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Ver 1.0 © Chua Kah Hean xmphysics 1
XMLECTURE
07 GRAVITATION NO DEFINITIONS. JUST PHYSICS.
7.1 Gravitational Force ...................................................................................................................... 2
7.1.1 Satellite Motion ..................................................................................................................... 3
7.1.2 Geostationary Orbit .............................................................................................................. 4
7.1.3 Comparisons of Orbits .......................................................................................................... 5
7.2 Gravitational Field Strength ......................................................................................................... 7
7.2.1 Gravitational Field of a Point Mass ....................................................................................... 7
7.2.2 Earth’s Gravitational Field Strength ...................................................................................... 9
7.2.3 True g and Apparent g ........................................................................................................ 11
7.3 Gravitational Potential and Energy ............................................................................................ 13
7.3.1 Escape Velocity .................................................................................................................. 14
7.4 Gravitational Potential ............................................................................................................... 15
7.4.1 Gravitational Potential of Point Mass .................................................................................. 16
7.5 Potential Energy Gradient and Potential Gradient ..................................................................... 18
7.6 PCOE Problems ....................................................................................................................... 20
7.7 Energy of Satellites ................................................................................................................... 22
Appendix A: Space Travel .............................................................................................................. 24
Online resources are provided at https://xmphysics.com/grav
Ver 1.0 © Chua Kah Hean xmphysics 2
7.1 Gravitational Force
According to fake news, an apple dropped on Isaac Newton’s head, and out came the universal law
of gravitation: two point masses M1 and M2, separated by a distance d, exert a mutual gravitational
force of attraction Fg on each other.
1 2
2g
M MF G
d
where 11 2 26.67 10 N kg mG is called the gravitational constant.
Worked Example 1
Calculate the gravitational attraction between the two bowling balls shown below.
Solution
1 2
2
11
2
8
3.6 7.2(6.67 10 )
0.22
3.57 10 N
g
M MF G
d
A few things to note:
Non-point masses can often be treated as point masses at their centres of masses. So d is the
centre-to-centre distance.
1 2
2g
M MF G
d is consistent with N3L. As an action-reaction pair, the two balls, despite their different
masses, exert the same magnitude of gravitational pull on each other.
(watch video at xmphysics.com)
3.6 kg 7.2 kg
radius = 11 cm
3.6 kg 7.2 kg
22 cm
Ver 1.0 © Chua Kah Hean xmphysics 3
Thanks to 11 2 26.67 10 N kg mG , gravitational forces between everyday objects of ordinary
masses are negligibly small. On the other hand, gravitational forces involving massive bodies
such as planets and stars are large and of huge significance.
7.1.1 Satellite Motion
An object orbits around a massive body in circular motion if the gravitational force Fg matches the
required centripetal force 2
c
vF m
r .
Worked Example 2
Ignoring air resistance, calculate the required speed v0 for an object to go into orbit just above the
Earth’s surface.
Earth’s radius 6370 km
Earth’s mass 245.97 10 kg
watch video at xmphysics.com
Solution
Gravitational pull provides the required centripetal force.
2
2
11 24
3
1
(6.67 10 )(5.97 10 )
6370 10
7910 m s
g cF F
Mm vG m
rr
GMv
r
vo
Ver 1.0 © Chua Kah Hean xmphysics 4
7.1.2 Geostationary Orbit
see animation at xmphysics.com
If a satellite were to appear stationary in the sky, it must “rotate” at the same rate as the Earth. This
is achieved by parking the satellite at an altitude of 36,000 km. This so-called geostationary orbit
1. has orbital period of 24 hrs,
2. lies in the equatorial plane, and
3. orbits in a west to east direction.
Worked Example 3
Calculate the altitude of the geostationary orbit.
Earth’s radius 6370 km
Earth’s mass 245.97 10 kg
Solution
GEO satellites have orbital period of 24 hours.
5 12 27.27 10 rad s
24 60 60T
The gravitational pull provides the required centripetal force for circular motion.
2
2
3 2
11 24 3 5 2(6.67 10 )(5.97 10 ) (7.27 10 )
42235 km
GMmmr
r
GM r
r
r
The altitude 42235 6370 35900 km
T=24 hrs
T=24 hrs fixed spot
Ver 1.0 © Chua Kah Hean xmphysics 5
7.1.3 Comparisons of Orbits
Is it better to park a satellite at high or low altitude? Well, they both have their own pros and cons.
Low altitude
Can take higher resolution images of Earth.
Transmits and receives signals/messages with shorter delay.
Requires less fuel and cheaper to launch into orbit.
High altitude
Wider coverage because can maintain direct line of sight with a large portion of Earth’s surface
at any one time
Suffers less atmospheric drag and requires less frequent orbital boosts.
Low Earth Orbits (LEO)
The majority of satellites are LEO satellites. At altitudes of only a few hundred kilometres, they have
orbital periods of about 90 minutes. A few celebrity satellites in the LEO orbits are the International
Space Station (ISS) and the Hubble Telescope.
see video at xmphysics.com
Middle Earth Orbits (MEO)
A famous occupant of the MEO space is The Global Positioning
System (GPS), a constellation of >24 satellites parked at altitude
of about 20,000 km. From this higher altitude, each satellite has
a larger coverage area. They have orbital periods of 12 hours.
see animation at xmphysics.com
Ver 1.0 © Chua Kah Hean xmphysics 6
Geostationary Orbits (GEO)
The GEO orbit is at a very high altitude which requires very powerful and expensive rockets to be
launched. Nevertheless, it is a popular orbit for communications satellites.
Geostationary satellites stay at a fixed point above the Earth, which is convenient for
communication satellites because they appear stationary when viewed from Earth (if not ground
antenna will have to continuously track the satellites as they move across the sky)
Unfortunately, being constrained to the equatorial plane also means geostationary satellites
cannot serve the polar regions (because from the polar regions they are below the horizon and
cannot be sighted.)
see video at xmphysics.com
Polar Orbit
The main attraction of polar orbits is that they are able to “raster scan” the entire earth’s surface,
including the polar regions.
see animation at xmphysics.com
polar orbit
geostationary orbit
Ver 1.0 © Chua Kah Hean xmphysics 7
7.2 Gravitational Field Strength
If a mass m is situated in a gravitational field, it experiences a gravitational force gF mg , where g is
the magnitude of the field strength at that location. You can think of g as the “gravitational force per
unit mass” (of that location).
By “coincidence”, the gravitational field strength is also the gravitational acceleration. One can easily
work out that the acceleration of an object due to gravity is F mg
a gm m
. So gravitational field
strength and gravitational acceleration are the exact same thing. In fact, N kg-1 and m s-2 are
equivalent units.
Worked Example 4
A 580 kg asteroid travelling in outer space experiences a gravitational force of 810 N.
a) Calculate the gravitational field strength at the asteroid position.
b) Calculate the gravitational force experienced by a 290 kg asteroid at the same location.
c) Calculate the acceleration of the asteroids.
Solution
a) 1810 580 1.40 N kggF
gm
b) 290 1.40 405 NgF mg
c) 1.40 m s-2
7.2.1 Gravitational Field of a Point Mass
Consider a tiny satellite of mass m orbiting at radius r around a massive astronomical body of mass
M. What is an expression for the gravitational force Fg experienced by tiny m?
(GM/r2)m gm
M m m
g
Ver 1.0 © Chua Kah Hean xmphysics 8
If you think of the mutual gravitational pull between M and m, you would say 2g
GMF m
r . If you think
of m sitting in the gravitational field g produced by M, you would say gF gm . Putting these two
together, we realize the magnitude of the field strength is given by the formula
2
GMg
r
As for the direction, it is always radially inward, since gravitational field is always attractive.
see video at xmphysics.com
A few things worth nothing:
1. Just like Fg, the magnitude of g also obeys the inverse-square law.
2. It is quite common for the formula to be written with a negative sign, i.e. 2
GMg
r . Direction-
wise, the negative sign reminds us that g is directed inward (in opposite direction to r). But if we
are concerned with only the magnitude, it is ok to drop the negative sign.
3. Strictly speaking, this formula is valid for point masses only. However, most massive astronomical
bodies (stars, planets, etc) are spherical in shape and have a spherical symmetry. Outside of the
sphere, they are equivalent to a point mass situated at the centre of the sphere. So this formula
accurately calculates the field strength for any point outside the sphere.
g ∝ 1/r2
Ver 1.0 © Chua Kah Hean xmphysics 9
7.2.2 Earth’s Gravitational Field Strength
The gravitational field we are most concerned with is that of our own planet Earth, which is roughly a
sphere of mass 245.97 10 kgM and radius 6370 kmER . Using the formula 2
GMg
r , we can
calculate the field strength of our planet Earth at different radius r away from the centre of the Earth.
Unless you are a space traveler, you are not going to experience the radially inward (towards Earth)
gravitational field that weakens according to the inverse-square law as you travel outward. Instead,
as a terrestrial dweller, you are going to be spending your entire life close to the surface of the Earth,
where the gravitational field is always vertically downward and the magnitude is practically constant.
r/m
0
|g| / N kg-1
0
RE= 6370 km
RE+0 km: 9.81 N kg-1
RE+400 km: ISS, 8.70 N kg-1
RE+600 km: HBT, 8.21 N kg-1
42,0
00 k
m: G
EO
, 0.2
25 N
kg
-1
27,0
00 k
m: G
PS
, 0.5
74 N
kg
-1
380,0
00 k
m: M
oon 0
.0027 N
kg
-1
2RE 3RE 4RE 5RE 7RE RE 6RE
M = 5.97 x 1024 kg
g ∝ 1/r2 constant g = 9.81 N kg-1
Ver 1.0 © Chua Kah Hean xmphysics 10
Using the formula 2
GMg
r , the gravitational field strength on the surface of the Earth can be
calculated to be
11 241
2 3 2
(6.67 10 )(5.97 10 )9.81 N kg
(6370 10 )E
GMg
R
.
This is of course the acceleration of free fall that we are so familiar with. No surprise here, gravitational
field strength is equal to the gravitational acceleration.
If the Earth is a non-rotating perfect sphere of uniform density, then acceleration of free fall would be
exactly equal to 9.81 N kg-1 at all points on its surface. But the Earth is rotating, not spherically
symmetric, and bulges at the equator. Consequently, there are slight deviations in the value of g
across its surface. The reasons are elaborated below:
1) Non-uniform density
The Earth is not uniform in composition. An oil field would cause the local g to be lower, but a mineral
ore would cause the local g to be higher.
2) Non-uniform radius
The Earth is not exactly a sphere. The radius of Earth is 6,378 kilometers at the equator and 6,356
km at the poles - a difference of 22 km. The bulge at the waistline results in a general trend of g
decreasing from the poles to the equator.
3) Earth’s rotation
The Earth is rotating. This means that objects which seem to be stationary are actually moving along
circular paths. The radius of circular motion depends on the latitude; Objects on the equator undergo
circular motion with the largest radius and speed. The North and South poles are the only two location
where there is no circular motion.
Remember that a centripetal force is required for circular motion? A small fraction of the gravitational
force is “used up” to provide the required centripetal force for circular motion. As a result, the value of
g is measured to be lower at the equator than at the poles.
It is worth noting that the Earth’s rotation does not change the gravitational field strength at the equator.
It only causes the field strength to “appear” weaker at the equator. This is basically the difference
between the true g and the apparent g, which we will discuss in greater detail in the next section.
Ver 1.0 © Chua Kah Hean xmphysics 11
7.2.3 True g and Apparent g
Let’s consider a mass m “resting” on the Equator. We know that this mass is actually in circular motion.
And the radius of the circle is the radius of the Earth, R.
As such, the two forces acting on the mass, namely the upward N and the downward mg, cannot be
exactly equal in magnitude. In fact, N has to be slightly smaller, in order for a net centripetal force.
Applying N2L on m, we have
2
2
( )
( )
netF ma
mg N mR
N m g R
If you think about it, N is actually how we “feel” our weight. If we divide N by m, we obtain
2'N
g g Rm
The second term in the equation is the centripetal acceleration at the Equator, which can be calculated
to be
2 3 2 22(6400 10 )( ) 0.03 m s
24 3600R
So 19.81 N kgg is the true g whereas 1' 9.81 0.03 9.78 N kgg is the apparent g. Based on
these numbers, a 1-kg mass weighs 9.81 N at the poles but 9.78 N at the equator. Similarly, all
masses free fall at 9.81 m s-2 at the poles but 9.78 m s-2 at the equator. It is only 0.3% difference. So
it’s not a big deal.
m
mg
N
N mg
radius = R
ω
m
Earth Earth
Ver 1.0 © Chua Kah Hean xmphysics 12
see video at xmphysics.com
The difference between the true and apparent g on the ISS however it HUGE. The ISS is the
International Space Station orbiting at about 400 km above the Earth’s surface. The Earth’s
gravitational field strength at that altitude is about 8.71 N kg-1. You can calculate it using 2
GM
r. Note
that it is NOT zero! If it were zero, there would be no gravitational pull on the ISS and the ISS would
not be orbiting Earth!
But the centripetal acceleration of the ISS is also exactly 8.71 m s-2. Because
2
2
GMm vm
rr is how
the circular orbit is achieved in the first place. So
2 2' 8.71 8.71 0 m sg g R
That’s why to the astronauts in the ISS, g is zero. Even though the true g is 8.71 m s-2, all of it is “used”
to keep the ISS (and all its contents) in circular motion, leaving an apparent g of 0. That’s why when
a ball is dropped in the ISS, the ball does not “fall”. It simply hovers in the ISS. The acceleration of
free fall is excatly zero! When a 1-kg mass is placed on a weighing scale, the scale reads zero.
Everything is weightless in the ISS! Even though true g is 8.71 m s-2 there.
see video at xmphysics.com
Ver 1.0 © Chua Kah Hean xmphysics 13
7.3 Gravitational Potential and Energy
If you pull two magnets apart and then let go, they attract each other and snap back together. There
is energy stored in the magnetic field, which was converted into KE as the magnets race towards
each other.
Likewise, if you snap a planet into two pieces, and pull them apart (imagination pls) and let go of them,
gravitational force will accelerate them towards each other. So the act of pulling two masses apart is
doing work to store energy in the form of GPE, which is converted back into KE when the holding
forces are removed.
see video at xmphysics.com
The amount of GPE stored between two masses M1 and M2 at a distance d apart, is given by the
formula
1 2M MU G
d
A few things to note:
GPE is zero when d is ∞. We have chosen the reference point for GPE to be zero when the two
masses are infinitely far apart.
GPE = 0 is the highest possible GPE. The closer together the two masses, the lower (the more
negative) the GPE.
GPE is always negative.
Ver 1.0 © Chua Kah Hean xmphysics 14
7.3.1 Escape Velocity
“What goes up must come down.” So says the adage. But wait. The Voyager 1 and Voyager 2, two
space probes launched in the 1970s, have gone up and have not come down yet. Not only have they
escaped from Earth, they have escaped from the Sun. Having left the Solar System, they have entered
interstellar space.
Because the Earth’s gravitational field decreases with distance (from Earth), it is possible to “jump”
out of the Earth forever. If the launch speed is fast enough. This minimum launch speed is called the
escape velocity.
see video at xmphysics.com
Worked Example 5
Ignore the effects of air resistance. Calculate the escape velocity from the Earth’s surface.
Earth’s radius 6370 km
Earth’s mass 245.97 10 kg
Solution
By PCOE,
2
11 24
3
1
( ) at launch ( ) at infinity
1( ) 0 0
2
2
2(6.67 10 )(5.97 10 )
6370 10
11.2 km s
e
E
e
E
KE GPE KE GPE
GMmmv
R
GMv
R
Ver 1.0 © Chua Kah Hean xmphysics 15
7.4 Gravitational Potential
For terrestrial earthlings, height is an obvious thing. When you gain height, you gain GPE. When you
lose height, you lose GPE. For space travelers, however, “height” is more abstract concept. Take for
example the journey from the Earth to the Moon. At the beginning as the spacecraft lifts itself away
from Earth’s gravitational field, it is and gaining GPE and thus moving to “higher height”. When it nears
Moon and makes its landing, it is losing GPE and thus moving to “lower height”. The maximum GPE
actually occurs at some point between the Earth and the Moon. Wouldn’t it be nice to have a number
to mark out the “gravitational height” along the journey?
This “gravitational height” is called the gravitational potential Φ (pronounced phi). Basically, you can
think of Φ (of a location) as the “gravitational potential energy U per unit mass” (of that location).
U
m
Alternatively, if a mass m is situated in a gravitational field where the gravitational potential is Φ, then
its GPE (at that location) would be
U m
Worked Example 6
A 580 kg asteroid is travelling at 1600 m s-1 through outer space. From location A where it started
with GPE of −900 MJ, it arrived at location B which has gravitational potential of -2.1 MJ kg-1.
a) Calculate the gravitational potential at location A.
b) Calculate the KE of the asteroid at location B.
Solution
a) 1900 580 1.55 MJ kgU m
b)
2 6 6
6 6 6
( ) at A ( ) at B
1(580)(1600) ( 900 10 ) (580)( 2.1 10 )
2
742.4 10 900 10 1218 10
1060 MJ
B
B
B
KE GPE KE GPE
KE
KE
KE
Ver 1.0 © Chua Kah Hean xmphysics 16
7.4.1 Gravitational Potential of Point Mass
Consider a tiny satellite of mass m orbiting at radius r around a massive astronomical body of mass
M. What is an expression for the gravitational potential energy U of the tiny m?
If you think of the GPE stored between M and m, you would say GM
U mr
. If you think of m sitting
at location where the gravitational potential is , you would say U m . Remember that U m .
This means that the gravitational potential produced by M must be GMm
mr
. So
GM
r
Things to note:
Just like GPE, is a scalar quantity. If we join all the points with the same gravitational potential,
we get concentric circles centred about the M.
−(GM/r)m
m
M m m
∝ 1/r
see video at xmphysics.com
Ver 1.0 © Chua Kah Hean xmphysics 17
The potential is always negative. It is inversely proportional to r, and approaches zero at the point
at infinity.
Strictly speaking, this formula is valid for point masses only. However, most massive astronomical
bodies (stars, planets, etc) are spherical in shape and have a spherical symmetry. Outside of the
sphere, they are equivalent to a point mass situated at the centre of the sphere. So this formula
accurately calculates the gravitational potential for any point outside the sphere.
It is useful to have the mental picture that the Earth sets up a gravitational potential well centred
about itself. To move away from Earth is akin to climbing out of a potential well, which requires
energy to accomplish.
see video at xmphysics.com
The potential gradient is the (negative of) field strength. (elaborated in the next section)
r/m
Φ / MJ kg-1
2RE 3RE 4RE RE
62.5
Ver 1.0 © Chua Kah Hean xmphysics 18
7.5 Potential Energy Gradient and Potential Gradient
An increase in GPE comes from positive work done by an external force Fext that overcomes the
gravitational force Fg.
extU F x
The potential energy gradient is thus the external force.
ext
dUF
dx
Since ext gF F , the negative of the potential energy gradient is the gravitational force.
g
dUF
dx
Since gF mg and U m , it follows that the negative of the potential gradient is the field strength.
( )d mmg
dx
dg
dx
Ver 1.0 © Chua Kah Hean xmphysics 19
Take for example the gravitational field strength and the gravitational potential due to Earth. The
gradient of the Φ-r graph is related to g, while the area under the g-r graph is related to Φ.
r
Φ
R
r
g
R
Ver 1.0 © Chua Kah Hean xmphysics 20
7.6 PCOE Problems
Escape from Potential Energy Well
A projectile m is launched vertically with KEi from the surface of a planet of mass M and radius R. The
graphs below illustrate how the KE, GPE and TE of the projectile varies along the projectile’s journey.
Things to note:
The initial GPE, i
GMmGPE
R
i
GMmTE GPE KE KE
R is constant. (Effects of air resistance is ignored)
As the projectile moves away from Earth, it gains GPE at the expense of losing KE. So GPE graph
is mirror image of KE graph.
At maxr r , KE drops to zero. This is the furthest point reached by the projectile before it drops
back to Earth.
To escape from Earth’s gravitational field, the projectile must be launched with large enough KEi
such that 0i
GMmKE
R
r/m
Φ / MJ kg-1
KE
TE
GPE
rmax
KEi
GPEi
R
Ver 1.0 © Chua Kah Hean xmphysics 21
Gravitational Well Hopping
Worked Example 7
The graph below shows the total gravitational potential along the line between planet A and moon B.
a) Describe the significance of position Y.
b) Calculate the minimum speed a projectile must be launched from the surface of moon B in order
for it to arrive at planet A. (Ignore effects of air resistance)
c) Calculate the minimum landing speed on planet A.
Φ / MJ kg-1
ΦX=-102.5 MJ kg-1
r/m
ΦR
ΦA
ΦB
ΦZ=-35.0 MJ kg-1
ΦY=-22.5 MJ kg-1
X Z Y A B
Ver 1.0 © Chua Kah Hean xmphysics 22
Solution
a)
Y is the gravitational null point where the resultant gravitational force is zero. On the left of Y, the
gravitational field is towards planet A. On the right of Y, the gravitational field is towards moon B.
b)
The projectile must start with enough KE to arrive at Y. (Upon reaching Y, the satellite can just fall to
Earth)
By PCOE:
2
2 6 6
1
( ) at Z ( ) at Y
10
2
1( 35.0 10 ) 22.5 10
2
5000 m s
Z Y
KE GPE KE GPE
mv m m
v
v
c)
By PCOE:
2 2
2 6 2 6
1
( ) at Z ( ) at X
1 1
2 2
1 1(5000) ( 35.0 10 ) 102.5 10
2 2
12649 m s
Z Z X X
X
X
KE GPE KE GPE
mv m mv m
v
v
7.7 Energy of Satellites
Consider a satellite of mass m in a circular orbit of radius r around the Earth. Since the orbital speed
depends on the orbital radius, the KE of the satellite is also a function of r.
2
2
2
( )
1 1
2 2
1
2
netF ma
GMm mv
rr
GMmmv
r
GMmKE
r
Ver 1.0 © Chua Kah Hean xmphysics 23
Needless to say, the GPE of the satellite is also a function of r.
GMmGPE
r
Finally the total energy of the satellite is also a function of r.
1( ) ( )2
1
2
TE KE GPE
GMm GMm
r r
GMm
r
What can you interpret from this graph?
2 2GPE KE TE
TE of a satellite is different at different altitudes. Even though a high altitude satellite has lower
KE than a low altitude one, it still has the higher TE because of the higher GPE.
To move a satellite from a lower orbit to a higher orbit involves an increase in the TE. The increase
in GPE is twice the decrease in KE.
energy
orbital radius, r
GPE
TE
KE
Ver 1.0 © Chua Kah Hean xmphysics 24
Appendix A: Space Travel
21st century looks set to be a very exciting time for space travel. SpaceX just achieved the major
milestone of sending astronauts to the ISS. NASA plans to go to the Moon again, and beyond. What
are you going to see during your lifetime? 10-min space vacations? Orbiting space hotels? Humans
on Mars? Meanwhile, we can binge on those SpaceX launch webcasts (and nod your head sagely
when the presenters spew terms like 1st orbital speed, MECO, boostback burn.)
see video at xmphysics.com