fysbook SPACE Bryan Maher

Space
created 13.75 billion years ago

HSC Topic 1 – Focus 1 – Gravitational PE


1. The Earth has a gravitational field that exerts a force on objects
both on it and around it
Students learn to: Students:
 define weight as the force on an object  perform an investigation and gather
due to a gravitational field information to determine a value for
acceleration due to gravity using
pendulum motion or computer-
 explain that a change in gravitational
assisted technology and identify
potential energy is related to work
reason for possible variations from the
done
value 9.8 ms-2
 gather secondary information to
predict the value of acceleration due
 define gravitational potential energy as
to gravity on other planets
the work done to move an object from
a very large distance away to a point in  analyse information using the
a gravitational field expression: F mg
to determine the weight force for a
m m
Ep G 1 2 body on Earth and for the same body
r on other planets


Weight and the Gravitational Field
Every massive body has an associated gravitational field surrounding it,
extending out to infinity but weakening with distance. The field due to a body
can be defined as the region of space surrounding it where other bodies will
feel a force due to it.

Thus, if a second mass enters that
field, it will experience a force of
attraction -

- and, in turn, it will exert a force of
attraction on the first mass.
Why?


This gravitational force, Fg, is the weakest of the four fundamental natural
forces.

The gravitational field due to a body extends to infinity, so the gravitational
force is infinite in range, although it becomes very weak at large distances as
it is an inverse square law.

What does an “inverse
square law” mean?

What does an “inverse
square law” look like
graphically?


The gravitational field can be visualised in terms of lines of force, or field lines –
with the direction of the field lines indicating the direction of the gravitational
force, and the relative spacing of field lines giving an indication of the
gravitational field strength.

The gravitational field strength g is thus a vector, and
the combination of the vectors at all points describes
the gravitational field.

For a spherical object such as the planet Earth, the
field lines are as shown, indicating a radially inward
field which weakens with distance from the centre
of the Earth.

Why do the field lines
never cross?


Close to the Earth’s surface, the gravitational
field is effectively uniform.

Note that a line (or more correctly, a surface) perpendicular to the field
lines and joining places of equal gravitational field strength also represents
a constant level of gravitational potential energy.

How does this link with
what you know already
about GPE?


The gravitational force acting on an object is defined as its weight, W.

The strength of the gravitational field is defined as the force per unit mass it
exerts on a mass within the field.

That is, the size of the gravitational force acting on a mass m defines the
strength of the gravitational field, g, at a point, by

g = Fg = W
m m

The units of g are N/kg.

What else is defined as
force per unit mass?


Notice that the strength of the gravitational field at a point does not depend
on the size of the mass m placed in it – but only on the size and location of
the masses which create the field.

Isaac Newton
I knew that – have a look at my equation for gravitational force.

Like .Comment .Share about 320 years ago


Calculation of W
Planet g at surface Weight at surface
Nkg-1 N
Mercury 3.78

Venus 8.60

Earth 9.78 978

Mars 3.72

Jupiter 22.9

Saturn 9.05

Uranus 7.77

Neptune 11.0


The gravitational field strength for a
uniform spherical body of mass M is
given by

g = GM
r2

where G is the universal gravitational
constant = 6.67 10-11 Nm2kg-2 and
r is the distance from the centre of
mass of the body.

Where does this
equation come from?


Calculation of g
Planet Mass Diameter g at surface
kg km Nkg-1
Mercury 3.34 1023 4 880 3.78

Venus 4.87 1024 12 100 8.60

Earth 5.98 1024 12 800 9.78

Mars 6.40 1023 6 790 3.72

Jupiter 1.90 1027 143 000 22.9

Saturn 5.69 1026 120 000 9.05

Uranus 8.67 1025 51 800 7.77

Neptune 1.03 1025 49 500 11.0


Notice that for a freely falling object
of mass m,
Fg
then a = Fg
m
= mg
m
=g

Albert Einstein
That is, the magnitude of the acceleration of a freely falling object
is equal to the gravitational field strength at that point - and so...


For example…..
1. What would a (very) accurate set of scales indicate as the weight of Liam at
New York (g = 9.803 ms-2) as compared to the equator (g = 9.780 ms-2)?
2. What are some reasons for the variation in g at different points on the Earth’s
surface?
3. How would you expect g at the North pole to be different compared to that at
the equator?
4. What weight force would Phoebe experience on the surface of Jupiter
(g = 24.8 ms-2)?
5. If the Neptunian moon Triton has a mass of 2.14 1023kg and a radius of
1.35 106m, determine the intensity of the gravitational field at the surface
and at an altitude of 100 km.
6. There exists between the Moon and the Earth a “parking space” for spacecraft
where the gravitational field is effectively zero. This is known as the
Langrangian point, and is shown as….

 Questions SET 1.1


Lagrangian
point

How does this come
about?


3.82 108m

Lagrangian
point

MM = 7.36 1022kg
ME = 5.98 1024kg
RM = 1.74 106 m
RE = 6.37 106 m

Solution


Gravitational Potential Energy
The energy an object has as a result of its position in space relative to
other massive objects is its gravitational potential energy.

When work is done on an object to move it against a gravitational force,
the gravitational potential energy of the object is increased.

Final GPE (greater)

Initial GPE


Gravitational Potential Energy

Work done against the field = Ep
= Fd
= mg h

Final GPE (greater)

h
F
Initial GPE


Likewise, if the object moves freely under the influence of the gravitational
field, its gravitational potential energy is decreased (and its kinetic energy
is increased).

That is,

Work done by field = - EP = EK

What if the mass
moves horizontally ? Does the actual path
matter?

Energy changes with height.xls


The previous analysis involved the special case where it was assumed the
gravitational field was uniform over the distance the mass was moved.

As this is not necessarily the case, a better analysis uses the definition
that the gravitational potential energy (EP) of an object of mass m1 at a
distance r from the centre of a mass of m2 is defined as the work done in
moving m1 from infinity to a distance r from the centre of m2.

This gives the result that

EP = - G m1m2
r

Isaac Newton
Solving problems like this was one of the reasons I invented calculus!



r=
EP = 0

m1 m2

This equation assumes the gravitational potential energy to be zero at an
infinite separation of the masses..
Why?

r<
EP decreases, so is < 0

m1 m2

Thus, since a mass released at infinity will lose potential energy and gain
kinetic energy as it accelerates under the influence of the gravitational
field, it must have increasingly negative values for its gravitational
potential energy.
So how can the change
in GPE be positive?


EP
rE r

Plot of gravitational potential
energy against distance from the
centre of Earth.
It is valid only for points beyond the
radius of the Earth, rE

Why?


EP
rE r
rf ri
GPEi

GPEf

Since PE = KE = Work done

So what is happening if, for example, a
satellite is lifted up to a higher geostationary
orbit?


If gravitational force per unit mass is plotted against distance from the
centre of a planet….

Fg
m
Why is this curve
“upside down”
9.8 Nkg-1 compared to the
previous one?

Is this curve just
“upside down”
compared to the
previous one?

rE r


…the area under the curve represents the work done per unit mass
either by or against the field in varying the distance of the mass from
the centre of Earth.

Fg
m

rE r


Gravitational force on a 1000 kg satellite at varying distance from Earth's
centre

10000
9000
8000
7000
6000
Force (N)

5000
4000
3000
2000
1000
0
0 10000 20000 30000 40000
Radius (km)


For example…
1. Use the formula to determine the GPE of a 100kg object at the surface of
the Earth, and at a height of 1000m.
2. What KE would result from the object falling from 1000m to the surface?
3. Compare this to the value obtained using PE = mg h.
4. Using the plot of Fg vs r from the previous slide, estimate the work
needed to lift a 1000kg satellite from an orbit of radius 10000km to one of
20000km.


For a body moving freely in a gravitational field, the total energy remains
constant. Thus
1 mm
ET mv 2 G 1 2
2 r

Hence, satellites in circular
orbits have constant EK
and EP , while those in
elliptical orbits vary their
EK and EP.

Why?

Johannes Kepler
I told you so….



For example…
1. By considering the Law of Conservation of Energy, explain why the
sign for gravitational potential energy is negative.
2. Explain the difference in kinetic energy for a satellite at aphelion
compared to perihelion.
3. Calculate the change in potential energy if a 100 kg satellite is moved
from a height of 200 km above the Earth's surface to a height of
3400 km.
4. A 1kg particle is traveling radially in toward Earth at 10ms-1 at an
elevation equal to the Earth’s radius. If air resistance is
neglected, with what speed does the particle strike the Earth’s
surface?

 Questions SET 1.2


“A more modern view on this topic was presented by Albert Einstein. In a far
more complex description, dealing with curved space, mass-energy tells
space-time where to bend and vice versa. Obviously, the effects on everyday
life are negligible. For the sake of completeness, it should be remarked that
there are indeed observable relativistic effects, such as the trajectory of light
being bent by the sun's mass. To summarise, Einstein's relativistic description
of gravity is more accurate, far more complicated, of negligible effect on
everyday life, and still incomplete!”


Determination of g
Method 1: Using computer assisted technology

By measuring values of displacement at different times, plot a curve of s/t
vs t and determine a value for acceleration due to gravity.


Method 2: Using motion of a pendulum

1 period

l
By measuring values of period, T, and using T 2
g
plot a curve of T vs l and determine a value for acceleration due
to gravity.

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