gaff gaffney ney-bil -bills ls (gg23 gg2365 657) 7) – tes test #2 Gravit vitati ation – tub tubma man n – (111 (11121 213) 3) This This prin printt-ou outt shou should ld hav have 17 ques questi tion ons. s. MultipleMultiple-cho choice ice questions questions may contin continue ue on the next column or page – find all choices before answering. 001 10.0 0.0 poin points ts If Spacecraft X has has twice the mass of Spacecraft Y craft Y ,, then what is true about X about X and Y and Y ?? I) On Earth, X Earth, X experiences experiences twice the gravitational force that Y that Y experiences; II) On the Moon, X Moon, X has has twice the weight of Y ; Y ; III) When When both are in the same same circul circular ar orbit, orbit, X has twice the centripetal acceleration of Y . Y .
1
re = 150 million km , ra = 468 million km .
and
From Kepler’s laws, T e2 T a2 = 3 re3 ra T a = =
ra re
3/2
T e
468 million km 150 million km
3/2
(1 year)
= 5.51102 year . 003 (part 2 of 2) 10.0 points What is the orbital velocity of the asteroid? Assume there are 365 days in one year.
1. I 1. I only 2. III 2. III only
Correct answer: 16919. 16919.5 m/ m /s.
3. I, 3. I, II, and III
Explanation:
4. I 4. I and II only correct only correct
Explanation: I) gravitational force mass. II) weight mass. III) The centripetal acceleration is determined by v2 ac = , r so X so X and Y and Y should should have the same centripetal acceleration when they are in the same circular orbit.
∝
2 π ra T a 2 π (4. (4.68 1011 m) 1 y = 5.51102 year 365 d 1d 1h 24 h 3600 s = 169 16919 19..5 m/ m /s .
va =
5. II 5. II and III II I only
×
∝
002 (part 1 of 2) 10.0 points The period of the earth around the sun is 1 year and its distance distance is 150 million km from the sun. An asteroid asteroid in a ccircul ircular ar orbit around around the sun is at a distance 468 million km from the sun. What is the period of the asteroid’s orbit?
×
004 (part 1 of 3) 10.0 points A satellite is in a circular orbit just above the surface of the Moon. What What is the the satel atelllite ite’s acce accele lerratio ation? n? The The value alue of gra gravita vitattiona ionall cons consta tan nt is 2 2 −11 6.67259 10 Ncdotm /kg and the mass of the moon is 7. 7.36 1022 kg and its radius is 2620..2 km . 2620
×
×
Correct answer: 0. 0.715325 m/ m/s2 . Explanation:
Correct answer: 5. 5.51102 year. Explanation: Let : T e = 1 year ,
Let : G = 6.67259 10−11 Ncdotm2 /kg2 , M moon 1022 kg , and moon = 7.36 rmoon = 2620. 2620.2 km = 2. 2.6202 106 m .
×
×
×
gaffney-bills (gg23657) – test #2 Gravitation – tubman – (111213) F c = m ac = F G , so G m M moon r2 G M moon ac = r2 = (6.67259 10−11 Ncdotm2 /kg2) 7.36 1022 kg (2.6202 106 m)2
m ac =
×
× × ×
= 0.715325 m/s2 .
2
C. They were deduced by Kepler from Brahe’s observations. D. They were derived by Newton from his gravitational and second laws. E. They were derived by Newton using his gravitational and second law together with Brahe’s data. 1. C and E 2. A, B, and C 3. B and E
005 (part 2 of 3) 10.0 points What is the satellite’s speed?
4. A and E
Correct answer: 1369.05 m/s.
5. C and D correct
Explanation: 6. B and D v2 ac = r v = ac r
7. A and D
√
=
Explanation:
(0.715325 m/s2 )(2.6202
6
× 10
m)
= 1369.05 m/s . 006 (part 3 of 3) 10.0 points What is the period of the satellite’s orbit? Correct answer: 3.34036 h.
008 10.0 points Energy is required to move a 1420 kg mass from the Earth’s surface to an altitude 2.77 times the Earth’s radius R E . What amount of energy is required to accomplish this move? The acceleration of gravity is 9.8 m/s2 . Correct answer: 6.51317
Explanation:
10
× 10
J.
Explanation:
2πR v 2 π (2.6202 106 m) = 1369.05 m/s
T =
×
×
Let :
1h 3600 s
= 3.34036 h . U = 007 10.0 points Which are the correct statements regarding Kepler’s laws? A. They were obtained first by Tycho Brahe. B. They were obtained by Kepler from Brahe’s observations combined with Newton’s second law.
m = 1420 kg , h = 2.77 RE , and g = 9.8 m/s2 .
E − G m RM E and g = G M , so the R 2
E
change in the potential energy of the massEarth system is ∆U = U f
− U i
= G m M E
1 Ri
−
1 Rf
gaffney-bills (gg23657) – test #2 Gravitation – tubman – (111213) = G m M E
− − 1 RE
m G M E = 1 RE 2.77 = m g RE 3.77
1 3.77 RE 1 3.77
escape from it, it appears black. Suppose a mass approximately the size of the Earth’s mass 7.08 1024 kg is packed into a small uniform sphere of radius r.
×
Use:
10
× 10
6
× 10
m)
2.77 3.77
J .
009 10.0 points Given: G = 6.67259 10−11 N m2 /kg2 A 1370 kg meteor comes in from outer space and impacts on the a moon’s surface (this moon is not necessarily the Earth’s moon). The mass of this moon is M = 7 1022 kg, and the radius of this moon is R = 1.5 106 m. How much work is done by the Moon’s gravitational field?
×
×
The speed of light c = 2.99792 108 m/s . The universal gravitational constant G = 6.67259 10−11 N m2/kg2 . Hint: The escape speed must be the speed of light. Based on Newtonian mechanics, determine the limiting radius r0 when this mass (approximately the size of the Earth’s mass) becomes a black hole.
×
= (1420 kg) (9.8 m/s2 ) (6.37 = 6.51317
×
Correct answer: 0.0105127 m. Explanation: Basic Concepts: Energy conservation
×
Correct answer: 4.26601
9
× 10
J.
Explanation: Using the formula relating the work W done by the Moon’s gravitational field on the meteor with the change ∆U in the potential energy of the meteor-Moon system, W = =
3
−∆U − (U − U ) G m M − 0 , =− − R final
initial
where M is the mass of the Moon, we obtain G m M W = R = (6.67259 10−11 N m2/kg2 ) (1370 kg) (7 1022 kg) 1.5 106 m = 4.26601 109 J .
×
×
×
×
×
010 10.0 points A black hole is an object so heavy that neither matter nor even light can escape the influence of its gravitational field. Since no light can
E =
− G mr M + K .
At minimum escape velocity, E = 0 (the pro jectile has just enough initial kinetic energy to overcome the gravitational potential). Solution: Technically speaking, in a region where gravity is extremely intense, Newton’s mechanics cannot be used. Rather, one needs to apply the “general theory of relativity” developed by Albert Einstein. Knowing this is the case, we still would like to see what Newtonian mechanics tells us. Setting vesc = c, the limiting radius is given by 1 1 G m M 2 = m c2 = m vesc , 2 2 r0 or
2 G M . c2 It turns out that the theory of relativity gives the same expression for this limiting radius, referred to as the “Schwarzschild radius”. In Newtonian mechanics, however, the description of what happens to objects attracted to this sphere is often highly inaccurate. But in any case, r0 =
2 G M c2 = 2 (6.67259
r0 =
11
× 10−
N m2/kg2)
gaffney-bills (gg23657) – test #2 Gravitation – tubman – (111213)
×
(7.08 1024 kg) (2.99792 108 m/s)2
× ×
4
1. Cannot be determined
013 (part 1 of 2) 10.0 points When it orbited the Moon, the Apollo 11 spacecraft’s mass was 9140 kg, and its mean distance from the Moon’s center was 2.15028 106 m. Find the orbital speed of the spacecraft. Assume its orbit to be circular and the Moon to be a uniform sphere of mass 7.36 1022 kg. The gravitational constant is 6.67259 10−11 N m2 /kg2.
2. smaller
Correct answer: 1511.26 m/s.
= 0.0105127 m . 011 10.0 points If the Earth shrank in size, with all other factors remaining the same, how would the escape velocity from its surface change?
3. greater correct 4. the same Explanation: The gravitational field at the surface of the Earth would increase if the Earth shrank; escape velocity would be correspondingly greater.
×
×
×
Explanation: The gravitational force supplies the centripetal force: GM m v2 = m d2 d G M v = d =
012 10.0 points What is the kinetic energy of a satellite of mass m that orbits the Earth of mass M in a circular orbit of radius R? 1 GM m 4 R 1 GM m 2. K = 2 R2
GM m R2 1 GM m 5. K = correct 2 R Explanation: The gravitational force on the satellite provides the centripetal force needed to keep it in circular orbit: GMm v 2 = F G = F c = m R2 R G M m m v2 = , so R 4. K =
1 1 GM m K = m v 2 = . 2 2 R
11
6.67259
2
× 10− N · m /kg 7.36 × 10 kg × 2.15028 × 10 m
2
22
6
= 1511.26 m/s .
1. K =
3. K = 0
·
014 (part 2 of 2) 10.0 points What is the minimum energy required for the craft to leave the orbit and escape the Moon’s gravitational field? Correct answer: 1.04374
10
× 10
J.
Explanation: The total energy is 1 G m M m v2 2 d 1 G M G m M G m M = m = 2 d d 2d = 6.67259 10−11 N m2 /kg2 (9140 kg)(7.36 1022 kg) 2(2.15028 106 m) = 1.04374 1010 J .
E = K + U =
− × −
−
− × ×
−
× ×
·
gaffney-bills (gg23657) – test #2 Gravitation – tubman – (111213) The minimum energy required for the craft to leave the orbit and escape the Moon’s gravitational field is = 1.04374 E min = E
× 10
10
J .
015 (part 1 of 3) 10.0 points What is the kinetic energy of a satellite of mass m which is in a circular orbit of radius 3 Re about the earth? 1. K = 2. K = 3. K = 4. K = 5. K =
G M e m 3 Re G M e m correct 6 Re m v2 6 Re G M e m 3 Re m v2 3 Re
−
1. E =
G M e m Re
1 2. E = m g Re + m v 2 2 3. E = 3 G M e m G M e m 3 Re Gm 5. E = Re G M e m 6. E = 6 R2e G M e m 7. E = correct 6 Re G M e m 8. E = 3 Re G M e m 9. E = 3 m g Re + 3 Re Explanation: The potential energy of the satellite is 4. E =
− − −
6. K = 3 m g Re U = 7. K = 3 G M e m G M e m 8. K = Re2
G M e m G M e m 6 Re 3 Re G M e m = . 6 Re
E = K + U =
and the kinetic energy is K =
1 G M e m m v2 = . 2 6 Re
016 (part 2 of 3) 10.0 points What is the total energy of the satellite?
−
−
−
m v2 G M e m F = m ac = = 3 Re (3 Re)2 G M e m m v2 = e Re
− G3M Reem
so the total energy is
9. K = m g Re G M e m 10. K = 6 Re2 Explanation: The acceleration of the satellite in circular v2 orbit of radius 3 Re is a c = , so the force 3 Re on the satellite is
5
017 (part 3 of 3) 10.0 points How much work must an external force do on the satellite to move it from a circular orbit of radius 2 Re to 3 Re, if its mass is 2000 kg? The universal gravitational constant 6.67 10−11 N m2 /kg2 , the mass of the Earth 5.98 1024 kg and its radius 6.37 106 m.
×
×
·
×
Correct answer: 1.04361 Explanation: Let :
G = 6.67 M e = 5.98 Re = 6.37
10
× 10 11
J.
2
2
× 10− N · m /kg × 10 kg , and × 10 m . 24 6
,
gaffney-bills (gg23657) – test #2 Gravitation – tubman – (111213) The work done by an external force to move the satellite from the closer orbit to the further orbit will be the work against gravity (a positive number which yields the change in potential energy) plus the change in kinetic energy (a negative number since the kinetic energy is smaller in the orbit with the greatest radius): W = E f
− E i = −
G M e m 12 Re 1 = (6.67 12 (5.98
G M e m 6 Re
− −
G M e m 4 Re
=
11
2
2
× 10− N · m /kg ) × 10 kg)(2000 kg) × 6.37 × 10 m = 1.04361 × 10 J .
24
6
10
6
