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Space: 1. The Earth’s gravitational field
| Syllabus reference (October 2002
version) |
|
1. The Earth has a gravitational field that exerts a
force on objects both on it and around it
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Students learn to:
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Students:
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Extract from Physics Stage 6 Syllabus (Amended
October 2002). © Board of Studies, NSW.
[Edit: 21 Aug 08]
Prior learning: Preliminary modules 8.4
(subsection 2), 8.5 (subsection 4)
Background: A gravitational field
surrounds all masses. The strength of the field is defined by
the vector g (units: N
kg-1), since g = F/m.
Because of the way field strength is defined, for the
earth’s gravitational field,
g has the same numerical value as
the acceleration experienced by a free falling object,
ag (units: m
s-2).

define
weight as the force on an object due to a gravitational
field
- Weight is the force on an object in a gravitational
field. It is a vector quantity and the measurement unit is
the Newton (N).

explain
that a change in gravitational potential energy is related to
work done
-
Consider the work done in moving an object from the
Earth's surface to a height, h metres.
- W = F.d
- Therefore, W = Fg.d (where Fg is the weight of the
object)
- Fg = mg (as weight of the object = mass x
acceleration due to gravity)
- Therefore, W = mg.d
- Therefore, W = mgh (h is the distance the object
has been moved)
-
Gravitational Potential Energy
Zona Land, Edward A. Zobel

perform an
investigation and gather
information to determine a value for acceleration due to
gravity using pendulum motion or computer assisted technology
and identify reason for possible variations from the value
9.8 ms-2
- You may be performing an investigation
that has been planned by your teacher. There are several
suitable investigations that will achieve this purpose. One
suitable investigation is described here.
A procedure for determining a value for
acceleration due to gravity
A value for acceleration due to gravity can easily and
accurately be measured by observing the motion of a
pendulum.
- Construct a pendulum at least one metre long,
attached at its top to a support (such as a clamp
connected to a retort stand) and with a small mass tied
to its lower end to act as the pendulum bob.
- Measure the length (l) of
your pendulum, from its point of attachment to the centre
of mass of its bob.
-
Pull the pendulum aside and release it so that it
starts swinging. Using a stopwatch (or other device for
measuring time), begin timing at an extreme of the
pendulum’s motion and time ten full swings (one
swing = back and forth) of the pendulum. Divide this
time by ten to get a value for the average period
(T) of the motion. Using this
averaging technique tends to minimise random errors.
The period of a pendulum depends upon the length
(l) and the value of
acceleration due to gravity (g), as described in the
following equation:
Rearranging this equation gives an expression that can
be used to calculate g.
- Substitute your values for
l and
T into this equation to
determine a value for g.
- As you gather information during your
investigation, you may need to carry out repeat trials to
confirm the reliability of your results. Also, you may want
to use other, more accurate, timing devices or procedures
to minimise the effect of random errors.
-
You need to be able to identify reasons
for possible variations from the value 9.8
ms-2. Variations can be as a result of two
general factors; either from experimental errors or from
actual variations in the value of acceleration due to
gravity as an effect of local variations in the nature of
the Earth’s crust.
Experimental errors might occur as a result of things
like the accuracy of equipment and human reaction time in
the use of equipment.
The value of acceleration due to gravity at the surface
of the Earth varies from the usually accepted value of
9.8 m s-2, due to a number of factors:
- The Earth’s lithosphere varies in structure,
thickness and density. Thickness variations are a
product of the source and history of the material.
Oceanic crust is thinner than continental crust.
Continental crust is thickest under mountain ranges.
Density variations occur due to the presence of
concentrated and large mineral deposits or petroleum
gas and related liquids trapped in sedimentary rocks
and structures. All of these variations can influence
local values of g.
- The Earth’s globe is flattened at the poles.
This means that the distance of the surface from the
centre of the Earth is less at the poles, which
increases the local value of
g.
- The spinning Earth also affects the value of
g. At the equator, the spin
effect is greatest resulting in a lowering of the value
of g. As you travel from the
equator to the poles, the spin effect on
g shrinks to zero.
- As a result of the above, the value of
g at the surface of the Earth
varies between 9.782 m s-2 at the equator
and 9.832 m s-2 at the poles.
- The value of g reduces
with altitude above the surface of a planet, becoming
zero only at an infinite distance. At low Earth orbit
altitude, the value of g is
approximately 8.9 ms-2.

gather
secondary information to predict
the value of acceleration due to gravity on other
planets
-
Try to gather information from astronomy
reference books or authoritative web sites. You will be
looking for information that will assist you to predict
the value of acceleration due to gravity at the surface
of other planets. Note that the gas giants Jupiter,
Saturn, Uranus and Neptune, do not possess a surface upon
which you could stand, so that the value of
g for those planets will be
theoretical only. There are, however, many moons within
our solar system that are large enough to possess a
significant gravitational field.
You may be able to find information about the radius
(r) and mass
(m) of planets, and use that
information to calculate a predicted value of
g by using the following
equation:
(G is the universal gravitational constant)
- Once you have gathered the
information, summarise and collate the information in a
table, such as the following. Some sample values have been
included.
| Planet |
Value of g |
| Mercury |
|
| Venus |
8.9 |
| Earth |
9.8 |
| Mars |
|
| Jupiter |
24.8 |
| Saturn |
|
| Uranus |
|
| Neptune |
|
| Pluto |
|

analyse
information using the expression F = mg to
determine the weight force for a body on Earth and for the
same body on other planets
- The magnitude of the weight force on an object can be
calculated using a slightly altered form of Newton’s
second law: F = mg. The direction of the weight
force is the same as the direction of the gravitational
field at the location of the object.
-
The value of acceleration due to gravity at the surface
of the Earth is 9.8 m s-2. The weight of a 65
kg person can therefore be calculated as follows:
- The table below shows the acceleration due to gravity
at the surface of other rocky planets in the solar system,
plus the Moon. Analyse the data by
calculating the weight force of the same 65 kg person at
each location. Click on the link below to check your
answers.
| Planet surface |
Value of g |
Weight force |
| Mercury |
4.0 |
|
| Venus |
8.9 |
|
| Mars |
3.7 |
|
| Pluto |
0.6 |
|
| the Moon |
1.6 |
|
Answers

define
gravitational potential energy as the work done to move an
object from a very large distance away to a point in a
gravitational field 
- The gravitational potential energy of an object at some
point within a gravitational field is equivalent to the
work done in moving the object from an infinite distance to
that point.
- It can be shown mathematically that the gravitational
energy, Ep, of an
object with mass, m1,
a distance, r, from the centre of
a planet of mass, m2,
is given by:
- A graph of the Ep
surrounding a planet looks like this:
When lifting an object against a gravitational field,
e.g. launching a rocket, work is done on the object, that
is, energy is transferred to the object. The object’s
gravitational potential energy,
Ep, that is, the
energy it has due to its position within the gravitational
field, increases as a result.
When an object moves toward the source of the
gravitational field, such as when dropping a stone, energy
due to position in a field is transformed into kinetic
energy (the stone speeds up).
Hence the position of lowest Ep in the
gravitational field surrounding a planet is at the surface
of the planet.
An object only has zero Ep when it is no
longer within the gravitational field, that is, a very
large distance away. (Mathematically, distance must be
infinite.)
