motion revision powerpoint

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

2013 Motion Outcome Dot Point 9

Gravitational Fields & Forces• apply gravitational field and gravitational

force concepts,

Question 1Work 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 10-11 Nm2kg-3 ME = 5.98 1024 kg re = 6.37 106m

g g g = 9.8299g = 9.83 Nkg-1

Question 2Work 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 = 18.69684F = 19 N

Question 3A 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 = mgF = 2 4F = 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

g = 4.0ms-2Weight = Fgrav

Question 3A 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-1 G = 6.67 10-11 Nm2kg-3, ME = 5.98 1024 kg

g r2 r r r = 9.9858 106 m r 1.0 107 m

Question 4In 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 10-7 Nkg-1 towards Jupiter

Saturn JupiterCassini

gsaturn = 2.50 10-7 gjupiter = 7.18 10-7

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 10-7 Nkg-1 G = 6.67 10-11 Nm2kg-3 r = 4.2 1011 m

g gr2 1.89888 1027 = M M 1.9 1027 kg

Question 5A 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 ThereF1 = 0.45N F2 = ?r1 = 4.0 107m r2 = 1.2 108mG = G G = GM = M M= Mm = m m = m

Since r × 3, F × So F2 = 0.45 = 0.050N

Question 5A 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 ThereF1 = 0.45N F2 = ?r1 = 4.0 107m r2 = 1.0 107mG = G G = GM = M M= Mm = m m = m

Since r × , F × = 16So F2 = 0.45 = 7.2N

Question 5A 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 ThereF1 = 0.45N F2 = ?r1 = 4.0 107m r2 = 9.0 106m G = G G = GM = M M= Mm = m m = m

Since r × , F × = 19.75309

So F2 = 19.75309 0.45 = 8.888888 8.9N

Question 5A 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 ThereF1 = 0.45N F2 = ?r1 = 4.0 107m r2 = 6.37 106 + 500000 = 6870000G = G G = GM = M M= Mm = m m = m

Since r × , F × = 33.90053

So F2 = 33.90053 0.45 = 15.25523 8.9N

rE r2

500km

Whenever the height or distance above the surface of the Earth is mentioned it is useful to draw a diagram

to be clear about the situation

r1 = 4.0 107m

A

Question 5A 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 Therer1 = 4.0 107m r2 = ?F1 = 0.45N F2 = 16.2G = G G = GM = M M= Mm = 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

Question 6Given that g = 10 Nkg-1 on the surface of the Earth and rE represents the radius of the Earth , what is the gravitational field strength 4rE from the centre of the Earth. Here Thereg1 = 10 g2 = ?r1 = rE r2 = 4rE G = G G = GM = M M= M

Since r × 4 , g ×

g2 = 10 × = 0.625 Nkg-1

4rE

Question 7Given that g = 10 Nkg-1 on the surface of the Earth and rE represents the radius of the Earth, what is the gravitational field strength rE above the surface.

Here Thereg1 = 10 g2 = ?r1 = rE r2 = 2rE G = G G = GM = M M= M

Since r × 2 , g ×

g2 = 10 × = 2.5 Nkg-1

r1 = rE

rE r2 = 2rE

Whenever the height or altitude above the surface of the Earth is mentioned it is useful to draw a diagram


Question 8At a point above the Earth’s surface g = 0.40 Nkg-1 . Given that on the surface of earth g = 10 Nkg-1 how many Earth radii is the point above the surface of the Earth.

Here Therer1 = rE r2 = ?g1 = 10 g2 = 0.40G = G G = GM = M M= M

rE 5rE

4rE



Since g × , r × = 5r2 = 5 rE

So the point will be 4 rE above the surface

Question 9

(a) What will be the gravitational field strength on the 3r from the centre of the moon?

Here Thereg1 = g g2 = ?r1 = r r2 = 3r G = G G = GM = 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 = mgF = m F =

The gravitational field strength r metres from the centre of the moon is g Nkg-1

Question 10If 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 ThereF1 = F F2 = ?r1 = r r2 = G = G G = GM = M M= Mm = m m = m

Since r × , F × 100So F2 = 100 F = 100N

Question 11 – 2007 Q12Pluto and the dwarf planet Eris have roughly the same mass and orbit the sun.

Pluto orbits the sun at 6.0 × 1012 mEris orbits the sun at 10.5 × 1012 m

What is

Pluto ErisFP = ? 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

Question 11 – 2007 Q12Pluto and the dwarf planet Eris have roughly the same mass and orbit the sun.

Pluto orbits the sun at 6.0 × 1012 mEris orbits the sun at 10.5 × 1012 m

What is

Here (Pluto) There (Eris)FP = F FE = ?rP = 6 × 1012 rE = 10.5 × 1012 so

Since r × = 1.75 F × = 0.3265 FE = 0.3265 F = = = 0.3265

Alternative


Weight & Weightlessness• apply the concepts of weight (W=mg), apparent

weight (reaction force, N), weightlessness (W=0) and apparent weightlessness (N=0)

Question 1A 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 = mgW = 30 × 3.2W = 96N

Question 1A 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.

Question 2Astronauts 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.

Question 3What would be the weight of a 20kg object at rE above the surface? Here ThereF1 = 200 F2 = ?r1 = rE r2 = 2rE G = G G = GM = M M= Mm = m m = m

F Since r × 2 , g ×

F2 = 200 × = 50N

r1 = rE

rE r2 = 2rE



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 kg rEarth

= 6.37 × 106 m Radius of ISS orbit: 6.72 × 106 m G = 6.67 × 10–11 N m2 kg–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


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?

The Examiners said: Since Helen was moving up at a constant speed the net force acting on her was zero. Therefore the normal reaction force (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.

Karen60kg

+

W = 60 × 10 = 600N

N

Since speed is constant N = 600N so apparent weight = 600N


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?

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 commentI 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.

Fnet= N – W

ma = N – 60060 × -2 = N – 600-120 = N – 600480 = NSo apparent Weight = 480N

Karen60kg

+

N = 60 × 10 = 600N

N

Question 7 – 2009 Q9Helen 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.

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.


Satellite Motion• model satellite motion (artificial, moon,

planet) as uniform circular orbital motion ()

Question 1A 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-1 r = 6720km = 6.37 × 106 mg = T2 = T=T=T = T = 5459.709sT= 90.9951T = 91 minutes

Question 1A 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 = 578.5W 579 N

Question 1A 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

ORv = ? T = 5459.709s r = 6720km = 6.37 × 106 m

v = v = v = 7733.56 ms-1

Question 1A 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.

Question 1A 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

Question 2Calculate the mass of the Earth from the following data involving the Moon’s orbit around the Earth.Period of Moon orbit = 28 days Rmoon orbit = 3.8 × 108 mG = 6.67 × 10-11 Nm2kg-2 M = ? Rmoon orbit = 3.8 × 108 m G = 6.67 × 10-11 Nm2kg-2 T = 28 24 3600 = 2419200

= M = M =

M = 5.549338 1024 kgM 5.5 1024 kg

A

Question 3A 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

Can also get an answer through the formulae strings

= (or = )v2 = v =

rE r

360kmB



Question 4A 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 mv = ? r = 6.4 106 G = 6.67 × 10-11 ME = 6.0 × 1024kg m = 200t = 200000kg

Ek

Ek = Ek = 6.253125 × 1012 JEk 6.3 × 1012 J

Question 5Two 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 1 Satellite 2

T1 = ? T2 = ? m1 =10m m2 =mr1 = r r2= r G = G G = GM=M M=M

v1 = v2 = = = 1

AlternativeSince 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

Question 5Two satellites are in orbit at the same height around the Earth. Satellite 1 has a mass ten times greater than the satellite 2.(b)

TSatellite 1 Satellite 2

T1 = ? T2 = ? m1 =10m m2 =mr1 = r r2= r G = G G = GM=M M=M

T1 T2

= = 1

AlternativeSince 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

Question 5Two 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 1 Satellite 2T1 = ? T2 = ? m1 =10m m2 =mr1 = r r2= r G = G G = GM=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.

Question 5Two 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 1 Satellite 2T1 = ? r1 = r G = G M=M m1 =10m T2 = ? r2= r G = G M=M m2 =m

Ek1 Ek

= = 10

Question 6What is correct units for G in terms of m, s & kg.

a =

= =

= kg-1m3s-2

Question 7Calculate the altitude of a satellite in a geo-stationary orbit. Data: G = 6.67 × 10-11 Nm2kg-2 ME = 6.4 × 1024 kg RE = 6.4 × 106 mr = ? G = 6.67 × 10-11 Nm2kg-2 ME = 6.4 × 1024 kg 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

Question 8Show 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 T is dependent on r and independent m so if two satellites have the same orbital radius they will have the same period.

Question 9 - 2004A 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

Question 9 - 2004A 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 = 7.0 103 ms-1

Question 10 – 2008 Q15The 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.

motion revision powerpoint

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