Motion Revision PowerPoint

1. Motion Revision PowerPoint This is a good resource to go over quickly as a mental quiz on the major areas of motion in the mid-year exam

2. 2013 Motion Outcome Dot Point 9Gravitational Fields & Forces apply gravitational field and gravitational force concepts,

3. Question 1 Work out the gravitational field strength on the surface of the Earth using the data: G = 6.67  10-11 Nm2kg-3, ME = 5.98  1024 kg, re = 6.37  106m g = ? G = 6.67  10-11Nm2kg-3 ME= 5.98  1024 kg re= 6.37  106m g g g = 9.8299 g= 9.83 Nkg-1

4. Question 2 Work out the gravitational force on a 3.0kg object 8.0  106m from the centre of the (data: G = 6.67  10-11 Nm2kg-3, ME = 5.98  1024 kg) F= ? m = 3.0kg G = 6.67  10-11 Nm2kg-3 ME= 5.98  1024kg r = 8.0  106m F F F = 18.69684 F= 19 N

5. Question 3 A 2.0kg piece of space junk is in a region above the earth where the gravitational field strength is 4.0Nkg-1. (a)What is the gravitational force on the object? F = ? m = 2.0kg g = 4.0Nkg-1 F = mg F = 2  4 F = 8.0 N (b)What is the weight of the object? W = 8.0 N (c)Describe the motion of the piece of space junk? The space junk will be accelerating at 4.0ms-2 towards the Earth Weight = Fgrav g = 4.0ms-2

6. Question 3 A 2.0kg piece of space junk is in a region above the earth where the gravitational field strength is 4.0Nkg-1. d)How far is the space junk from the centre of the Earth? Data: G = 6.67  10-11 Nm2kg-3, ME = 5.98  1024 kg r = ? g = 4.0Nkg-1G = 6.67  10-11 Nm2kg-3, ME = 5.98  1024 kg g r2 r r r = 9.9858  106 m r 1.0  107m

7. Question 4 In 2002 the space probe Cassini was directly between Jupiter and Saturn. Its mission was to deliver a probe to one of Saturn’s moons, Titan, and then to orbit Saturn. When Cassini is at a particular position between the two planets the gravitational field strengths are as follows gsaturn = 2.50  10-7 Nkg-1 and gjupiter= 7.18  10-7 Nkg-1 (a)What is the net gravitational field strength at Cassini? g = 7.18  10-7– 2.50  10-7 = 4.68  10-7Nkg-1 towards Jupiter Jupiter Saturn Cassini gjupiter = 7.18  10-7 gsaturn = 2.50  10-7

8. Question 4 (b)Given that gjupiter = 7.18  10-7 Nkg-1 and the fact that Cassini is 3.9  1011 m from Jupiter, what is the mass of Jupiter? M = ? g = 7.18  10-7Nkg-1G = 6.67  10-11 Nm2kg-3r = 4.2  1011 m g gr2 1.89888  1027 = M M1.9  1027 kg

9. Question 5 A small meteor approaching the earth has a gravitational force of 0.45N on it when it is 4.0  107m from the centre of the earth. (a) What was the gravitational force on the meteor when it was is 1.2  108m from earth? Here There F1= 0.45NF2= ? r1= 4.0  107m r2= 1.2  108m G = G G = G M = M M= M m = m m = m Since r × 3, F × So F2 =  0.45 = 0.050N

10. Question 5 A small meteor approaching the earth has a gravitational force of 0.45N on it when it is 4.0  107m from the centre of the earth. (b) What was the gravitational force on the meteor when it is 1.0  107m from earth? Here There F1= 0.45NF2= ? r1= 4.0  107m r2= 1.0  107m G = G G = G M = M M= M m = m m = m Since r × , F × = 16 So F2 =  0.45 = 7.2N

11. Question 5 A small meteor approaching the earth has a gravitational force of 0.45N on it when it is 4.0  107m from the centre of the earth. (c) What will be the gravitational force on the meteor when it is 9.0  106m from earth? Here There F1= 0.45NF2= ? r1= 4.0  107m r2= 9.0  106m G = G G = G M = M M= M m = m m = m Since r × , F × = 19.75309 So F2 = 19.75309  0.45 = 8.888888  8.9N

12. Whenever the height or distanceabove the surface of the Earth is mentioned it is useful to draw a diagram to be clear about the situation Question 5 A small meteor approaching the earth has a gravitational force of 0.45N on it when it is 4.0  107m from the centre of the earth. (d) What was the gravitational force on the meteor when it is when it is 500km above the surface of the earth given than re = 6.37  106m? Here There F1= 0.45NF2= ? r1= 4.0  107m r2= 6.37  106 + 500000 = 6870000 G = G G = G M = M M= M m = m m = m 500km r1 = 4.0  107m rE r2 A Since r × , F × = 33.90053 So F2 = 33.90053 0.45 = 15.25523  8.9N

13. Question 5 A small meteor approaching the earth has a gravitational force of 0.45N on it when it is 4.0  107m from the centre of the earth. (e) At what distance from the centre of the earth will the gravitational force on the meteor be 16.2N Here There r1 = 4.0  107m r2 = ? F1= 0.45NF2= 16.2 G = G G = G M = M M= M m = m m = m If the question had asked what height above the surface of the earth will the gravitational force be 16.2 then you would have had to subtract the radius of the earth Since F × , r × = 0.16666667 So F2 = 0.1666667 4.0  107 = 666666667  6.7  106m

14. Question 6 Given that g = 10 Nkg-1on the surface of the Earth and rE represents the radius of the Earth , what is the gravitational field strength 4rEfrom the centre of the Earth. Here There g1 = 10 g2 = ? r1 = rEr2= 4rE G = G G = G M = M M= M Since r × 4, g × g2 = 10 × = 0.625Nkg-1 4rE

15. Whenever the height or altitudeabove the surface of the Earth is mentioned it is useful to draw a diagram to be clear about the situation Question 7 Given that g = 10 Nkg-1on the surface of the Earth and rE represents the radius of the Earth, what is the gravitational field strength rEabove the surface. Here There g1 = 10 g2 = ? r1 = rEr2= 2rE G = G G = G M = M M= M Since r × 2 , g × g2 = 10 × = 2.5 Nkg-1 r2 = 2rE rE r1 = rE

16. Whenever the height or altitudeabove the surface of the Earth is mentioned it is useful to draw a diagram to be clear about the situation Question 8 At a point above the Earth’s surface g = 0.40 Nkg-1. Given that on the surface of earth g = 10 Nkg-1how many Earth radii is the point above the surface of the Earth. Here There r1 = rEr2= ? g1 = 10 g2 = 0.40 G = G G = G M = M M= M 4rE rE 5rE Since g × , r × = 5 r2 = 5rE So the point will be 4 rE above the surface

17. Question 9 The gravitational field strength r metres from the centre of the moon is g Nkg-1 (a) What will be the gravitational field strength on the 3r from the centre of the moon? Here There g1 = gg2= ? r1 = r r2 = 3r G = G G = G M = M M= M Since r × 3 , g × so g2 = g× = Nkg-1 (b) What will be the gravitational force on a mass m at a distance of 3r from the centre of the moon? F = ? m = m g = F = mg F = m  F =

18. Question 10 If the gravitational force on an object r metres from the centre of a planet is F, what would be the Weight of the object metres from the centre of the planet? Here There F1= F F2= ? r1= rr2= G = G G = G M = M M= M m = m m = m Since r × , F × 100 So F2 = 100  F = 100N

19. Question 11 – 2007 Q12 Pluto and the dwarf planet Eris have roughly the same mass and orbit the sun. Pluto orbits the sun at 6.0 × 1012 m Eris orbits the sun at 10.5 × 1012 m What is Pluto Eris FP= ? r = 6.0 × 1012 G = G M=M m=m FP = ? r = 6.0 × 1012 G = G M=M m=m Fp = FE= =  = = 0.32653  0.326

20. Question 11 – 2007 Q12 Alternative Pluto and the dwarf planet Eris have roughly the same mass and orbit the sun. Pluto orbits the sun at 6.0 × 1012 m Eris orbits the sun at 10.5 × 1012 m What is Here (Pluto)There (Eris) FP = F FE = ? rP = 6 × 1012rE = 10.5 × 1012 so Since r × = 1.75 F × = 0.3265 FE = 0.3265 F = = = 0.3265

21. 2013 Motion Outcome Dot Point 10Weight & Weightlessness apply the concepts of weight (W=mg), apparent weight (reaction force, N), weightlessness (W=0) and apparent weightlessness (N=0)

22. Question 1 A 30kg object is seen by a camera to be floating around inside an orbiting satellite where g = 3.2 Nkg-1. (a) What is the weight of the floating object? W = ? m = 30kg g = 3.2 Nkg-1 W = mg W = 30 × 3.2 W = 96N

23. Question 1 A 30kg object is seen by a camera to be floating around inside an orbiting satellite where g = 3.2 Nkg-1. (b) Why is the object floating when there is a gravitational force on it? Since the only force acting on the satellite and 30kg object is gravity they are in freefall and objects in free fall appear to be floating relative to each other.

24. Question 2 Astronauts working in capsules orbiting the Earth are said to be weightless. Is this an accurate description? The astronauts are not weightless since they are in a gravitational field and hence there is a gravitational or weight force on them. The astronauts are instead experiencing apparent weightlessness because they are in freefall (along with the capsule) and they are not experiencing any normal reaction forces.

25. Whenever the height or altitudeabove the surface of the Earth is mentioned it is useful to draw a diagram to be clear about the situation Question 3 What would be the weight of a 20kg object at rE above the surface? Here There F1= 200F2= ? r1 = rEr2 = 2rE G = G G = G M = M M= M m = m m = m F Since r × 2 , g × F2= 200 × = 50N r2 = 2rE rE r1 = rE

26. Question 4 – 2010 Q18 The International Space Station (ISS) is currently under construction in Earth orbit. It is incomplete, with a current mass of 3.04 × 105 kg. The ISS is in a circular orbit of 6.72 × 106 m from the centre of Earth. m ISS = 3.04 × 105 kg Mearth = 5.98 × 1024 kgrEarth = 6.37 × 106 m Radius of ISS orbit: 6.72 × 106 m G = 6.67 × 10–11 N m2kg–2 What is the weight of the international space station? F = ? m ISS = 3.04 × 105 kg G = 6.67  10-11 Nm2kg-3 ME= 5.98  1024kg r = 8.0  106m F F = 2685109 F ≈ 2.69 × 106N

27. Question 5 – 2009 Q7 + N Karen 60kg A ride in an amusement park allows a person to free fall without any form of attachment. A person on this ride is carried up on a platform to the top. At the top, a trapdoor in the platform opens and the person free falls. Approximately 100 m below the release point, a net catches the person. Helen, who has a mass of 60 kg, decides to take the ride and takes the platform to the top. The platform travels vertically upward at a constant speed of 5.0 m s–1 What is Helen’s apparent weight as she travels up? W = 60 × 10 = 600N Since speed is constant N = 600N so apparent weight = 600N The Examiners said: Since Helen was moving up at a constant speed the net force acting on her was zero. Therefore the normal reactionforce(apparent weight) equalled the gravity force = mg = 60 x 10 = 600 N. Some students did calculations which assumed there was an acceleration. Others derived an apparent weight of zero, seemingly confusing the normal reaction force with the net force. Another incorrect assumption was that there was a net force of mv2/R.

28. The Examiners said: Helen was travelling up but slowing. By applying Newton’s second law and taking down as the positive direction,mg – N = ma. Therefore 600 – N = 60 x 2, which led to a normal reaction (apparent weight) of 480 N. The main errors students made involved mixing up positive and negative signs for the forces acting. Some students were confused about actual forces and the net force, while others introduced a net force of mv2/R. RJ comment I prefer to have the positive direction in the direction of initial motion (just like an object thrown upwards). This also made sense because deceleration makes me think of a negative acceleration and this only works if positive is in direction of initial motion. Question 6 – 2009 Q8 + As the platform approaches the top, it slows to a stop at a uniform rate of 2.0 m s–2. (mKaren = 60kg) What is Helen’s apparent weight as the platform slows to a stop? N Karen 60kg N = 60 × 10 = 600N Fnet= N – W ma = N – 600 60 × -2 = N – 600 -120 = N – 600 480 = N So apparent Weight = 480N

29. Question 7 – 2009 Q9 Because the only force on Helen is the gravitational weight force she is in freefall meaning that she is accelerating downwards. Because there is no normal reaction force on her body she will experience apparent weightlessness. Helen drops through the trapdoor and free falls to the safety net below . Ignore air resistance. During her fall, Helen experiences ‘apparent weightlessness’. In the space below explain what is meant by apparent weightlessness. You should make mention of gravitational weight force and normal or reaction force. The Examiners said: Apparent weight is the normal reaction force. Helen was in free fall, accelerating at the value of the gravitational field, so the normal force was zero. Since there was a gravitational field, there was still a weight force acting on Helen. It was common for students to incorrectly state that the normal force equalled the gravitational force, thereby cancelling each other out and creating a feeling of weightlessness. Others referred to Helen reaching terminal velocity and equated this to apparent weightlessness. Another common approach was to explain apparent weightlessness in terms of an astronaut in orbit in a spaceship; however this did not relate to the question.

30. 2013 Motion Outcome Dot Point 11Satellite Motion model satellite motion (artificial, moon, planet) as uniform circular orbital motion ()

31. Question 1 A shuttle orbits the Earth at a distant 6720km from its centre with female astronaut from Australia. The gravitational field strength at the orbit is 8.9N kg-1. (a) Calculate the period of the shuttle in minutes. T = ? g = 8.9Nkg-1r = 6720km = 6.37 × 106 m g = T2= T= T= T = T = 5459.709s T= 90.9951 T = 91 minutes

32. Question 1 A shuttle orbits the Earth at a distant 6720km from its centre with female astronaut from Australia. The gravitational field strength at the orbit is 8.9N kg-1. (b) The aussienaut on board the shuttle has a mass of 65kg. What is her weight in orbit? W = ? m = 65kg g = 8.9Nkg-1 W = mg W = 65  8.9 W = 578.5 W  579 N

33. Question 1 A shuttle orbits the Earth at a distant 6720km from its centre with female astronaut from Australia. The gravitational field strength at the orbit is 8.9N kg-1. (c) What is the speed of the shuttle as it orbits the Earth? v= ? g = 8.9 Nkg-1 r = 6720km = 6.37 × 106 m v = v = v = 7733.56 ms-1 OR v = ? T = 5459.709s r = 6720km = 6.37 × 106 m v = v = v = 7733.56ms-1

34. Question 1 A shuttle orbits the Earth at a distant 6720km from its centre with female astronaut from Australia. The gravitational field strength at the orbit is 8.9N kg-1. (d) Whilst in orbit, she is outside the shuttle, securely attached to the shuttle by a safety line. While repairing a malfunction in the external gismo meta-operative jim jams, the safety line snaps. When she is detached from the shuttle, will she plummet to earth (explain your answer)? No. Because she is travelling at the same speed as the shuttle and will be experiencing the same gravitational acceleration, she will continue to orbit at the same orbital radius as the shuttle.

35. Question 1 A shuttle orbits the Earth at a distant 6720km from its centre with female astronaut from Australia. The gravitational field strength at the orbit is 8.9N kg-1. (e) People in Earth-orbit vehicles (such as space shuttles) are often described as being ‘weightless’ Which of the following is the best description of this experience? A. They are far enough away from the Earth that gravity is greatly reduced. B.The effects of circular motion forces cancel out any gravity forces. C. They are in free fall towards the centre of the Earth. D.The orbiting vehicles have technology to cancel gravity forces

36. Question 2 Calculate the mass of the Earth from the following data involving the Moon’s orbit around the Earth. Period of Moon orbit = 28 days Rmoonorbit = 3.8 × 108 m G = 6.67 × 10-11 Nm2kg-2 M = ? Rmoonorbit = 3.8 × 108m G = 6.67 × 10-11 Nm2kg-2 T = 28  24  3600 = 2419200 = M = M = M = 5.549338  1024 kg M  5.5  1024 kg A

37. Whenever the height or altitudeabove the surface of the Earth is mentioned it is useful to draw a diagram to be clear about the situation Question 3 A space shuttle vehicle is in orbit at a height of 360km above the surface of the Earth. The radius of the Earth is 6.4 × 106m. Take G = 6.67 × 10-11 (in SI units). Take the mass of the Earth as 6.0 × 1024kg. Calculate the speed of the shuttle. Show your working. v = ? r = 6.4  106 + 360000 = 6860000 G = 6.67 × 10-11 ME = 6.0 × 1024kg v = v = v= 7637.944ms-1 v= 7.7  103ms-1 B 360km rE r Can also get an answer through the formulae strings = (or = ) • v2= • v =

38. Question 4 A space shuttle of mass 200t is in circular orbit around the Earth, at a height of 200km. Calculate the kinetic energy of the space shuttle in this orbit. Data: G = 6.67 × 10-11 Nm2kg-2 ME= 6.0 × 1024 kg RE= 6.4 × 106 m v = ? r = 6.4  106G = 6.67 × 10-11 ME = 6.0 × 1024kg m = 200t = 200000kg Ek Ek = Ek = 6.253125 × 1012 J Ek 6.3 × 1012 J

39. Question 5 Two satellites are in orbit at the same height around the Earth. Satellite 1 has a mass ten times greater than the satellite 2. (a) v= Satellite 1Satellite 2 T1= ? T2= ? m1=10m m2=m r1= r r2= r G = G G = G M=M M=M v1= v2= = = 1 Alternative Since there is no mass in the v formula and everything else is the same for both satellites, they must have the same velocity so = 1

40. Question 5 Two satellites are in orbit at the same height around the Earth. Satellite 1 has a mass ten times greater than the satellite 2. (b) T Satellite 1Satellite 2 T1= ? T2= ? m1=10m m2=m r1= r r2= r G = G G = G M=M M=M T1T2 = = 1 Alternative Since there is no mass in the T formula and everything else is the same for both satellites, they must have the same period so = 1

41. Question 5 Two satellites are in orbit at the same height around the Earth. Satellite 1 has a mass ten times greater than the satellite 2. (c) Ek Satellite 1Satellite 2 T1= ? T2= ? m1=10m m2=m r1= r r2= r G = G G = G M=M M=M Ek1 Ek = = 10 Alternative Ek = = 10 When you are working out ratios that use the same formula you can simply write out the proportionality variables since the constants in each formula cancel each other out.

42. Question 5 Two satellites are in orbit at the same height around the Earth. Satellite 1 has a mass ten times greater than the satellite 2. (c) Ek Satellite 1Satellite 2 T1= ? r1= r G = G M=M m1 =10mT2= ? r2= rG = G M=M m2 =m Ek1 Ek = = 10

43. Question 6 What is correct units for G in terms of m, s & kg. a = = = = kg-1m3s-2

44. Whenever the height or altitudeabove the surface of the Earth is mentioned it is useful to draw a diagram to be clear about the situation Question 7 Calculate the altitude of a satellite in a geo-stationary orbit. Data: G = 6.67 × 10-11 Nm2kg-2 ME= 6.4 × 1024kg RE= 6.4 × 106 m r = ? G = 6.67 × 10-11 Nm2kg-2 ME = 6.4 × 1024kg RE= 6.4 × 106 m T = 24 h = 24  3600 = 86400 r = 2.75088  1011 r = 2.8  1011m Altitude = 6.4 × 106+ 2.75088  1011 =2.75094  1011 2.8  1011m A B

45. Question 8 Show that two satellites of different mass with the same radius orbit about the Earth must have the same speed and period. The formula for the velocity of a satellite v = is dependent on r and independent m so if two satellites have the same orbital radius they will have the same velocity. The formula for the period of a satellite Tis dependent on r and independent m so if two satellites have the same orbital radius they will have the same period.

46. Question 9 - 2004 A spacecraft of mass 400kg is placed in a circular orbit of period 2.0 hours about the Earth. (a) Show that the space craft orbits at a height of 1.7 × 106m above the surface of the earth. Data: G = 6.67 × 10-11 Nm2kg-2 ME= 5.98 × 1024kg rE= 6.37 × 106 m r = ? m = 400 T = 2 h = 7200s G = 6.67 × 10-11 Nm2kg-2 ME= 5.98 × 1024kg rE = 6.37 × 106 m r = r = 8060786 So the height above surface = 8060786 – 6.37 × 106= 1690786  1.7 × 106m A

47. Question 9 - 2004 A spacecraft of mass 400kg is placed in a circular orbit of period 2.0 hours about the Earth. (b) Calculate the speed of the spacecraft given that the previous answer determined the orbital radius as 8.1 × 106 m (more accurately 8060786m). Other Data: G = 6.67 × 10-11 Nm2kg-2 ME= 5.98 × 1024kg rE= 6.37 × 106 m v = ? T = 2 h = 7200s r = 8060786 v = 7034.362 ms-1 v = 7.0  103ms-1

48. Question 10 – 2008 Q15 The figure opposite shows the orbit of a comet around the Sun. Describe how the speed and total energy of the comet vary as it moves around its orbit from X to Y The speed decreases from X to Y because kinetic energy is converted to potential energy but the total energy remains constant. The examiners said: Some students said that the speed changed, but did not say where it was greater. Others suggested that the potential energy increased and the kinetic energy decreased but did not relate this to the total energy.