Homework 34

# Homework 34 - homework 34 BAUTISTA ALDO Due 4:00 am Version...

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homework 34 – BAUTISTA, ALDO – Due: Apr 24 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 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 accelera- tion of the satellite is given by 1. a c = g r R 2 2. a c = g 2 3. a c = g R r 4. a c = 1 2 g r R 2 5. a c = 1 2 g R r 2 6. a c = g R r 2 correct 7. a c = 1 2 g R r 8. None of these 9. a c = g Explanation: Basic Concepts: The Universal Law of Gravity is F = G m 1 m 2 r 2 . Gravitational potential energy is U = - G m 1 m 2 r and if we set U = 0 at r = . Solution: The force of gravity acts as cen- tripetal force, F c = F G , or m a c = G M m r 2 so a c = G M r 2 . (1) One might already recall the expression for g ; however, if we don’t, for an object of mass m 0 on the surface of the Earth m 0 g = G M m 0 R 2 so that g = G M R 2 . (2) (The actual value, 9.8 m/s 2 , differs somewhat from this value due to other effects, such as the rotation of the Earth). Therefore our a c is a c = G M r 2 = G M R 2 R 2 r 2 = g R r 2 . (3) Question 2 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. Δ E = G M m R r 2 2. Δ E = G m 2 4 r 3. Δ E = G M m 3 r 4. Δ E = G M m r 4 R 2 5. Δ E = 2 G M m 3 r 6. Δ E = G M m 2 r correct 7. Δ E = G M m 4 r 8. Δ E = G M 2 3 r 9. Δ E = G M m r 10. Δ E = 3 G M m 4 r

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homework 34 – BAUTISTA, ALDO – Due: Apr 24 2006, 4:00 am 2 Explanation: If we insert the expression a c = v 2 r into our force equilibrium equation (1) in Part 1, we can solve for the velocity squared v 2 v 2 = G M r and consequently the kinetic energy of the satellite is K = 1 2 m v 2 = G M m 2 r . Also, the potential energy is U = - G M m r , so the total energy of the satellite is E = K + U = G M m 2 r - G M m r = - G M m 2 r . Note: This could also be seen from ready- made formulae. 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 = G M m 2 r . Question 3 Part 1 of 1. 10 points. For this problem, we assume that we are on Planet-I. The radius of this planet is R =4360 km, the gravitational acceleration at the surface is g I =6 . 82 m / s 2 , and the grav- itational constant G = 6 . 67 × 10 - 11 N m 2 /kg 2 in SI units. The mass of Planet-I is not given. Not all the quantities given here will be used. Suppose a cannon ball of mass m = 4790 kg is projected vertically upward from the sur- face of this planet. It rises to a maximum height h =21320 . 4 km above the surface of the planet.
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