = 2πC G = 2π(25.138 kg/m2 ) × (6.6726 × 10−11 N · m2 /kg2 ) g= = 1.05391 × 10−8 N/kg . 006 (part 1 of 2) 10 points A satellite with mass m is orbiting around the Earth on a circular path with a radius r. Denote the mass and the radius of the Earth by M and R, respectively, and the gravitational acceleration at the surface of the Earth by g. The magnitude of the centripetal acceleration of the satellite is given by 1. ac = g g 2. ac = 2 ³ r ´2 R 3. None of these µ ¶ 1 R 4. ac = g 2 r µ ¶2 R 5. ac = g correct r 6. ac = g 1 ³ r ´2 7. ac = g 2 R 2 U = −G m1 m2 r and if we set U = 0 at r = ∞. Solution: The force of gravity acts as centripetal force, Fc = FG , or m ac = G Mm r2 so GM . (1) r2 One might already recall the expression for g; however, if we don’t, for an object of mass m0 on the surface of the Earth ac = m0 g = G so that g= M m0 R2 GM . R2 (2) (The actual value, 9.8 m/s2 , differs somewhat from this value due to other effects, such as the rotation of the Earth). Therefore our ac is GM G M R2 ac = 2 = =g r R2 r 2 µ ¶2 R . r (3) 007 (part 2 of 2) 10 points The minimum increment of energy needed for the satellite to escape (i.e., to leave its orbit and move to infinity) is 1. ∆E = GM m r Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin GM m correct 2. ∆E = 2r GM m 3. ∆E = 4r GM m 4. ∆E = 3r GM mR 5. ∆E = r2 2GM m 6. ∆E = 3r GM mr 7. ∆E = 4 R2 G m2 8. ∆E = 4r G M2 9. ∆E = 3r 3GM m 10. ∆E = 4r Explanation: If we insert the expression ac = v2 r into our force equilibrium equation (1) in Part 1, we can solve for the velocity squared v 2 v2 = GM r and consequently the kinetic energy of the satellite is K= 1 GM m m v2 = . 2 2r Also, the potential energy is U =− GM m , r so the total energy of the satellite is E =K +U GM m GM m = − 2r r GM m =− . 2r Note: This could also be seen from readymade formulae. 3 Now we just have to remember that we set the potential energy at infinite distance to be zero. The least energy with which an object escapes the gravitational attraction is when it has no velocity after escaping. Thus, we need to bring E up to zero. To do this we must add an increment of energy ∆E so that E + ∆E = 0 . Therefore ∆E = −E = GM m . 2r 008 (part 1 of 3) 10 points Given: G = 6.67259 × 10−11 N m2 /kg2 After a supernova explosion, a star may undergo a gravitational collapse to an extremely dense state known as a neutron star, in which all the electrons and protons are squeezed together to form neutrons. A neutron star having a mass of 1.7 × 1030 kg (about equal to that of the Sun) would have a radius of about 9.62 km. Find the free-fall acceleration at its surface. Correct answer: 1.22573 × 1012 m/s2 . Explanation: The free-fall acceleration at the surface of the neutron star is given by GM r2 = (6.67259 × 10−11 N m2 /kg2 ) (1.7 × 1030 kg) × (9620 m)2 = 1.22573 × 1012 m/s2 . g= 009 (part 2 of 3) 10 points Find the weight of a 83.9 kg person at its surface. Correct answer: 1.02838 × 1014 N. Explanation: The weight of the person is given by w = mg = (83.9 kg) (1.22573 × 1012 m/s2 ) = 1.02838 × 1014 N . Cunningham, Drew – Homework 30 – Due: Apr 14 2006, 4:00 am – Inst: Florin 4 Ei = Ef (neglecting air friction), so 010 (part 3 of 3) 10 points Find the energy required to remove a particle of mass 1.65 × 10−27 kg from its surface to infinity. Correct answer: 1.94559 × 10−11 J. Explanation: The energy required to remove the particle from the surface of the star to infinity is the negative of the potential energy of the particle at the surface, so we have G M Mp E= r = (6.67259 × 10−11 N m2 /kg2 ) (1.7 × 1030 kg) (1.65 × 10−27 kg) × (9620 m) −11 = 1.94559 × 10 J. 011 (part 1 of 1) 10 points Given: G = 6.67259 × 10−11 N m2 /kg2 . A projectile is fired vertically upward from the surface of planet Moorun of mass 3 × 1024 kg and radius 3 × 106 m. If this projectile is to rise to a maximum height above the surface of Moorun equal to 6 × 106 m, what must be the initial speed of the projectile? Correct answer: 9.43228 km/s. Explanation: Basic Concept: Total energy (Kinetic and Gravitational Potential) is E= GM m 1 m v2 − . 2 r Solution: Denote the mass of the projectile by m. The initial total energy (before takeoff) is 1 GM m Ei = m v 2 − . 2 R The final total energy (when the projectile is at the maximum height) is Ef = 0 − GM m , R+h since the projectile has zero velocity at its highest point. Energy conservation implies 1 GM m GM m m v2 − =− . 2 R R+h Solving for v, we have s 2M h v= G R (R + h) ½ = (6.67259 × 10−11 N m2 /kg2 ) 2 (3 × 1024 kg) (6 × 106 m)h × (3 × 106 m) (9 × 106 m) = 9432.28 m/s = 9.43228 km/s , (1) ¾ 12 where R + h = (3 × 106 m) + (6 × 106 m) = 9 × 106 m . Comment: Notice that if we had applied ordinary kinematics, we would have utilized the familiar formula v12 − v02 = 2 a (x1 − x0 ) , which in our problem, is 0 − v 2 = 2 (−g)(h − 0) , so that the initial velocity to rise to height h would be p v = 2gh (2) q = 2 (9.8 m2 /s2 ) (6 × 106 m) = 10.8444 km/s , which is NOT the right answer. This is because the attraction of gravity decreases as we go further away from the planet, requiring less initial velocity than if the attraction had been constant (which our simple kinematics formula assumes). It is, however, a good approximation when we are close to a planet’s surface. Try picking a smaller height h and see when the simpler description (2) presents a good approximation to the velocity of formula (1) !