This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Husain, Zeena – Exam 4 – Due: Dec 3 2003, 8:00 pm – Inst: H L Berk 1 This printout should have 14 questions, check that it is complete. Multiplechoice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points Given: G = 6 . 67259 × 10 11 Nm 2 / kg 2 The acceleration of gravity on the surface of a planet of radius R = 6100 km is 8 . 6 m / s 2 . What is the period T of a satellite in circular orbit h = 13054 km above the surface? Correct answer: 29443 . 5 s. Explanation: Since the gravity on the surface of a planet is given by g = GM/R 2 , we can solve for the mass of the planet: M = gR 2 G = (8 . 6 m / s 2 )(6 . 1 × 10 6 m) 2 6 . 67259 × 10 11 Nm 2 / kg 2 = 4 . 79583 × 10 24 kg . From Kepler’s Third Law, we have: T 2 = µ 4 π 2 GM ¶ ( h + R ) 3 T = µ 4 π 2 GM ¶ 1 / 2 ( h + R ) 3 / 2 = 29443 . 5 s 002 (part 1 of 1) 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. The minimum increment of energy needed for the satellite to escape ( i.e. , to leave its orbit and move to infinity) is 1. Δ E = GM m 4 r 2. Δ E = GM m 3 r 3. Δ E = GM 2 3 r 4. Δ E = GM mR r 2 5. Δ E = GM m r 6. Δ E = GM mr 4 R 2 7. Δ E = GM m 2 r correct 8. Δ E = 3 GM m 4 r 9. Δ E = 2 GM m 3 r 10. Δ E = Gm 2 4 r Explanation: The total energy of the satellite is E = GM m 2 r . 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 Δ E so that E + Δ E = 0 . Therefore Δ E = E = GM m 2 r . 003 (part 1 of 1) 10 points For this problem, we assume that we are on PlanetI. The radius of this planet is R =3860 km, the gravitational acceleration at the surface is g I =5 . 74 m / s 2 , and the grav itational constant G = 6 . 67 × 10 11 N m 2 /kg 2 in SI units. The mass of PlanetI is not given. Not all the quantities given here will be used. Suppose a cannon ball of mass m = 4070 kg is projected vertically upward from the sur face of this planet. It rises to a maximum height h =9225 . 4 km above the surface of the planet. Caution: Here the gravitational accelera tion decreases as the cannon ball travels away from PlanetI. Husain, Zeena – Exam 4 – Due: Dec 3 2003, 8:00 pm – Inst: H L Berk 2 R h Determine the kinetic energy (in Joules) of the cannon ball immediately after it is fired off. Correct answer: 6 . 35758 × 10 10 J....
View
Full
Document
This test prep was uploaded on 03/23/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

Click to edit the document details