Testing for an inverse-square law - AS-A2-Physics11... · 1 Advancing Physics Testing for an inverse-square law Question 10W: Warm-up Exercise 1. 500 / 2000 = 1 / 4 and 20 / 80 =

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


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


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


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.


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


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


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.


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


.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


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:


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.


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


.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


.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


.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

Documents

Testing for an inverse-square law - AS-A2-Physics11... · 1 Advancing Physics Testing for an inverse-square law Question 10W: Warm-up Exercise 1. 500 / 2000 = 1 / 4 and 20 / 80 =