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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

https://xmphysics.com/grav


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

http://www.xmphysics.com/7.1



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

http://www.xmphysics.com/7.1.1



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.



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




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.






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.)


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.


polar orbit

geostationary orbit




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



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.


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



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



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.


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




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.





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.


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.




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.


Worked Example 5

Ignore the effects of air resistance. Calculate the escape velocity from the Earth’s surface.



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




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


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





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.


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



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



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


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



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


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



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


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.)


http://www.xmphysics.com/7.A

http://www.xmphysics.com/7.A

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