Download IJC2013 Tut 7 Gravitational Field Tutorial (Q and A)

2013 H2Physics JC1
Name: …………………………………………….…
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Date: __ May 2013
Tutorial 7 : Gravitational Field
Gravitational constant G = 6.67  10-11 N m2 kg-2
Gravitational potential  = - GM/r
(iii) If m2 is doubled,
𝐹′ =
Gravitational Force and Field Strength
1.
F
m1
𝐹′ =
F2
m2
(i)
The mass m2 is doubled.
(ii)
The mass m1 is doubled.
𝐹′ =
(5)  (1)
(i)
(ii)
(iii)
(iv)
(v)
F1
2F
2F
4F
¼F
16F
F2
2F
2F
4F
¼F
16F
…… (Eqn 1)
1
4
= (2 × 2) ÷ ( ) = 16
http://www.youtube.com/watch?v=uGBANgbRkws
Note that the forces acting on the bodies are
action-reaction pair and so must have the
same magnitude but opposite in direction.
Let |F1|=|F2|= F
If m2 is doubled,
𝐺𝑚1 2𝑚2
…… (Eqn 2)
𝑟2
𝐹′
=2
𝐹
…… (Eqn 5)
The Earth and Moon actually orbit about their
barycentre (common centre of mass), which
lies about 4600 km from Earth's centre (about
three quarters of the Earth's radius).
Newton’s Universal Gravitational Law gives
(2)  (1)
1
2
( 𝑟)2
(b) We now apply the law to the Earth-Moon
system where m1 is taken to be the mass of
the Earth and m2 to be that of the Moon.
Explain why it is wrong to regard the
separation x as the radius of orbit of the Moon.
Comment on the values of F1 and F2 with respect
to the changes to the two masses and their
separation.
𝐹′ =
𝐹′
𝐹
𝐺(2𝑚1 )(2𝑚2)
The gravitation force is directly proportional
to the product of the two masses and inversely
proportional to the square of their separation.
Write down your values for the above in the table
below.
(i)
 F’ = ¼ F
F1 and F2 are always equal (and opposite) as
they are action and reaction pair.
Both the masses m1 and m2 are doubled
and the separation x is halved.
𝐺𝑚1 𝑚2
𝑟2
1
=4
 F’ = 16 F
(iii) Both masses m1 and m2 are doubled.
(iv) The separation x is doubled.
𝐹=
𝐺𝑚1 𝑚2
…… (Eqn 4)
(2𝑟)2
(iv) If masses are double and x halved,
What are the magnitudes of F1 and F2 in
terms of F for each of the following
conditions treated separately?
(v)
𝐹′
𝐹
(4)  (1)
x
(a)
 F’ = 4F
(iv) If x is doubled,
Gm1m2
x2
F1
𝐹′
=4
𝐹
(3)  (1)
Based on Newton’s universal gravitational law, the
force F between two point masses m1 and m2
separated by a distance x is given by
𝐺(2𝑚1 )(2𝑚2 )
…… (Eqn 3)
𝑟2
 F’ = 2F
(ii) Similarly when m1 is doubled,
 F’ = 2F
1
2. [N10/I/15]
The neutral point in the gravitational field between
the Sun, the Earth and the Moon is the point at
which the resultant gravitational field due to the
three bodies is zero. The mass of the Earth is about
80 times the mass of the Moon. At what position is it
possible for the neutral point to be?
(The diagram is not drawn to scale.)
4. [Hutching Pg 126]
The mass of the Earth is 6.0 × 1024 𝑘𝑔 and the mass
of the Moon is 7.4 × 1022 𝑘𝑔 . The distance between
their centres is 3.8 × 108 𝑚.
The position of the gravitational neutral point X, lies
between the Earth and the Moon.
Show that the gravitational neutral point X divides
the distance between the Earth and the Moon
roughly in the ratio 9:1 being closer to the Moon.
Calculate the distance between the neutral point X
and the centre of the Moon.
[3.8×10 7 m]
Neutral
Point, X
ME
R
Answer : B
If we consider the Earth-Moon system only,
the neutral point would be at point A.
Thus,
W’
h
R
W
𝑊=
𝑅2
𝑅
𝑟
=
𝐺𝑀𝑀
𝑟2
6.0×1024
𝑀
= √ 𝑀𝑀 = √7.4×1022=9.00
𝐸
𝑅 = 9.0𝑟
D 0.96 W
since R=d-r
𝑑−𝑟
= 9.0
𝑟
𝐺𝑀𝑚
𝑟2
𝑊∝
𝐺𝑀𝐸
𝑅 2 𝑀𝑀
=
𝑟 2 𝑀𝐸
3. [N92/I/8]
On the ground, the gravitational force on a satellite
is W. What is the gravitational force on the satellite,
𝑅
when at a height of where R is the radius of the
50
Earth?
C 0.98 W
Moon
At the neutral point X, the net gravitational force of
gravitational field strength at that point equals zero.
Learning point:
The vector sum of the gravitational field
strengths at the neutral point is zero.
The neutral point lies always closer to the
smaller mass.
B 1.02 W
r
Let R = distance between the neutral point and centre of
Earth
Let r = distance between the neutral point and centre of
Moon.
However, when we consider the
gravitational attractive force of the Sun,
the neutral point is shifted slightly to B.
A 1.04 W
d
Earth
MM
𝑑
− 1 = 9.0
𝑟
1
𝑟2
2
𝑊′
𝑅
1 2
=(
)
𝑅) = (
𝑊
1.02
𝑅 + 50
𝑑
= 10.0
𝑟
𝑊 ′ = 0.96𝑊
𝑟=
Answer : D
Learning point
The variation of the weight with distance from
Earth’s centre obeys the inverse-square law.
Even at a great height above the Earth, the
weight dropped only 4%. Does this contradict the
weightlessness experienced by the astronaut at
this altitude?
𝑑
3.8 × 108
=
= 3.8 × 107 𝑚
10.0
10.0
Learning point
The net force is zero at the neutral point X as the
grav.force due to the Earth is equal and opposite
that due to the Moon. This is the consequence of
Newton’s 1st Law.
The neutral point is always closer to the smaller mass.
2
(ii) . N kg-1 is equivalent to m s-2 in base units.
5. [N94/III/Q1part / modified]
(a) The gravitational field strength of the Earth at
its surface is 9.81 N kg-1.
Show that
(i) the acceleration of free fall at the surface of
the Earth is 9.81 m s-2.
(ii) N kg-1 is equivalent to m s-2 in base units.
[4]
[𝑔] =
(b)
(i) This question serves to test your concept of g.
The gravitational field strength g is defined as
the force acting per unit mass.
Thus, on the Earth’s surface 𝑔 =
𝐹
𝑚
[𝐹] [𝑚][𝑎]
=
= 𝑚 𝑠 −2
[𝑚]
[𝑚]
Use the value for the gravitational field
strength of Earth quoted in (a), together with
the value of G, the gravitational constant, and
radius of the Earth (6.38 × 106 m), to calculate
the mass of the Earth.
[4]
[5.99×1024 kg]
𝑔=
(9.81 N kg-1)
From Newton’s 2nd law, we have F=ma
9.81 =
If the force F on an object of mass m is due to the
gravitational force W of the Earth on the object,
we then have
(c)
where a = acceleration of free fall at the Earth’s
surface ( 9.81 m s-2)
𝑚𝑎
𝑚
= 𝑎 = 9.81
1
𝑟2
𝑔′
𝑟 2
𝑟 2
=( ) =(
)
𝑔
𝑟′
𝑟+ℎ
Examiner’s Report
𝑔′
6.38 × 106
=(
)
9.81
6.50 × 106
Too little explanation. It was strange how few
candidates started with the definition of
gravitational field strength as the force acting on
unit mass. Many simply stated F=mg without stating
what they meant, and it was impt here to distinguish
carefully between the symbols ‘a’ and ‘g’ as ‘a’ could
have been just any generalized acceleration in force =
mass×acceleration or it could have been acceleration
of free fall.
2
𝑔′ = 9.45 𝑁 𝑘𝑔−1
Examiner’s Report
Parts (b) and (c) were done well.
(d)
It was surprising to find candidates quoting g=GM/r2 as
the definition of the gravitational field strength. They
are not good at distinguishing between a defining
equation for a quantity and an equation which contains
that quantity. All too often this is what was seen which
is NOT what is wanted.
Explain briefly why an astronaut in a satellite
orbiting at this altitude may be described as
weightless.
[2]
Both the satellite and the astronaut are falling
freely with the same acceleration. The only force
acting on the astronaut is his gravitational force
due to the Earth. As such, his apparent weight is
zero.
𝐺𝑀𝑚
𝑟2
𝐺𝑀
∴𝑔= 2
𝑟
𝑚𝑔 =
Examiner’s Report
It was amazing the number, who just having worked
out that g has fallen from 9.81 N kg-1 to 9.45 N kg-1,
followed this up by saying that the force on the
astronaut was negligibly small when in orbit. The idea
that both the astronaut and the satellite are falling
freely with the same acceleration was very rarely stated.
More frequent unfortunately this was often said to be
because the gravitational force was balanced by the
centripetal force.
𝐹 = 𝑚𝑎
F = mg
𝐺𝑀𝑚
= 9.8𝑚
𝑟2
∴ 𝑔 = 9.8 = 𝑎
Calculate the Earth’s gravitational field
strength at a height of 0.12×106 m above the
Earth’s surface.
[3]
[9.45 N kg-1]
Since, 𝑔 ∝
Thus, the gravitational field strength and the
acceleration due to free fall on the Earth’s
surface is numerically equal.
∴𝐹=
(6.67 × 10−11 )(𝑀)
(6.38 × 106 )2
𝑀 = 5.99 × 1024 𝑘𝑔
W= F = ma
Substituting, we have 𝑔 =
𝐺𝑀
𝑟2
Shown
3
(e)
The satellite is said to be in a state of free fall
to the Earth. What is the meant by the term
free fall? How can the height of the satellite
stay constant if it is in free fall?
(b) Use the expression in (a) to calculate the gain
in the potential energy of the satellite of mass
3000 kg between its launch and when it is at a
height of 0.12×106 m above the Earth’s surface.
[3500 MJ] [4]
(This question was added to the actual one)
‘Free fall’ refers to condition when the object is
subjected to the gravitational force only. There
are no other forces acting on the satellite other
than its own weight.
Initial distance of the satellite from Earth’s
centre = radius of Earth = r = 6.38 × 106 m
Final distance of satellite from Earth’s centre
=r+h
= (6.38 × 106) + (0.12×106) = 6.50 × 106 m
‘Free fall’ should not be taken to be literally an object
‘falling freely’. For example, when an object is thrown up, it
is still considered a case of ‘free fall’.
Using gravitational potential
𝐺𝑀
𝜙=−
The satellite has a tangential speed that makes it
fall around the Earth rather than directly toward
the Earth. The satellite’s curved path of trajectory is
parallel to the Earth’s curved surface, thus the
altitude remains constant.
𝑟
Change in 𝜙 = Final 𝜙 – Initial 𝜙
Δ
Gravitational Potential and GPE
(−
6. [continued from Q7-N94/III/Q1part]
=
𝐺𝑀
𝑟
)=(
=𝐺𝑀(
(a) The value of the gravitational potential  at a
point in the Earth’s field is given by the
equation
𝐺𝑀
𝜙=−
𝑟
𝐺𝑀
𝑟
1
𝜙𝑓 − 𝜙𝑖 = (−
)−(
𝐺𝑀
𝑟+ℎ
)
1
× 1024 ) [
Δ = 1.156 × 106 J kg-1
Since Δ =
)−
− )
𝑟+ℎ
𝑟
−11 )(5.99
=(6.67 × 10
where M is the mass of the Earth and r is the
distance of the point from the centre of the
Earth. (r is greater than the radius of the Earth)
𝐺𝑀
𝑟+ℎ
1
6.38×106
−
1
6.5×106
]
(positive means gain)
∆𝑈
𝑚
ΔU = m (Δ)
= (3000) (1.156 × 106) = 3.47×109 J
= 3500 MJ
(i) Explain what is mean by the term
gravitational potential,
(ii) Explain why the potential has a negative
value.
The satellite gains GPE by 3500 MJ
[3]
(i) Gravitational potential at a point in the
gravitational field is the work done per
unit mass by an external agent in
bringing the mass from infinity to the
point.
Examiner’s Report
Some confused potential and potential energy. 1.1 MJ or
1.1 MJ/kg were very common final answers.
Some substitute 3000 kg into an equation involving M,
when the mass which was required here was the mass of
the Earth, not the mass of the satellite.
(ii) The potential at infinity is zero and so
at a distance less than that the value for
the potential is negative.
The potential by its definition has a
negative value because the work done
by the external agent is negative as the
displacement is opposite to the applied
force by the external agent.
The gravitational force is attractive
and so negative work must be done by
the external agent when the object is
brought closer to the Earth.
Comments
Incomplete answers. Some forgot that GPE is taken
to be zero at infinity and that gravitational forces are
attractive. Many students simply referred to ‘negative
work done in moving from infinity, thus potential is
negative’, without making clear the agent doing the
work and meaning of negative work.
4
7.
A comet of mass m is moving around the Earth in
an elliptical orbit. It is moving with a speed v when
it is at P at a distance r from the Earth’s centre.
Mass of the Earth is M.
v
Earth
Mass ME
Sun
Mass
Ms
Satellite mass
m
r
P
Comet
R
Not to scale
The satellite rotates in a circle around the Sun
once a year and therefore moves around the Sun
with the Earth, both have the same angular
velocity . Which force = mass × acceleration
equation applies for the satellite?
r
Earth
What is the total energy of the comet at P?
A GMm  1 mv 2
r
2
B  GM  1 mv 2
r
2
C  GMm
r
D  GMm  1 mv 2
r
2
Answer : D
A
𝐺𝑀𝑆 𝑚
B
𝐺𝑀𝐸 𝑚
C
𝐺𝑀𝐸 𝑚
D
𝐺𝑀𝑆 𝑚
Total Energy = GPE + KE
𝑅2
𝑅2
𝑟2
𝑅2
= 𝑚 × (𝑅𝜔2 )
= 𝑚 × (𝑟𝜔2 )
−
−
𝐺𝑀𝑆 𝑚
= 𝑚 × (𝑅𝜔2 )
𝑅2
𝐺𝑀𝐸 𝑚
= 𝑚 × (𝑅𝜔2 )
𝑟2
Satellites and their orbits
8. [N2012/I/14]
An unpowered artificial (solar) satellite called SOHO
has been placed in a stable orbit around the Sun at a
distance of 0.99R from the Sun, where R is the
orbital radius of the Earth.
The satellite is always on the radial line from the
Sun to the Earth so that it has a period of 1.0 year,
the same as the Earth.
Which statement relating to the satellite helps to
explain how it is possible for its period to be 1.0
year?
Sun
10. [N2011/III/6 part ]
(a) The Earth may be assumed to be an isolated
uniform sphere with its mass M concentrated
at its centre. A satellite of mass m orbits the
Earth in a circular path of radius R.
Earth
R
For the satellite in its orbit, show that
satellite
(i)
Not to scale
Its kinetic energy EK is given by
𝐸K =
A The gravitational field of the Sun at the
satellite is reduced by the presence of the
Earth.
B The gravitation force of the Earth on the
satellite balances the gravitational force of the
Sun on the satellite.
C The mass of the satellite is much less than the
mass of the Earth.
D The resultant gravitational force on the
satellite is less than the gravitational force the
Sun exerts on it.
‘A’ is not the answer because the g of the sun is
unchanged but the resultant g at the point is changed
due to the presence of the Earth.
Answer : D
See Question 9 for follow-up
𝐺𝑀𝑚
2𝑅
Solution
When the satellite is in circular orbit (radius R)
round the Earth, the gravitational force provides
the centripetal force for the satellite.
𝐹𝐺= 𝐹𝐶
(try not to write this short form; you must write the
above statement)
𝐺𝑀𝑚 𝑚𝑣 2
=
𝑅2
𝑅
Thus,
𝑚𝑣 2 =
𝐺𝑀𝑚
𝑅
1
9.
As kinetic energy is given by 𝑚𝑣 2
[N2011/I/16]
The diagram below shows a solar satellites, mass
m, positioned directly between the Earth, mass ME
and the Sun, mass MS. The satellite is a distance r
from the Earth and a distance R from the Sun.
2
1 𝐺𝑀𝑚
KE of the satellite = (
2
5
𝑅
)=
𝐺𝑀𝑚
2𝑅
Examiner’s Comment
The derivation frequently lacked explanation. In deriving, it
is essential to explain the physics of the situation. A
statement of “the gravitational force of attraction
provides the force necessary for circular motion” is
required.
(ii) Determine quantitatively whether the
satellite could be in a geostationary orbit.
[4]
𝑣 = 𝑅𝜔
2𝜋
𝑣 = 𝑅( )
𝑇
(ii) Its total energy ET is given by
2𝜋
7455 = (7.2 × 106 ) ( )
𝐺𝑀𝑚
𝐸T = −
2𝑅
𝑇
T = 6068 s = 1.7 h
The satellite’s total energy is the sum of its kinetic
and gravitational potential energies.
Since a satellite in geostationary orbit must have a
period of 24 h, the satellite cannot be in
geostationary orbit.
Total Energy = Ep + Ek
=−
𝐺𝑀𝑚
𝐺𝑀𝑚
𝑅
2𝑅
+
𝐺𝑀𝑚
=−
2𝑅
=
Comments
There were several different approaches to this problem.
Most candidates were successful in making a comparison
of either the period or linear speed or the angular speed or
the radius of the orbit with that of a geostationary satellite.
𝐺𝑀𝑚 1
𝑅
( − 1)
2
Comments
Again, there was a marked tendency to write down
algebraic expressions without explaining the physics
concepts.
(b)
(c) The satellite in (b) gradually loses energy due
to small resistive forces. Suggest why many
such satellite eventually ‘burn up’ in the
Earth’s atmosphere.
[4]
The Earth has radius 6.4×106 m and
mass 6.0×1024 kg.
A satellite has mass 850 kg and orbital radius
7.2×106 m.
(i)
When the satellite loses energy, its orbital
radius will progressively decrease as it
spirals towards the Earth.
As it loses altitude, the progressive loss of
its gravitational potential energy will
result in a corresponding gain in its
kinetic energy.
Use an expression in (a) to determine the
speed of the satellite.
[7.5×103 m s-1] [3]
The kinetic energy of the satellite is given
by
𝐸𝐾 =
1
𝐺𝑀𝑚
𝑚𝑣 2 =
2
2𝑅
𝑣2 =
𝑣=√
On reentering the Earth’s atmosphere, the
the drag (or resistive) force due to the
atmosphere will increase even more as
it descends with increasing speed. This in
turn causes the satellite to lose even more
energy as the frictional force between the
air and satellite will cause the satellite to
heat up.
The rate of conversion of energy of the
satellite to thermal energy will be so
great that it will cause the satellite to
burn up.
𝐺𝑀
𝑅
𝐺𝑀
(6.67 × 10−11 )(6.0 × 1024 )
=√
𝑅
7.2 × 106
v = 7455 m s-1 = 7500 m s-1 (2 s.f.)
Comments
Many did not clearly state that the decreasing total energy
of the satellite leads to a reduction in the orbit’s radius.
Those who realise that the kinetic energy of the satellite
increase usually did not continue to argue that increasing
resistive forces give rise to an increasing rate of conversion
of energy of the satellite to thermal energy.
Comments
Some unnecessarily work out the calculation of kinetic
energy ½mv2. The most common error was to confuse the
radius of the satellite’s orbit with its altitude.
6
11. [N07/3/5] part
Many systems, such as the Global Positioning
System (GPS), use several satellites in low obits
that pass over the Earth’s poles. Suggest two
advantages of these low polar orbits and two
advantages of geostationary orbits.
[4]
(a) (i) Derive an expression, in terms of M, m
and R, for the kinetic energy of the
satellite. Explain your working.
[2]
The gravitational force provides the centripetal
force, Fc = mac
Advantages of low polar orbit
(1) Because of its lower altitude, the satellite is
capable of capturing close-up images of the
Earth’s surface with greater resolution (ie
with more detailed). However, because the
radius of rotation of the polar orbit is smaller,
these satellites would need to circle round the
globe several times a day (Period is about 2
hr). To ensure continuity in its function, a
chain or series of satellites need to be used.
GMm mv 2

r
r2
GMm
r
GMm
KE  ½mv 2 
2r
mv 2 
Examiner’s Comment
The vast majority of candidates scored full marks for this
part. Most candidates were familiar with the expression
mv2/r = GMm/r2.
(2) Because
of
its
low
altitude,
telecommunication signals are relayed to
other parts of the globe via a ‘chain’ of low
orbit satellites resulting in a reduction in the
delay between the time of transmission and
reception of the signal compared to those
relayed by geostationary satellite which lies
much further from the Earth’s surface.
(ii) Show that, for the satellite in orbit, the
ratio
gravitational potential energy of satellite
kinetic energy of satellite
is equal to -2.
[1]
(ii)
Advantages of geostationary orbit
1. A geostationary satellite is able to relay
telecommunication signal continuously
(24-7) between the transmitter and the
reception since it is always positioned at
the same spot with respect to the Earth’s
surface.
2. As it is situated further away from the
Earth, it spans over a wider surface area of
the Earth. Thus, it can take wider angle of
view of the Earth’s surface compared to
the low orbit satellite which can view only
smaller sections of the Earth’s surface for
short periods of time.
GMm
r
GMm

GPE
r
 2

GMm
KE
2r
GPE  
Examiner’s Comment
This part was generally well answered
(b) The variation with orbital radius R of the
gravitational potential energy of the satellite is
shown in Fig 12.1
(i) On Fig 12.1, draw the variation with orbital
radius of the kinetic energy of the satellite.
Your line should extend from R = 1.5 RP to
R = 4 RP
[2]
Examiner’s Comment
Very few candidates scored full marks here. Common
misconception included either the need to supply energy to
keep the satellites moving in orbit or that the geostationary
satellite is in outer space. Many thought that the time to
transmit a message would be shorter, rather than any
delay between transmission and receipt. Very few realized
that polar satellites are, in general, required to make a
number of orbits in order to complete their functions.
Since U / K = -2
K = -U/2
for all the points.
Shape of KE graph for satellite
K is positive while U is negative.
K is minus half the value of U.
Examiner’s Comment
A significant number of candidates were not able to draw
the kinetic energy graph with the precision expected.
However, the majority drew a smooth curve in the correct
region and scored one mark. A minority drew the kinetic
energy curve with the same values as the potential energy
curve despite the –1/2 factor being given in (a).
12. [N08/2/3]
A satellite of mass m orbits a planet of mass M and
radius RP. The radius of the orbit is R. The satellite
and the planet may be considered to be point
masses with their masses concentrated at their
centres. They may be assumed to be isolated in
space.
7
(ii) When R = 4Rp,
KE = 1.25 × 109 J
½ (1600) v2 = 1.25 × 109
v = 1.250 × 103 m s-1
When R = 2Rp,
KE = 2.50 × 109 J
½ (1600) v’2 = 2.50 × 109
v' = 1.717 × 103 m s-1
Change in orbital speed, Δv
= v’ – v = +517 m s-1
There is an increased in speed moving from
4Rp – 2Rp.
Examiner’s Comment
This part proved to be a very good discriminator. The
majority were able to determine the correct values for the
kinetic energy at the two points. There were some who
then forgot the 109 factor.
A significant majority then went on to calculate the change
in speed using the expression ½ mΔv 2 = ΔEk giving a
maximum mark of 3.
A significant minority of candidates scored zero marks as
they equated the change in potential energy with the
change in kinetic energy.
The weaker candidates made no attempt to answer the
question or tried to solve various equations to derive the
change in speed rather than use energy values from the
graph.
Fig. 10.1
(ii) The mass m of the satellite is 1600 kg. The
radius of the orbit of the satellite is
changed from R = 4 RP to R = 2 RP.
Phew! The End
Use Fig 12.1 to determine the change in
orbital speed of the satellite.
[5]
[+517 m s-1]
8