Episode 402: Fields, field lines and field strength
This episode introduces the notion of a “field of force” and how we can diagrammatically represent such a field using field lines. The actual strength of the field at a point in space is defined by the field strength.
Summary
Discussion: Fields and field lines. (10 minutes)
Discussion: Field strength, g. (10 minutes)
Worked examples: Field strength. (20 minutes)
Spreadsheet exercise: Analysing data from the Apollo 11 mission. (20 minutes)
Discussion:
Fields and field lines
When you pick up an object such as a pen, there is direct contact between you and the pen. This direct contact exerts a force on the pen, causing it to move in the way that it does. However, the pen also has a weight due to its presence in the Earth’s gravitational field. How is this force exerted, even when there is no direct contact between the Earth and the pen? A force is exerted on the pen from the Earth because the pen is in the Earth’s gravitational field. We can define the field due to a body as the region of space surrounding it where other bodies will feel a force due to it.
What extent do gravitational fields have? (The gravitational force is infinite in range, although it becomes very weak at large distances as it is an inverse square law. The gravitational field due to a body is thus also infinite.)
We cannot see or touch this field, but we can try to model it using field lines or lines of force. In a field line diagram, the direction of the field line at a point gives the direction of the force of attraction that would be felt by a small mass placed there. The relative density of field lines on the diagram is an indication of the strength of the field. (Compare this with the more familiar magnetic field patterns.)
Thus for a spherical mass, like the Earth, we would have the following diagram:
(resourcefulphysics.org)
The field lines are directed radially inwards, because at any point in the Earth’s field, a body will feel a force directed toward the centre of the Earth. The field lines become more spread out as the distance from the Earth increases, indicating the diminishing strength of the field. Note that the field is really 3dimensional, but of course on paper, we can only take a 2dimensional slice of it. This is a radial or spherical field.
Close to the surface of the Earth, the field lines look like:
(resourcefulphysics.org)
They are directed downwards (the direction in which a body near the Earth’s surface would feel a gravitational force), and they are parallel and equidistant indicating that the field is constant, or uniform.
A couple of important points to note:
1) Field lines do not start or stop in empty space (even though on diagrams they have to stop somewhere!). They end on a mass and extend back all the way to infinity.
2) Field lines never cross. (If they did, then an object placed at the point where they crossed would feel forces in more than one direction. These forces could be resolved into one direction – the true direction of the field line there.)
(The concepts of field and field lines will be used again when we look at electric fields and magnetic fields).
Discussion:
Gravitational field strength, g.
As you’ve already seen, you pull on the Earth with exactly the same force as it pulls on you. However, your gravitational field is intrinsically much, much weaker than that of the Earth. To try to get a handle on the intrinsic strength of the gravitational field of a body, we need to define field strength at a point in the field.
We define field strength at a point in a body’s field as the gravitational force exerted on an object placed at that point, per kg of the object’s mass. In other words, it is just the number on newtons of attractive force acting per kg of the object’s mass. Since the attractive force is simply what we call weight, we can write this as:
g = W/m
where W = weight in newtons. Thus g has units N/kg.
We can use this definition to get an equation for g using Newton’s Law of Universal Gravitation. The attractive force of a mass M (causing the field) on a mass m a distance r away is simply GMm/r^{2}. Thus the attractive force per kg of mass of the object (mass m) is (GMm/r^{2})/m.
Thus, g = GM/r^{2}
This gives an expression for the field strength at a point distance r from a (point or spherical) mass M.
A couple of points to note:
1) The gravitational field strength at a point in a field is independent of the mass placed there – it is a property of the field. Thus, two objects of different mass placed at the same point in the field will experience the same field strength, but will feel different gravitational forces.
2) In some texts a minus sign is present in this equation, so it reads g = GM/r^{2}; this is from the more correct, but off syllabus, vector equation for g. It is better to calculate the magnitude of the field strength using g = GM/r^{2} and then the direction is given by the fact that gravity is always attractive (i.e. the field always acts towards the gravitating body).
Worked examples:
Field strength
Calculations involving field strength.
TAP 402:1 Field strength – students’ sheet
Spreadsheet exercise
Here, students can try analysing data from the Apollo 11 mission.
TAP 4022: Analysing data from the Apollo 11 mission to the Moon
TAP 4023: 'Data from the Apollo 11 mission'
TAP 402:1 Field Strength – students’ sheet
Data required: G = 6.67 10^{11} N m^{2 }kg^{2}, mass of the Earth = 6.0 10^{24} kg,
radius of the Earth = 6.4 10^{6} m.
1) Treating the Earth as a perfect sphere, show that the field strength at the Earth’s surface is around 9.8 N kg^{1}.
2) What is the strength of the gravitational field due to you at a distance equal to the radius of the Earth?
3) In view of your answers to the two questions above, your field strength at a distance of 6400 km from you (where the Earth’s centre is) is MUCH weaker than the Earth’s field strength at a distance of 6400 km from its centre (where you are right now). Can you explain why is then that you and the Earth exert the same size of force on each other?
4) Assume the Earth has an average density . Rewrite the expression g = GM/r^{2} to include [The volume of a sphere is given by V = (4/3) r^{3}.] Now, assume that the Moon has the same average density as the Earth. On the surface of the Moon, the gravitational field strength is about 1.6 N kg^{1}. Use this to estimate the radius of the Moon.
5) In fact the radius of the Moon is 1.7 10^{6} m. In view of your answer to question 4, what does this suggest about the average density of the Moon compared with that of the Earth?
Practical advice
Remember that a lot of students have difficulties in using standard form correctly on a scientific calculator. You may need to help them in the use of the EXP button
Answers and worked solutions
1 g = GM/r^{2} = (6.67 x 10^{11} x 6.0 x 10^{24})/ (6.4 x 10^{6})^{2} = 9.8 N kg^{1}
2 What is the strength of the gravitational field due to you at a distance equal to the radius of the Earth?
Each student will have to know/estimate their own mass in kg. The calculation below is based on a student mass of 70 kg.
g = GM/r^{2} = (6.67 x 10^{11} x 70)/ (6.4 x 10^{6})^{2} = 1.1 x 10^{22} N kg^{1}
Note, that while this is very very small, it is not zero, again showing that the gravitational force falls off quickly with distance, but is always there (infinite in range).
3 Because the actual force acting on an object at a point in a field is the product of the field strength and the mass of the object placed there. This product is the same in both situations mentioned in the question – the product of the Earth’s field (9.8 N kg^{1}) and the student’s mass (70 kg) is the same as the product of the student’s field (1.1 x 10^{22} N/kg) and the Earth’s mass (6.0 x 10^{24 }kg). In both cases this product is 686 N, and this can equally be viewed as the student’s weight in the Earth’s gravitational field, or the Earth’s weight in the student’s gravitational field
4 g = GM/r^{2}, but M = x V = (4/3) r^{3}.
Therefore g = [G x (4/3) r^{3}/ r^{2} = (4/3) Gr.
(Thus, for bodies of the same density, the field strength at their surface is proportional to their radius).
The surface field strength of the Moon is (1.6 / 9.8) times smaller than that of the Earth. Therefore its radius must be smaller by the same factor.
Therefore, radius of the Moon is (1.6 / 9.8) x 6.4 x 106 = 1.0 x 10^{6 }m
This could also have been calculated by finding the average density of the Earth numerically from g = (4/3) Gr (using g = 9.8 N kg^{1}), and proceeding from there)
If you have students of high enough ability (and you have enough time!) you might like to ask if
g = (4/3) Gr violates the inverse square law. It now looks distinctly like a direct proportion between g and r. Why is this? It’s because we are dealing with the field strength on the surface of a planet. As the radius of a planet increases, its mass increases with the radius^{3} (for constant density). Looking at g = GM/r^{2} this r^{3} dependence of the mass in the numerator will dominate the r^{2} in the denominator, giving overall a direct proportion between g and r
5 Since g = (4/3) Gr, if the radius is bigger than we expected, the density must be smaller than we assumed. Thus the average density of the Moon is less than that of the Earth. In fact, because the Moon’s radius is 1.7 times more than expected, its average density must be around (1/1.7) times that of the Earth (which is about 60% of the Earth’s density.
TAP 4022: Analysing data from the Apollo 11 mission to the Moon
A largescale experiment in space
Detailed data are kept of every space flight. These data, giving speeds and distances of Apollo 11 as it went to and returned from the Moon, are extracted from data supplied by NASA. They can be thought of as the results of an experiment to probe Earth’s gravitational field.
You will need

computer running a spreadsheet

spreadsheet data table
TAP 4023 'Data from the Apollo 11 mission' is provided to go with this activity.
Finding the gravitational field at different distances
The spreadsheet just provides raw data. To get information from it you will have to select and process data from it.
The flight data have a number of values of velocity and distance, arranged in pairs measured 10 minutes (600 seconds) apart. You can see how far Apollo 11 travelled in 10 minutes. You can see that Apollo 11 is decelerating on the way out and accelerating on the way home. You can find the acceleration and so the gravitational field at different distances:
1. Calculate the changes in velocity between each pair of distances.
2. Calculate the average distance for each pair.
3. Calculate the average acceleration at each distance, from the change in velocity and the time interval (600 s).
4. Plot one or more graphs to test whether the acceleration varies as the inverse square of the distance. Take trouble to find the graphs which best tell the message in these data. Give the graphs a caption saying what they have to tell.
Warning: the velocities given point approximately away from the centre of the Earth, but the space flight does change direction. So a simple calculation of the acceleration is only roughly an indication of the Earth’s gravitational field at that point.
Practical advice
These data are selected from a huge printout supplied by NASA. More data is provided here so that it can be looked at and discussed, and so that further analyses can be done. We think that there is also some interest in imagining the whole mission, which is why times from throughout the 200hour mission are included.
The exercise also provides students with further opportunities to think about the most effective graph to plot. It is worth insisting that graphs have captions that convey their intended message.
Alternative approaches
You may, especially with a class lacking in confidence, find it useful to print out the data for everyone, and look through them together, talking about what is happening at each stage.
The exercise can be done without a spreadsheet but can be very tedious. If a computer is not available try thinking along these lines:
1 Calculate the average acceleration between each pair of points with no rocket burn on the outward journey.
2 Without further calculations deduce an estimate for the gravitational field strength at the mid point of each interval
3 Plot a graph that enables you to check whether g does vary as.
Answers and worked solutions
1
Mean distance = (29.0 + 26.3) x 10^{6} / 2 = 27.7 x 10^{6} m
Time for interval = 600 s
Acceleration = (vu)/t = (5374  5102) / 600 = 0.4553 m s^{2} = g
Mean distance
r / 10^{6} m

Mean acceleration
Acceleration / m s^{2}

27.7

0.4553

55.4

0.122

96.5

0.042

170.4

0.013

The –ve sign indicates towards the Earth
2 The values of the gravitational field strength are identical to the acceleration
3 A graph of g vs. gives a straight line
Social and human context
NASA’s raw data were actually provided with distances in nautical miles above the surface of the Earth, and with velocities in knots.
External references
This activity is taken from Advancing Physics chapter 11, 240S
Questions 14 in alternative approaches are taken from an adaptation of Revised Nuffield Advanced Physics question 24L
TAP 4023: 'Data from the Apollo 11 mission'

Events

Ground elapsed time / hours:minutes:seconds


Distance r from centre of Earth / 10^{6} m

Velocity v / m s^{–1}


launch from surface of Earth

00:00:00





ignition to inject into coasting orbit to Moon (5 m 20 s burn)

02:44





coasting begins with no rocket burn

03:08:00

1

11.054

8406








10 minute =

03:58:00

2

26.306

5374


600 s interval

04:08:00

2A

29.030

5102








10 minute =

05:58:00

3

54.356

3633


600 s interval

06:08:00

3A

56.368

3560








10 minute =

09:58:00

4

95.743

2619


600 s interval

10:08:00

4A

97.242

2594








10 minute =

19:58:00

5

169.900

1796


600 s interval

20:08:00

5A

170.945

1788








no rocket burn until this time

26:44:57.92

6

209.228

1531.56


3.55 second burn

26:45:01.47

7

209.232

1527.16








10 minute =

32:58.00

8

240.624

1356


600 s interval

33:08:00

8A

241.417

1352


Landing on Moon. Moon walk.






Rocket burn to return

150:28





Coasting back to Earth






10 minute =

166:38:00

9

241.637

1521


600 s interval

166:48:00

9A

240.740

1524








10 minute =

172:18:00

10

209.722

1676


600 s interval

172:28:00

10A

208.737

1681








10 minute =

178:28:00

11

170.891

1915


600 s interval

178:38:00

11A

169.766

1923








10 minute =

187:58:00

12

96.801

2690


600 s interval

188:08:00

12A

95.241

2715








10 minute =

191:48:00

13

56.368

3626


600 s interval

191:58:00

13A

54.310

3699








10 minute =

193:48:00

14

28.427

5201


600 s interval

193:58:00

14A

25.640

5486








10 minute =

194:38:00

15

13.311

7673


600 s interval

194:48:00

15A

10.036

8854


Rocket burn on reentry

195:03




Practical advice
These data are selected from a huge printout supplied by NASA. More of the data is provided here so that it can be looked at and discussed, and so that further analyses can be done. We think that there is also some interest in imagining the whole mission, which is why times during the nearly 200 hour mission are included.
The exercise also provides students with further opportunities to think about the most effective graph to plot. It is worth insisting that graphs have captions that convey their intended message.
Alternative approaches
You may, especially with a class lacking in confidence, find it useful to print out the data for everyone, and look through them together, talking about what is happening at each stage.
Social and human context
NASA’s raw data were actually provided with distances in nautical miles above the surface of the Earth, and with velocities in knots.
External reference
This activity is taken from Advancing Physics chapter 11, 160T
