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1 Advancing Physics
Testing for an inverse-square lawQuestion 10W: Warm-up Exercise
1. 500 / 2000 = 1 / 4 and 20 / 80 = 1 / 4 so yes, the data appear to obey an inverse-square law.
2. k = 2.0 107 W in all cases, so the data obey an inverse-square law.
3. The graph is a straight line graph through the origin, confirming an inverse-square law.
4. The relation here is one of inverse proportion:
Vp
1
(Boyle’s law) so tests for an inverse-square law will fail.
5.
2
4
d
l
A
lR
so tests for an inverse-square law between R and d should succeed.
6. This is an exponential decay, so tests for an inverse-square law will fail.
Orbital velocities and accelerationQuestion 20W: Warm-up Exercise
1. t = ½ T
2. v = 2 v
3. a = 4 v / T
4. T = 2 r / v
5. a 0.64 v2 / r
6. t = 1/4 T
7. v = 21/2 v
8. a 0.90 v2 / r
9. t = 1/6 T
10. v = v
11. a 0.95 v2 / r
12. t = ( / 2) T
13. v P Q = v
14. a = v2 / r
2 Advancing Physics
15. Acceleration is at right angles to motion.
16. Centripetal acceleration = v2 / r
17. Centripetal force = m v2 / r
Radians and angular speedQuestion 70W: Warm-up Exercise
1. 2 / 2
2. Angular speed
2min s 60
min rev 151-
-1
= 1.6 10–4 radians s–1
3.
.sradian107.8seconds)60602(
radians2
secondsintime
radiansinanglespeedAngular 14
4.
.sm017.0m20s107.8 114
v
rv
they may just be able to perceive it but it is unlikely – they would see the skyline move at lessthan 2 cm each second.
Newton’s gravitational lawQuestion 80W: Warm-up Exercise
1. For the values estimated in the answers:
.N106.1
m5.0
kg100kg60)mkgN1067.6( 62
2211
2
r
GMmF
2. Pull on the 2.0 kg mass
.N103.3
m200.0
kg0.2kg10)mkgN1067.6( 92
2211
2
r
GMmF
The pull on the 10 kg mass will be equal but opposite in direction.
3.
.m27.0N100.2
mkgN1067.6kg150
5
22112
F
Gm
F
Gmr
3 Advancing Physics
4.
.N100.2
m108.3
kg)1034.7(kg)1097.5()mkgN1067.6( 2028
22242211
2
r
GMmF
.N23
m102.4
kg100kg)1097.5()mkgN1067.6(27
242211
2
r
GMmF
.N100.5
m100.8
kg80kg)1097.5()mkgN1067.6( 226
242211
2
r
GMmF
5. .kgsmmkgsmkgmkgN 12322222
6. Moon
.N103.1
m1064.1
kg)1034.7(kg72)mkgN1067.6( 226
222211
2
r
GMmF
Earth
.N101.7
m1037.6
kg)1097.5(kg72)mkgN1067.6( 226
242211
2
r
GMmF
7. Sun–Moon
.N104.4
m105.1
kg)1034.7(kg)100.2()mkgN1067.6( 20211
22302211
2
r
GMmF
Earth–Moon
.N100.2
m108.3
kg)1034.7(kg)1097.5()mkgN1067.6( 2028
22242211
2
r
GMmF
2.2N 100.2
N 104.4sattraction of ratio
20
20
8. The Moon does of course orbit the Sun, as part of the Earth–Moon system. You can think of theMoon’s orbit of the Earth as superimposed on its orbit of the Sun.
Change in momentum as a vectorQuestion 40W: Warm-up Exercise
1. P = 140 kg m s–1 P1 = 210 kg m s–1
P2 = 350 kg m s–1
P1 = 70 kg × 3 m s–1 = 210 kg m s–1
P2 = 70 kg × 5 m s–1 = 350 kg m s–1
2.
4 Advancing Physics
p = –24 kg m s–1
p1 = 32 kg m s–1
p1 = 8 kg m s–1
3.
P = –2100 g m s–1
–P1 = –1200 g m s–1 P2 = 900 g m s–1
4.
P2 = 6000 kg m s–1
P = 60002 + 60002 kg m s–1 = 8490 kg m s–1
= tan–1 (6000) = tan–1 (1) = 45°6000
5.
= tan–1 ( 500 ) = 5.7°5000
P = 500 kg m s–1 N
P1 = 5000 kg m s–1
P2 = 5020 kg m s–1
bearing 275.7°
Pole vaultingQuestion 200W: Warm-up Exercise
1. Kinetic energy as she runs, elastic strain energy as the pole bends, gravitational potential energyas she reaches the maximum height.
2.
5 Advancing Physics
.kJ7.1
2
sm5.7kg60
2
212
mv
Ek
3.
mghmv
2
2
so
.m9.2sm8.92
sm5.7
2 2
12
g
vh
4. While the pole vaulter is standing, her centre of mass is approximately 1 m above the ground.The calculation shows she should be able to raise her centre of mass by 2.9 m, giving a total near3.9 m. However, not all the kinetic energy can be changed to gravitational potential energy.Competing at top levels it will be essential to use jumping techniques which allow the centre ofmass to pass below the bar, and to pull downwards to bend the pole before take off.
Centripetal forceQuestion 30S: Short Answer
Solutions1. Potential energy = m g h = 200 kg 9.8 N kg–1 45 m = 88 200 J.
2. Assume there is no work done against friction or air resistance, so loss of potential energy equalsgain in kinetic energy:
–1
2
2
s m 7.29
kg 200
J 882002
J 882002
1
v
v
mv
3.
2–2–12
s m 6.17m 50
s m 7.29 onaccelerati lcentripeta
r
v
4.
kN 53.3s m 6.17kg 200 force lcentripeta 2-2
r
mv
In addition, the weight of the passengers = 200 kg 9.8 N kg–1 = 1.96 kN. Total force = 5.49 kN.This is 5488 N /1960 N = 2.8 times the weight of the passengers.
Circular motion – more challengingQuestion 40S: Short Answer
6 Advancing Physics
1. The speed of the ride at 60 s is 10.8 m s–1. From the graph:
.6.20minuteperrevs
s60
m52minutepersrevolutionofnumbersm8.10 1
The ride is operating within safety limits.
2.
.N1586
kgN8.9kg7662cos 1
R
R
3.
2–2–12
s m 3.23m 0.5
s m 8.10 onaccelerati lcentripeta
r
v
4.
–1
2
2
s m 6.9
kg 76
m 5.0N 1400m 0.5
kg 7628cos N 1586
v
v
v
5.
.N1490
2kgN8.9kg762Weight 1
This is less than the 1586 N in question 2.
Finding the mass of a planet with a satelliteQuestion 110S: Short Answer
1.
.4
2
32
GT
RM
2.
.kg1063.8)s3600323()mkgN1067.6(
)m1082.5(44 2522211
382
2
32
GT
RM
3.
.kg108.5)s3600243.27()mkgN1067.6(
)m108.3(44 2422211
382
2
32
GT
RM
4.
7 Advancing Physics
.kg103.1)s3600244.6()mkgN1067.6(
)m109.1(44 2222211
372
2
32
GT
RM
5.
.kg100.2)s360024365()mkgN1067.6(
)m105.1(44 3022211
3112
2
32
GT
RM
6. A space probe sent from Earth can act as an artificial satellite, giving values for R and T.
Impulse and momentum in collisionsQuestion 150S: Short Answer
Solutions1.
kN 7.1–
s 5.0h s 3600
km m 1000h km 80–h km 50kg 1000
– –1
–1–1–1
t
uvmF
2.
kN 7.8–
s 0.4h s 3600
km m 1000h km 0.8h km 2.0–kg 1250
–1
–1–1–1
t
uvmF
3.
hmgEmv
E pk 2
2
so
.sm4.4m0.1)sm8.92(2 12 ghv
4.
–1–2 s m 1m 05.0s m 8.922 ghv
5.
N 270
s 0.001
s m 4.4s m 1.0–kg 050.0 –1–1
t
uvmF
6. Let the mass of the lighter wagon be m
–1
–1
–1–1
s m 8.0
7.2s m 2.2
7.2s m 5.00.1s m 17.1
v
vmm
vmmm
7. The centre of mass moves at constant velocity after impact. During the sloshing period, thismeans the wagons do not move steadily.
8.
8 Advancing Physics
–1
–1–1
–1
–1–1
s km 05.0
s km 95.6kg 401
s km kg 2788
kg 401s km kg 12)–(2800
kg 401s km 12–kg 0.1s m 7kg 400
v
v
v
v
9. From the viewpoint of the Russian module
.sm15.0kg106.24
sm2.0)kg1018(
kg106.618sm2.0kg)1018(
13
13
313
v
v
Collisions of spheresQuestion 160S: Short Answer
1. Momentum is conserved. Because the mass doubles, the velocity after joining together is onehalf of the initial velocity. The time taken to fall to the ground is the same in both cases, since thespheres fall freely under gravity. At any given time after collision, the horizontal distance travelledis halved.
2. A remains on the shelf. B follows the path A would have taken if B were not there.
3. No. For momentum to be conserved, this would require B to move with a greater velocity than Ahad on collision. This would require an increase in kinetic energy, which is clearly not possible.
Jets and rocketsQuestion 180S: Short Answer
Solutions1. Ns720000s 4N180000 Thrust Ft
–1s m 72
kg 10000Ns 720000
v
v
mumvFt
2. Xenon ions would provide more thrust. This is because there would be a greater momentumchange per second since they have a greater mass than krypton ions. Let m = mass of each ion,n = number of ions emitted per second, v = speed of ejection of ions.
Then, impulse, Ft = change in momentum = (m n) (v – 0): so
.s105.1)sm101.3()kg102.2(
s1N1.0 1191425
mv
tFn
3. The pump pushes the water forwards, which by Newton’s third law exerts a force of equal sizeback on the hose and the pump system. The hose, gripped by the firefighter, exerts a backward
9 Advancing Physics
force on the firefighter
.N60sm30s10
kg20 1
vt
m
t
pF
4. To eject the gas, the rocket exerts a forward force on the gas. By Newton’s third law, the gasexerts a force of equal size back on the rocket. This force is responsible for the deceleration ofthe rocket
kg.50sm5000
sm5kg000501-
1
gas
rocketrocketgasgas
rocketgas
m
vmvm
pp
5. m = mass of air brought to rest per second, v = initial speed of air, A = area of sail, = density ofair:
.N109.4sm5skg98
skg98sm5m15mkg3.1
211
1123
vt
m
t
pF
Avt
V
t
m
6. The kestrel pushes down on the air, giving it downward momentum. The air pushes back up onthe kestrel’s wing (by Newton’s third law). If this upward push equals the weight of the kestrel, thebird can hover at a constant vertical velocity of 0 m s–1.v = downward speed gained by air, = density of air, A = area of air column pushed down, mb =mass of bird:For the bird to hover, the push of the air must equal mb g, the weight of the bird.
–1
243-
–1
2
2
s m 0.5
)m 10600(m kg 1.3
kg N 9.8kg 2.0
kestrel on force upward air on force downward
so
but
air on force downward
v
v
A
gmv
gmAv
AvF
Avt
mt
mv
t
pF
b
b
W8.9s m 0.5kg N 9.8kg 2.0 power minimum –1–1 Fv
Gravitational potential energy and gravitational potentialQuestion 210S: Short Answer
10 Advancing Physics
Solutions1.
kJ2.1m0.2kgN8.9kg60 1
hmgEp
2.
–1–1 kg J 19.6m 0.2kg N 8.9
hgVg
3.
–1
–12
221
s m 3.6
kg J 6.192
v
v
Vmmv g
4. Same as question 3.
5. weight = mg = 24 kg × 9.8 N kg–1 = 235 N
6. Ep = mgh = 235 N × 3.0 m = 706 J
7.
–1
–12
221
221
s m 7.7
m 0.3kg N 8.92
v
v
hmgmv
Emv p
8.
MJ 29)100.1(kg N 9.8kg) 103( 3–13 PE
9. Most of the energy goes to warming up the brakes and the air which cools them.
10. The material properties of the brake shoe and lining will change at high temperature, so they mayfail.
11. At the top of a swing, the bob has maximum gravitational Ep and no kinetic energy; at the bottomit has maximum Ek and no gravitational Ep; the total Ek and gravitational Ep remains constant.
12. No. The bob will have zero kinetic energy when it returns to its starting position.
13.
.m0.5
kgN8.9/kgJ47
/
11
gVh g
14.
.J49
)kgJ492(kg5.0 1
gp VmE
11 Advancing Physics
15. 0 J. No work is done in moving along an equipotential surface, as there is no force component inthis direction.
16.
A
B
The field strength and potential at A and at B are the same for both paths.
17. 22.8 J kg–1
15.2 J kg–1
7.6 J kg–1
0 J kg–1
Summary questions for chapter 11Question 250S: Short Answer
1.
.m107.6
)]s6090/(2[
)kg100.6()skgm107.6(
)/2(
6
2
24213113
2Earth3
2satelliteEarth2
satellite
r
r
T
GMr
r
mGMrmF
2. It is not possible to have a satellite orbiting at a larger distance with the same orbital period. Thecentripetal force is the force of gravity and this force has a unique value at each distance, a value
12 Advancing Physics
which depends only on the distance to Earth’s centre. If, at each distance, there is only onecentripetal force available for any given satellite, then there is only one possible angular speed itcan have.
3.
.N0.2)m104.6(s360024
2kg60 6
22
rmF
The gravitational pull of Earth provides the centripetal force.
4. If Earth started spinning more rapidly, you would feel lighter, since the contact force between youand the ground would lessen. There would be less friction between you (and all standing objects)and the ground, because friction depends on the contact force. At a certain speed, all objectswould appear ‘weightless’ and would appear to float in the same way that objects in an orbitingspace station appear to float. (However, objects would still have a weight in the sense that Earthwould still be exerting a gravitational pull on them.) Also, day and night would alternate more andmore frequently.
5. At the North Pole, because it does not describe a circular motion, none of the strange weighteffects would be observed. The Sun would appear to move more quickly across the sky,however.
6. The mass of a family is approximately 300 kg:
7. 11 smkg600sm2kg300 vmp
8.
N600s1
smkg600 1
t
pF
9.
W80
s60
sm2kg)3008(2/
time
21212
t
mvEP k
CentrifugesQuestion 50C: Comprehension
1. No, the angular velocity is common throughout the apparatus.
2.
–1s rad 105s 60
radians 21000
3. Different masses require different forces to move in a circle of this radius (F is proportional to m).
4. For 1 g mass:
N 1.1
m 10.0)s rad (105kg 001.0 2–1
2
rmF
13 Advancing Physics
For 2 g mass:.N2.2F
5 If the force (on a less dense material) is too large, it will move in a spiral path towards the centre.If the force (on a more dense material) is too small, it will move in a spiral path outwards from thecentre. So less dense material moves to the top of the tube and more dense material moves tothe bottom of the tube.
How Cavendish didn’t measure G and Boys didQuestion 60C: Comprehension
1. Minimum distance apart = 2r . A 5 kg sphere of radius 5 cm has density 9.56 g cm–3 so lead andgold are possible materials.
2. 1 in = 0.0254 m so 39.14 in = 39.14 0.0254 = 0.9942 m; T = 2 s (a tock as well as tick):
g
lT 2
and so
.sm812.9
s2
m9942.044 22
2
2
2
T
lg
3. 1 ft = 0.304 m, giving R = 6370 km (same as the modern value); G = g R2 / ME.
kg 100.6m kg 1048.5m) 104.6( 24-333634 VM
2–21124
262
kg m N107.6kg 106.0
m) 104.6(8.9
M
gRG
Are there planets around other stars?Question 90C: Comprehension
1.
m 108.2 yrs 1016.3)s m 10(3.0 yearslight 30 17–17–18
2. Detect reflected starlight or infrared radiation from the planet. Detectors tend to be swamped bymuch higher emissions from the star being orbited.
3.
14 Advancing Physics
F1
F2
starplanet
F1 = –F2
4. From table 1 (with R in AU and T in years)
.142
3.0
5.3;140
0.1
2.5;146
0.3
6.7 3333
sM
R
5. For star mass 3.0M (with R in AU and T in years)
.0.36.61
6.22;0.3
12
6.72
3
2
3
For star mass 1.0M
.00.10.6
3.3;98.0
12
2.52
3
2
3
For star mass 0.3M (with R in AU and T in years)
.26.05.0
4.0;30.0
12
5.32
3
2
3
6. For small
radians105m104
m102tan 11
8
2
and
radians.108.4reesdeg360
radians2
degree
seconds3600
seconds1010arcofs10 11
6
7. For Ms = 3.0 M
AU
s10
AU6.22
s226;
AU
s10
AU6.7
s76
orbital
onperturbati
R
and for Ms = 1.0 M
.AU
s30
AU3.3
s99;
AU
s30
AU2.5
s157
15 Advancing Physics
Variations in gQuestion 130C: Comprehension
1.
.sm104.3
)m1037.6(s360024
2
22
62
2
rgg o
2.
.sm1074.2
m103789.6
1
m1037.6
1kg)1098.5()kgmN1067.6(
23
2
6
2
6242211
21
22
r
GM
r
GMg
3. Fm
FR
Fe
.reesdeg107.4m102
m1037.6
kg1098.5
)m106.1(mkg3000tantan 4
2
3
6
24
39311
e
m
F
F
4.
2
2
tan
tan
m
eme
m
e
e
m
e
m
R
RMM
R
R
M
M
F
F
Collision with spaceship EarthQuestion 170C: Comprehension
16 Advancing Physics
1.
111
skm30s360024365
)m105.1(22
T
rv
2.
J105.4
2
sm109.29kg010.0
26
2142
mv
Ek
3.
16
sm95kg1000
)J105.4(22
m
Ev k
To convert to miles per hour:
.mph212km
mile8/5hkm340
kmm1000
hs3600sm95 1
1
11
v
For comparison, the land speed record is currently held by the Thrust SSC, a jet-powered vehicleof mass 20.4 tonnes. It was driven by RAF pilot Andy Green at 759 mph in May 1999.
4. The fragments of clay pigeon would scatter but their centre of mass would still be travellingtowards you. You might well be hit by some of them.
5.
comet
mass ejection
Earth
Getting a satellite up to speedQuestion 90C: Comprehension
1. Each molecule has greater momentum, hence imparts greater forward momentum to the rocket.
2. Safety: liquid fuels are much more dangerous to transport and handle.
3. Mass of a golf ball 100 gNumber of balls = 4 000 000 g / 100 g 40 000.
17 Advancing Physics
4. Using the law of conservation of momentum and remembering that the lorry has 4 tonnes of golfballs on board: 0.1 kg 30 m s–1 = 6 000 kg v
v = 5 × 10–4 m s–1
5. Now the mass of the lorry is just 2 tonnes so:
0.1 kg 30 m s–1 = 2 000 kg v
v = 1.5 × 10–3 m s–1
6. The velocity increase is a little greater each time a golf ball is driven off, and the slope increasessteadily since the lorry has less and less mass.
time / hours11
33
7. This starts to resemble the molecular case in some respects. There are far more peas than golfballs; they have less mass so each impulse is smaller, and the pea shooter may not fire them sorapidly as the golf club. The overall effect is the same, but each step is smaller. It is like a limitingprocess where the finite m is in progress towards the infinitesimal m.
Using Kepler's third lawQuestion 10D: Data Handling
1.
2
3
2
3
h5.1h24
km20042 R
so
.km6650h24
h5.1km200423
2
23
R
Therefore
18 Advancing Physics
.km280km6370km6650 h
16hours 1.5
hours 24 orbits of number
2.
km 960 to km 280 is range
km 960km 6370–km 7330 height
km 7330minutes 101.8
)minutes 101.1()km 109.2(
minutes) 105(minutes) 90(
km) 6650(
23
243113105
2
3105
2
3
2105
3105
290
390
R
R
R
T
R
T
R
3. Low orbits give smaller image detail (is it a battlefield tank?); higher orbits give greater coverageand endurance (because there is less atmospheric friction).
4.
4 October1957
25 October1957
25 December1957
Orbital period / minutes 96.2 95.4 91.0
Minimum height / km 219 216 196
Maximum height / km 941 866 463
Mean height / km 580 541 330
Mean radius / km 6950 6911 6700
R3 / T2 two significant figures
36 106 36 106 36 106
The Kepler ratio for each case is the same; the deviation from the mean height decreases, so theorbit becomes more like a circle.
Average orbit time was 93.6 minutes. In 3 months (90 days) it made approximately
orbits 1400min 94
hr min 60day hours 24 days 90 –1–1
The gravitational field between the Earth and the MoonQuestion 120D: Data Handling
19 Advancing Physics
1. The force of attraction between the spacecraft and Earth is equal in magnitude but opposite indirection to the force of attraction between the spacecraft and the Moon. There is no net force, sono acceleration.
2. The gravitational force is a right angles to the direction of motion, so there is no change in speed.The direction, however, will change so that the force is always acting from the centre of thespacecraft to the Moon. Since there is a change in direction, there is a change in velocity,therefore there is an acceleration.
3. The thrust acts at right angles to the speed, since this does not change.
v
vv
As shown in the vector triangle the change in velocity v is v (for a small angle ), F = (mv) /t
s.10N00012
rad100/1ms6000kg2000 1
F
mv
F
vmt
4. If the thrust motors are not used, the only force acting on the spacecraft is the gravitational forceof attraction between the Earth and the spacecraft; this acts towards the centre of the Earth. F t =change in momentum and since F = – (G M m) / R2
.sN800s100
m104.6
kg2000)mkgN104(26
214
Ft
5. The measured distance from D to the Moon is three times the distance from D to the Earth. Thismeans that the gravitational force of attraction on the spacecraft due to the Moon (an inversesquare law) would be 1/9 that due to the Earth if the Earth and Moon had the same mass. But theMoon’s mass is smaller than Earth’s, and thus it is fair to ignore it in these calculations.
6. The change in gravitational potential energy =
2112
11
RRGMm
R
GMm
R
GMm
J.106.3
m102
1
m102
1kg2000)kgmN104( 10
781214
.Since gravitational potential is 0 at infinity and work needs to be done to reach infinity, thepotential energy is less at a distance closer to the centre of a gravitational field, in this case at E.Since potential energy is a scalar quantity and does not depend on the path taken, this value isthe same as if the motors had been used to move between C and E.
7. From conservation of mechanical energy (no work done against friction, motors not used), thedecrease in potential energy = increase in kinetic energy , 3.6 1010 J. At C, the kinetic energy
was already 0.5 2000 kg (6000 m s–1)2 = 3.6 1010 J so the total kinetic energy is 7.2 1010 J.
8. For a circular orbit:
20 Advancing Physics
J 102m) 102(2
)kg m N 100.4(kg 2000
2
so
107
–12142
21
K
2
2
2
R
mGMmvE
R
GMv
R
mv
R
GMmF
Gravitational potential difference, field strength and potentialQuestion 220D: Data Handling
1. v2 against 1/r gives a straight line graph, when probe a long way from Jupiter 1 / r = 0 and v2 =112 106 m2 s–2 so v = 11 km s–1.
2. If the total energy remains constant, GMm / r = ½ m v2. The straight line graph confirms this. Thekinetic energy gained by the spacecraft is equal to energy gained from falling through Jupiter’sfield.
3.
kg 109.1kg s m1067.62
s m 105.2
2s m 105.2gradient graph from
27–12–311
2–317
–2317
J
J
M
GM
4.
–1–1g
g
kg J 98m 10kg N 8.9
V
hgV
5.
–1
26
–1214
2g
kg J 98
m 106.4
m 10)kg m N 100.4(
hr
GMV
6. The potential gradient decreases with distance from Earth’s centre. So the weight of the rocketdecreases while the thrust force stays constant. The net force therefore increases.
7.
21 Advancing Physics
r / 106m
8. From the graph at r = 20 106 m Vg = 43 MJ kg–1
energy to lift 1000 kg = 1000 kg 43 MJ kg–1 = 43 GJ
9. From the graph at r = 36 106 m Vg = 51 MJ kg–1
energy to lift 200 kg = 200 kg 51 MJ kg–1 = 10.2 GJ
10. From the graph at r = 40 106 m Vg = 52.7 MJ kg–1
at r = 50 106 m Vg = 54.7 MJ kg–1
to move 1kg from 40 106 m to 50 106 m requires
54.7 MJ kg–1 – 52.7 MJ kg–1 = 2.0 MJ
11. Total energy = 54.7 MJ + 7.8 MJ = 62.5 MJ
12. Gravitational Ep:
.J10kgJ105.624.62kg1 516
gp VmE
13. W = F s so F = 105 J / 104 m = 10 N.
This is approximately the force at the Earth’s surface, though it decreases with height.
14. Gravitational Ep:
.J105.61kgJ105.621kg1 616 gVm
15. Because the force is much less than 10 N for most of the distance.
16. 62.5 MJ
17.
MJ5.622/2 mv
so
22 Advancing Physics
.skm2.11kg1
J105.622 16
v
18. Gravitational potential is the same on every kilogram.
19. Start with Vg at r = 10 106 m. Distances are double, 4 and 8, so Vg will be 1/4, 1/16 and 1/64
times the value.
r / 106 m
0
–10
–20
–30
–40
20 40 60 800
20. N kg–1; 0.25 N kg–1
21. They represent the magnitude of the fields strengths at those points.
22. The ratio is 4:1. Since the ratio of the distances is 2:1, this is consistent with an inverse-squarefield.
Changing orbitsQuestion 230D: Data Handling
1.
.skm07.3s606024
km200422 1
t
sv
2.
minutes5.90days0628.0day120042
6670
km20044
km300km6370
day1
23
3
2
2
T
T
so
23 Advancing Physics
.skm72.7s605.90
km66702 1
t
sv
3. E k depends on v2 so the kinetic energy is less in the geostationary orbit.
4.
km1044.22
km20042km6670 4
R
and the time for half an orbit will be
hr 5.3day 22.02
day 44.0day 1km 44200
km 1044.2
km 44200
km 1044.2
day 1
234
34
2
2
T
T
T
5. Upper line = kinetic energy; middle line = total energy; lower line = gravitational potential energy.
6. The diagram should reflect the times read from the graph 0.7 h in the parking orbit (half an orbit);5.3 h in elliptical transfer orbit (half ellipse); 6 h in geostationary orbit (1/4 of orbit). Total 1/2 + 1/2+ 1/4 = 1 1/4 orbits.
Why is a ‘black hole’ black?Question 240D: Data Handling
1. As there is no friction, there will be conservation of mechanical energy, Ek = Ep.
rGMmmv /2/2 re-arranges to
.2
r
GMv
2. Gravitational potential at infinity is zero. The decrease in kinetic energy used in reaching infinity isequal to the gain in potential energy to move from the surface of the Earth to infinity:
.sm20011m104.6
)kg100.6()mkgN107.6(2 16
242211
v
3. Density = mass / volume:
.kg100.2
)m107(3
4)mkg104.1(
30
3833
Vm
So
24 Advancing Physics
.sm102.6m107
)kg100.2()mkgN107.6(2 158
302211
v
4. 2/1)/2( RGMv
.sm105.12
sm100.3
2
sm104.1m1012
)kg100.2()mkgN107.6(2
1818
183
302211
c
v
5.
.km0.3
sm100.3
)kg100.2()mkgN107.6(2218
302211
R