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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
𝐺𝑀
𝐺𝑀𝑚
force concepts, 𝑔 = 2 and 𝐹𝑔 = 2
𝑟
𝑟
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  10-11 Nm2kg-3 ME = 5.98  1024 kg
g
g
𝐺𝑀
= 2
𝑟
6.67 ×10−11 × 5.98×1024
=
6.37×106 2
g = 9.8299
g = 9.83 Nkg-1
re = 6.37  106m
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
𝐺𝑀𝑚
= 2
𝑟
6.67 ×10−11 × 5.98×1024 × 3
=
8.0×106 2
F = 18.69684
F = 19 N
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
F = mg
F=24
F = 8.0 N
g = 4.0Nkg-1
(b) What is the weight of the object?
W = 8.0 N
Weight = Fgrav
g = 4.0ms-2
(c) Describe the motion of the piece of space junk?
The space junk will be accelerating at 4.0ms-2
towards the Earth
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-1 G = 6.67  10-11 Nm2kg-3,
𝐺𝑀
g= 2
𝑟
𝐺𝑀
2
r =
𝑔
r=
ME = 5.98  1024 kg
𝐺𝑀
𝑔
6.67 ×10−11 × 5.98×1024
4
r=
r = 9.9858  106 m
r  1.0  107 m
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
Saturn
gsaturn = 2.50  10-7 gjupiter = 7.18  10-7
Jupiter
Cassini
(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
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
g=
𝐺𝑀
𝑟2
gr2 = 𝐺𝑀
𝑔𝑟 2
𝐺
=𝑀
7.18 ×10−7 ×
3.9 ×1011
6.67 ×10−11
1.89888  1027 = M
M  1.9  1027 kg
2
=𝑀
r = 4.2  1011 m
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?
𝐺𝑀𝑚
F= 2
𝑟
Here
F1 = 0.45N
r1 = 4.0  107m
G=G
M=M
m=m
There
F2 = ?
r2 = 1.2  108m
G=G
M= M
m=m
F∝
1
𝑟2
Since r × 3, F ×
So F2 =
1
9
1
9
 0.45 = 0.050N
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?
𝐺𝑀𝑚
F= 2
𝑟
Here
F1 = 0.45N
r1 = 4.0  107m
G=G
M=M
m=m
There
F2 = ?
r2 = 1.0  107m
G=G
M= M
m=m
F∝
1
,
4
1
𝑟2
42
12
Since r × F ×
= 16
So F2 = 𝟏𝟔  0.45 = 7.2N
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?
𝐺𝑀𝑚
F= 2
𝑟
Here
F1 = 0.45N
r1 = 4.0  107m
G=G
M=M
m=m
There
F2 = ?
1
r2 = 9.0  106m
F∝ 2
𝑟
G=G
M= M
2
4.0×107
9.0×106
Since r ×
, F×
= 19.75309
m=m
4.0×107
9.0×106 2
So F2 = 19.75309  0.45 = 8.888888  8.9N
Question 5
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
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?
𝐺𝑀𝑚
F= 2
𝑟
Here
F1 = 0.45N
r1 = 4.0  107m
G=G
M=M
m=m
r1 = 4.0 
500km
107m
rE
There
F2 = ?
r2 = 6.37  106 + 500000 = 6870000 A
G=G
1
M= M
F∝ 2
𝑟
m=m
Since r ×
68700000
4.0×107
, F×
4.0×107
r2
2
6870000 2
= 33.90053
So F2 = 33.90053 0.45 = 15.25523  8.9N
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
𝐺𝑀𝑚
𝑟2
𝐺𝑀𝑚
𝑟2 =
𝐹
𝐹=
𝑟=
Here
r1 = 4.0  107m
F1 = 0.45N
G=G
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
𝐺𝑀𝑚
𝐹
There
r2 = ?
F2 = 16.2
G=G
M= M
m=m
𝑟∝
Since F ×
𝟏𝟔.𝟐
𝟎.𝟒𝟓
, r×
𝟎.𝟒𝟓
𝟏𝟔.𝟐
1
𝐹
= 0.16666667
So F2 = 0.1666667 4.0  107
= 666666667  6.7  106m
Question 6
Given 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.
𝑔=
𝐺𝑀
𝑟2
Here
g1 = 10
r1 = r E
G=G
M=M
There
g2 = ?
r2 = 4rE
G=G
M= M
𝑔∝
1
𝑟2
𝟏
Since r × 4 , g ×
𝟏𝟔
g2 = 10 ×
𝟏
𝟏𝟔
= 0.625 Nkg-1
4rE
Question 7
Whenever the height or altitude above the surface of
the Earth is mentioned it is useful to draw a diagram
to be clear about the situation
Given 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.
𝑔=
𝐺𝑀
𝑟2
Here
There
g1 = 10
r1 = rE
G=G
M=M
g2 = ?
r2 = 2rE
G=G
M= M
𝑔∝
1
𝑟2
𝟏
Since r × 2 , g ×
𝟒
𝟏
𝟒
g2 = 10 × = 2.5 Nkg-1
r2 = 2rE
rE
r1 = rE
Question 8
Whenever the height or altitude above the surface of
the Earth is mentioned it is useful to draw a diagram
to be clear about the situation
At 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.
𝐺𝑀
𝑟2
𝐺𝑀
𝑟2 =
𝑔
𝑔=
𝑟=
Here
r1 = rE
g1 = 10
G=G
M=M
4rE
𝐺𝑀
𝑔
rE
There
r2 = ?
g2 = 0.40
G=G
M= M
𝑟∝
5rE
1
𝑔
Since g ×
𝟎.𝟒
𝟏𝟎
, r×
𝟏𝟎
𝟎.𝟒
=5
r2 = 5 rE
So the point will be 4 rE above the surface
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?
𝐺𝑀
𝑔= 2
𝑟
Here
g1 = g
r1 = r
G=G
M=M
There
g2 = ?
r2 = 3r
G=G
M= M
𝑔∝
1
𝑟2
𝟏
Since r × 3 , g ×
𝟗
𝟏 𝒈
so g2 = g × =
𝟗 𝟗
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
F = mg
𝒈
F=m
F=
𝒎𝒈
𝟗
𝟗
g =
𝒈
𝟗
Question 10
If the gravitational force on an object r metres from the centre of
𝑟
a planet is F, what would be the Weight of the object metres
10
from the centre of the planet?
F=
Here
F1 = F
𝐺𝑀𝑚
𝑟2
There
F2 = ?
𝑟
10
r1 = r
r2 =
G=G
M=M
m=m
G=G
M= M
m=m
F∝
𝟏
,
𝟏𝟎
1
𝑟2
Since r ×
F × 100
So F2 = 100  F = 100N
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
𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑙 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑢𝑛 𝑜𝑛 𝐸𝑟𝑖𝑠
𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑙 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑢𝑛 𝑜𝑛 𝑃𝑙𝑢𝑡𝑜
F=
𝐺𝑀𝑚
𝑟2
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 =
𝐺𝑀𝑚
6.0×1012
𝐹𝐸
𝐺𝑀𝑚
=
𝐹𝑃
10.5×1012 2

FE =
2
6.0×1012
𝐺𝑀𝑚
2
=
6.0×1012
2
10.5×1012 2
𝐺𝑀𝑚
10.5×1012
2
= 0.32653  0.326
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
𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑙 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑢𝑛 𝑜𝑛 𝐸𝑟𝑖𝑠
𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑙 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑢𝑛 𝑜𝑛 𝑃𝑙𝑢𝑡𝑜
Here (Pluto)
FP = F
rP = 6 × 1012
𝐹=
Since r ×
𝐺𝑀𝑚
𝑟2
10.5×1012
6×1012
There (Eris)
FE = ?
rE = 10.5 × 1012
so
𝐹∝
1
𝑟2
= 1.75
F×
1
1.752
= 0.3265
FE = 0.3265 F
𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑙 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑢𝑛 𝑜𝑛 𝐸𝑟𝑖𝑠
𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑙 𝑎𝑡𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑢𝑛 𝑜𝑛 𝑃𝑙𝑢𝑡𝑜
=
𝐹𝐸
𝐹𝑃
=
0.3265𝐹
𝐹
= 0.3265
2013 Motion Outcome Dot Point 10
Weight & Weightlessness
• apply the concepts of weight (W=mg), apparent
weight (reaction force, N), weightlessness
(W=0) and apparent weightlessness (N=0)
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
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.
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.
Question 3
Whenever the height or altitude above the surface of
the Earth is mentioned it is useful to draw a diagram
to be clear about the situation
What would be the weight of a 20kg object at rE above the
surface?
𝐹=
𝐺𝑀𝑚
𝑟2
Here
There
F1 = 200
r1 = r E
G=G
M=M
m=m
F2 = ?
r2 = 2rE
G=G
M= M
m=m
F∝
1
𝑟2
𝟏
Since r × 2 , g ×
𝟒
𝟏
𝟒
F2 = 200 × = 50N
r2 = 2rE
rE
r1 = rE
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
𝐺𝑀𝑚
𝐹= 2
𝑟
6.67×10−11 × 5.98×1024 ×3.04×105
F=
(6.72×106 )2
F = 2685109
F ≈ 2.69 × 106N
Question 5 – 2009 Q7
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?
+
N
Karen
60kg
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 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.
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?
Fnet= N – W
ma = N – 600
60 × -2 = N – 600
-120 = N – 600
480 = N
So apparent Weight = 480N
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
weight)
+ reaction (apparent
N
of 480 N.
The main errors students made
involved mixing up positive
Karen and
negative signs for the60kg
forces acting.
Some students were confused about
actual forces and the net force, while
others introduced a net force
N =of60mv×2/R.
10
= 600N
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 7 – 2009 Q9
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.
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.
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.
2013 Motion Outcome Dot Point 11
Satellite Motion
• model satellite motion (artificial, moon,
planet) as uniform circular orbital motion
𝑣2
4𝜋𝑟 2
(𝑎 = = 2 )
𝑟
𝑇
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-1 r = 6720km = 6.37 × 106 m
𝟒𝝅𝟐 𝒓
g= 𝟐
𝑻
𝟒𝝅𝟐 𝒓
2
T =
𝒈
T=
𝟒𝝅𝟐 𝒓
𝒈
T=
𝟒𝝅𝟐 ×6.72 ×106
𝟖.𝟗
T = 𝟐𝟗𝟖𝟎𝟖𝟒𝟐𝟑
T = 5459.709s
T= 90.9951
T = 91 minutes
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
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 = 𝒈𝒓
v = 𝟖. 𝟗 × 6.72 × 106
v = 7733.56 ms-1
OR
v=?
r = 6720km = 6.37 × 106 m
T = 5459.709s
v=
v=
𝟐𝝅𝒓
𝑻
𝟐𝝅 × 6.72 ×106
90.9951
v = 7733.56 ms-1
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.
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
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
G = 6.67 × 10-11 Nm2kg-2
M=?
Rmoon orbit = 3.8 × 108 m
Rmoon orbit = 3.8 × 108 m
G = 6.67 × 10-11 Nm2kg-2
T = 28  24  3600 = 2419200 A
M=
𝐺𝑀 4𝜋2 𝑟
= 2
2
𝑟
𝑇
4𝜋2 𝑟 3
M=
𝐺𝑇 2
3
4𝜋2 × 3.8 ×108
6.67 ×10−11 × 24192002
M = 5.549338  1024 kg
M  5.5  1024 kg
Question 3
Whenever the height or altitude above the surface of
the Earth is mentioned it is useful to draw a diagram
to be clear about the situation
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
B
v=
𝐺𝑀
𝑟
v=
6.67 ×10−11 × 6.0 ×1024
6860000
v = 7637.944ms-1
v = 7.7  103ms-1
360km
rE
r
Can also get an answer through the
formulae strings
𝑚𝑣 2
𝑟
v2 =
v=
𝐺𝑀𝑚
𝑟2
𝐺𝑀
𝑟
𝐺𝑀
𝑟
=
(or
𝑣2
𝑟
=
𝐺𝑀
)
𝑟2
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  106 G = 6.67 × 10-11 ME = 6.0 × 1024kg m = 200t
= 200000kg
Ek=
Ek =
𝐺𝑀𝑚
2𝑟
6.67 ×10−11 ×6.0 ×1024 ×200000
2 × 6.4 ×106
Ek = 6.253125 × 1012 J
Ek  6.3 × 1012 J
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)
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟏
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟐
𝐺𝑀
𝑟
v=
Satellite 1
Satellite 2
T1 = ?
m1 =10m
r1 = r
G=G
M=M
v1 =
T2 = ?
m2 =m
r2= r
G=G
M=M
𝐺𝑀
𝑟
𝑣1
=
𝑣2
v2 =
𝐺𝑀
𝑟

𝑟
𝐺𝑀
=1
𝐺𝑀
𝑟
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
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟏
𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟐
=1
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.
𝒑𝒆𝒓𝒊𝒐𝒅 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟏
𝒑𝒆𝒓𝒊𝒐𝒅 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟐
(b)
T=
4𝜋2 𝑟 3
𝐺𝑀
Satellite 1
Satellite 2
T1 = ?
m1 =10m
r1 = r
G=G
M=M
T1 =
T2 = ?
m2 =m
r2= r
G=G
M=M
4𝜋2 𝑟 3
𝐺𝑀
𝑇1
𝑇2
==
T2 =
4𝜋2 𝑟 3
𝐺𝑀

𝒑𝒆𝒓𝒊𝒐𝒅 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟏
𝒑𝒆𝒓𝒊𝒐𝒅 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟐
4𝜋2 𝑟 3
𝐺𝑀
𝐺𝑀
4𝜋2 𝑟 3
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
=1
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.
𝒑𝒆𝒓𝒊𝒐𝒅 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟏
𝒑𝒆𝒓𝒊𝒐𝒅 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟐
(c)
Ek=
𝐺𝑀𝑚
2𝑟
Satellite 1
Satellite 2
T1 = ?
m1 =10m
r1 = r
G=G
M=M
T2 = ?
m2 =m
r2= r
G=G
M=M
𝐺𝑀10𝑚
𝐺𝑀𝑚
Ek=
2𝑟
2𝑟
𝐸𝑘1
𝐺𝑀10𝑚
2𝑟
==

= 10
𝐸𝑘2
2𝑟
𝐺𝑀𝑚
Ek1 =
Alternative
Ek∝ 𝑚
𝐸𝑘1
𝐸𝑘2
=
𝑚1
𝑚2
=
10𝑚
𝑚
= 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 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 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟏
𝑬𝒌 𝒐𝒇 𝒔𝒂𝒕𝒆𝒍𝒍𝒊𝒕𝒆 𝟐
Ek=
𝐺𝑀𝑚
2𝑟
Satellite 1
T1 = ?
r1 = r G = G
Ek1 =
Satellite 2
M=M m1 =10m
𝐺𝑀10𝑚
2𝑟
𝐸𝑘1
𝐸𝑘2
T2 = ?
r2= r G = G M=M m2 =m
Ek=
==
𝐺𝑀10𝑚

2𝑟
2𝑟
𝐺𝑀𝑚
= 10
𝐺𝑀𝑚
2𝑟
Question 6
What is correct units for G in terms of m, s & kg.
a=
𝐺𝑀
𝑟2
𝒂𝒓𝟐
𝑴
=𝑮
𝑮 =
𝑮 =
𝒂𝒓𝟐
𝑴
𝒎𝒔−𝟐 𝒎 𝟐
𝒌𝒈
𝑮 = kg-1m3s-2
Whenever the height or altitude above 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 × 1024 kg
RE = 6.4 × 106 m
r=?
G = 6.67 × 10-11 Nm2kg-2 ME = 6.4 × 1024 kg
T = 24 h = 24  3600 = 86400 A
𝑟=
𝑟=
3
𝐺𝑀𝑇 2
4𝜋2
3
6.67 ×10−11 ×6.0 ×1024 ×864002
4𝜋2
3
𝑟 = 7.567367 × 1022
r = 2.75088  1011
r = 2.8  1011m
RE = 6.4 × 106 m
B
Altitude = 6.4 × 106 + 2.75088  1011 = 2.75094  1011  2.8  1011m
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 =
𝐺𝑀
𝑟
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=
4𝜋2 𝑟 3
𝐺𝑀
is
dependent on r and independent m so if two satellites have
the same orbital radius they will have the same period.
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=
3
𝑟=
3
3
𝐺𝑀𝑇 2
4𝜋2
6.67 × 10−11 × 5.98 × 1024 × 7200
4𝜋 2
2
r = 5.23760 × 1020 A
r = 8060786
So the height above surface = 8060786 – 6.37 × 106 = 1690786  1.7 × 106m
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
2𝜋𝑟
𝑣=
𝑇
2 × 𝜋 × 8060786
𝑣=
7200
v = 7034.362 ms-1
v = 7.0  103 ms-1
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.