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## Gravitational fields

Lesson for 16-19

Students will have been aware of the effects of gravity since they first fell down, so it is probably the force they feel most familiar with. In this topic, they will extend their terrestrial experience to develop a much broader picture of the force of gravity, and its representation in the form of a gravitational field.

Students are often initially surprised by the fact that they pull up on the Earth with exactly the same force that the Earth pulls down on them. This topic will develop the idea that Newton’s Law of Universal Gravitation is precisely that: universal. Every single object with a mass in the universe gravitationally attracts every other object with mass.

## Episode 400: Preparation for gravitational fields topic

Teaching Guidance for 16-19

- Level Advanced

In this topic, students will extend their terrestrial experience to develop a much broader picture of the force of gravity, and its representation in the form of a gravitational field.

## Main aims of this topic

Students will:

- understand the concept of a field and its depiction with field lines
- understand and use Newton’s law of universal gravitation
- understand what is meant by an
*inverse square law* - understand and use the term
*field strength* - understand and use the terms
*energy*and*potential*and know that the zero of potential is defined to be at infinity - understand that the gradient of potential is related to field strength
- understand the motion of objects in gravitational fields (circular orbits only, including geosynchronous orbits)

## Prior knowledge

Students will have a vague notion of the concept of a field, and is representation by lines of force, from studies of magnetism. When developing the idea of a gravitational field, you can refer back to these earlier ideas.

In earlier studies, your students should have covered the following concepts:

mass; weight *W* = *m**g*
; *g* defined as field strength in N kg^{-1} or as gravitational acceleration in m s^{-2}; all bodies in free-fall (no resistive forces) accelerate at the same rate; work; energy; energy in the gravitational store close to the Earth’s surface = *m**g**h*
; conservation of energy; Newton’s laws of motion.

In their advanced studies, they should also have covered circular motion and centripetal force.

## Where this leads

The concept of a field is hugely important in physics (traditionally as a model to explain action-at-a-distance

effects). In classical physics the known forces (gravitation and electromagnetism) were modelled incredibly successfully as fields, with a great many applications to industry and society as a whole.

More recently, QED (quantum electrodynamics), often quoted as the most accurately tested theory known to man, is in essence nothing more than a theory about electric fields. Although the modern view of a field is quite different to the classical one, fields are an essential cornerstone of physics.

Gravitation is a great starting point for the subject of fields, introducing and developing concepts that are found in all field physics (such as field strength and potential), and it also gives an opportunity for revising Circular Motion and Energy.

### Up next

### Newton's law of universal gravitation

## Episode 401: Newton's law of universal gravitation

Lesson for 16-19

- Activity time 65 minutes
- Level Advanced

This episode introduces Newton’s law of universal gravitation for point masses, and for spherical masses, and gets students practising calculations of the force between objects. The meaning of inverse square law

is discussed.

Lesson Summary

- Discussion: Introduction to Newton’s law of universal gravitation (5 minutes)
- Discussion: Newton’s law of universal gravitation:
*F*_{gravity}=*G**M**m**r*^{ 2}
(10 minutes)
- Worked examples: Using
*F*_{gravity}=*G*×*m*_{1}×*m*_{2}*r*^{ 2}(25 minutes) - Student questions: More practice with
*F*_{gravity}=*G**m*_{1}*m*_{2}*r*^{ 2}(20 minutes)

## Discussion: Introduction to Newton’s law of universal gravitation

Here are some questions and answers which lead towards Newton’s law of universal gravitation.

What causes the weight that each student feels? (gravitational attraction by the Earth.)

What affects the size of the Earth’s pull on you? Why would you weigh a different amount on the Moon? (Your mass, and its mass.)

If the Earth is pulling down on you, then what else must be occurring, by Newton’s 3rd Law? (You must be pulling up on the Earth with a force equal to your weight.)

What happens to the strength of the pull of the Earth as you go further away from it? (It gets weaker – most students guess this correctly from the incorrect assumption that in space, astronauts are weightless!)

So, in summary the force depends upon the masses of the Earth and you, and weakens with distance. This is all embodied in Newton’s law of universal gravitation

## Discussion: Newton’s law of universal gravitation

Present the equation which represents Newton’s law of universal gravitation.

*F*_{gravity} = *G**m*_{1}*m*_{2}*r*^{ 2}

*F*_{gravity} is the gravitational force of attraction in newton

*m*_{1} , *m*_{2} are the interacting masses, in kilogram

*r* is the separation of the two masses in metre

*G* is known as the universal gravitational constant

( *not* to be confused with little

g). It sets the strength of the gravitational interaction in the sense that if it were doubled, so would all the gravitational forces.

*G* = 6.67 × 10^{-11} N m^{2} kg^{-2}

Show how the units can be worked out by rearranging the original equation.

This law applies between point masses, but spherical masses can be treated as though they were point masses with all their mass concentrated at their centre.

This force is *always* attractive. In some texts you will see a minus sign in the equation, so that *F*_{gravity} = − *G**m*_{1}*m*_{2}*r*^{ 2}. This minus sign is there purely to indicate that the force is attractive (it’s a relic from the more correct, but well beyond the syllabus, vector equation expressing Newton’s Law of universal gravitation). It’s simplest to calculate the magnitude of the force using

*F* = *G**m*_{1}*m*_{2}*r*^{ 2} , and the direction is given by the fact that the force is always attractive.

Every object with a mass in the universe attracts every other according to this law. But the actual size of the force becomes very small for objects very far away. For example, the Sun is about one million times more massive than the Earth, but because it’s so far away, the pull on us from the Sun is dwarfed by the pull on us from the Earth (which is around 1650 times greater). As the separation of two objects increases, the separation^{2} increases even more, dramatically. The gravitational force will decrease by the same factor (since separation^{2} appears in the denominator of the equation). This is an example of an inverse square law

, so called because the force of attraction varies in inverse proportion to the square of the separation.

## Worked examples: Using *F* = *G**m*_{1}*m*_{2}*r*^{ 2}

You can work through these examples, or you can set them as a task for your students if you feel they will be able to tackle them.

Episode 401-1: Worked examples; Gravitation forces (Word, 29 KB)

## Student questions: More practice with *F* = *G**m*_{1}*m*_{2}*r*^{ 2}

Episode 401-2: Newton’s gravitational law (Word, 53 KB)

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### Up next

### Fields field lines and field strength

## Episode 402: Fields, field lines and field strength

Lesson for 16-19

- Activity time 60 minutes
- Level Advanced

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.

Lesson 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:

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 3-dimensional, but of course on paper, we can only take a 2-dimensional slice of it. This is a *radial* or *spherical* field.

Close to the surface of the Earth, the field lines look like:

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:

- 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.
- 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* is the weight in newtons. Thus *g* has units N kg^{-1}.

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 *G**M**m**r*^{ 2}. Thus the attractive force per kg of mass of the object (mass *m*) is *G**M**m**m**r*^{ 2}.

Thus,
*g* = *G**M**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:

- 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.
- In some texts a minus sign is present in this equation, so it reads
*g*= −*G**M**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*=

*G*

*M*

*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.

Episode 402:1 Field strength – students’ sheet (Word, 31 KB)

## Spreadsheet exercise

Here, students can try analysing data from the Apollo 11 mission.

Episode 402-2: Analysing data from the Apollo 11 mission to the moon (Word, 29 KB)

Episode 402-3: Data from the Apollo 11 mission (Word, 85 KB)

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### Up next

### Orbital motion

## Episode 403: Orbital motion

Lesson for 16-19

- Activity time 85 minutes
- Level Advanced

In this episode, students will learn how to combine concepts learned in the study of circular motion with Newton’s Law of Universal Gravitation to understand the (circular) motion of satellites. While real orbits are very rarely circular, they should still find it satisfying to see the relative simplicity with which the laws they have learned lead to an explanation of Kepler’s 3^{rd} law, which had previously been only an empirical law.

Lesson Summary

- Discussion: Newton’s cannonball (10 minutes)
- Discussion: Kepler’s laws (20 minutes)
- Discussion: Geostationary orbits (5 minutes)
- Worked examples: Orbital motion (25 minutes)
- Student questions (15 minutes, 25 minutes)

We will start with a thought-experiment, often attributed to Newton, in which students will learn that astronauts orbiting the Earth are not actually weightless, but feel as though they are because they are in constant free-fall.

There are some applets available to aid discussion in this episode. If it is at all possible, students should try to play around with these applets themselves too.

## Discussion: Newton’s cannonball

What is the motion of a cannonball fired horizontally on the Earth’s surface, neglecting air resistance? (If you fire a cannonball horizontally (neglecting air resistance), it will travel some distance before it strikes the ground. If you fire it much faster (again horizontally) then it will travel much further before hitting the ground.)

(Newton asked what would happen if you fired faster again. He reasoned that the cannonball would maintain a constant height from the surface of the Earth – in other words, it would move in a circular orbit.)

The cannonball is fired horizontally from the top of a mountain at a user-definable initial speed. For speeds that are too slow, the cannonball falls towards the Earth’s surface. Above a certain speed (about 15 500 mph) the cannonball orbits the Earth (the applet stops it after one orbit). At a speed of just over 16 000 mph this orbit is circular – i.e. at a fixed height above the Earth’s surface.

How would you feel if you were orbiting like the cannonball? (Well, you would be falling the whole time, whilst moving horizontally at the same time. Imagine falling in a lift with the cable cut. You’d have butterflies in your stomach and if you held a ball out in front of you and let go it wouldn’t move relative to you. This is exactly what would happen if there were no gravitational field, and it is in this sense that you would feel weightless. When astronauts in an orbiting spacecraft are said to be weightless, it is not actually true – they are still in (a relatively strong part of) the Earth’s gravitational field, and so they have a weight, but they just don’t feel it because both they and the spacecraft surrounding them are constantly falling towards the centre of the Earth (whilst also moving tangentially around the orbit).)

So, launching a satellite into a circular orbit requires it being set off with the right tangential speed for the height at which you want it. The following applet allows you to play around with initial speeds/directions to see what would happen. It’s quite surprising to some students to realise that at a particular height, there is a range of speeds and directions that you can start the satellite at, for which it will be trapped in orbit (although elliptical, not circular).

## Discussion: Kepler’s laws

Kepler discovered three laws of planetary motion based on the painstakingly recorded observations of Brahe in the early 17^{th} century. These laws were however empirical – that is they fitted the patterns observed but had no physical explanation. Newton later managed to explain the laws with his laws of motion and Law of Universal Gravitation. Kepler’s three laws were:

All of the planets move in elliptical orbits with the Sun at one focus.

The radius vector (imaginary line between the Sun and the planet) sweeps out equal areas in equal times.

The square of the period of a planet’s orbit is directly proportional to the cube of its mean distance from the Sun.

These laws apply for all satellite motions (not just for the planets around the Sun). Remember that circular orbits are just a special case of elliptical orbits.

Each of these laws can be demonstrated with the applet on the webpage below (instructions on the webpage):

Keplar motion – NTNUJAVA Virtual Physics Laboratory

We shall concentrate on circular orbits at constant speed only (elliptical orbits are well beyond the scope of A-level courses). The first two of Kepler’s laws are trivial for circular orbits. An ellipse has two foci normally but in the case of a circle they are coincident and at the centre of the circle. Also if the planet moves at constant speed in a circle then the radius vector has to sweep out equal areas in equal times. So we shall concentrate on deriving the third law for circular orbits.

What is the only force acting on a planet orbiting the Sun? (The gravitational pull from the Sun, neglecting the negligible pulls from other objects in the solar system and beyond.)

And if the planet is moving at constant speed in a circle it must have centripetal force acting on it.

What is the expression for centripetal force? *m**v*^{ 2}*r*

For the gravitational force ? *G**M**m**r*^{ 2}

Equating these expressions, we have
*m**v*^{ 2}*r* = *G**M**m**r*^{ 2}
(with *M* being the mass of the Sun and *m* being the mass of the planet).

But the speed *v* can be calculated as distance travelled in one orbit (2 π *r* ) divided by the time taken, *T* :

*v* = 2 π *r**T*

Plugging this into the previous equation, and cancelling the m terms on both sides gives us:

*G**M**r*^{ 2} = 4 π ^{2}*r**T*

Rearranging again gives:
*T*^{ 2} = 4 π ^{2}*G**M**r*^{ 3}

Or in other words, since
4 π ^{2}*G**M*
is constant, the square of the period of orbit is proportional to the cube of the radius of the orbit – Kepler’s third law. A powerful result proved simply by the laws of motion and gravitation. This law shows that the further away a planet is, the longer its period of orbit (but in a non-linear way).

Although this law (as are all of Kepler’s laws) was originally discovered for the motion of the planets, it works equally well for all satellite motion – for example the motion of the Moon and artificial satellites around the Earth. We just need to realise that the actual *M* in the constant (4 * π *^{ 2}*G**M*) is the mass of the Earth in that case (or the mass of whichever body is being orbited).

## Discussion: Geostationary orbits

We can apply this idea to geostationary orbits. We know from Kepler’s third law that the further away a satellite is from the body it is orbiting, the longer its orbital period.

If an orbiting satellite had a period of 24 hours, and you saw it overhead at, say 10.00 am, when would you next see it overhead? (Because both the Earth would have completed one rotation in the same time it took the satellite to complete one orbit, it would next be overhead at 10.00 am the next day. Such a satellite is said to be *geosynchronous*.)

A difficult question – if you wanted the satellite to remain directly overhead (i.e. above a fixed point on the Earth) at all times (not just once per day) where on the Earth would you have to be? (All satellites (in circular orbits) orbit around the centre of the Earth. The only points on the Earth’s surface that orbit around the centre of the Earth are those on the equator. Thus, you would have to be on the equator.)

If a satellite has a period of 24 hours and orbits above the equator such that it always appears to be above one point on the equator, it is known as a *geostationary* satellite, and its orbit is a geostationary orbit. A geostationary orbit has a radius of around 42 000 km (over 6 times the radius of the Earth) – i.e. at a height above the Earth’s surface of around 36 000 km. (These figures will be calculated in the worked examples that follow).

**Image courtesy of the Celes Trak website**

Basics of the geostationary orbit – Celes Trak

The orbit labelled GEO above is geosynchronous, but not geostationary, because the satellite would appear from the equator to wander first north, and then south and then back again over a 24 hour period.

The orbit labelled GSO is geostationary.

Geostationary satellites are predominantly used for communications. Satellite TV companies use geostationary satellites to cover a constant area on the Earth’s surface – hence you point your satellite dish receiver in the direction of the geostationary satellite. 3 geostationary satellites placed into orbit 120 degrees apart above the equator would be able to cover the entire Earth (except for very near the poles). Because geostationary satellites have to be launched so high (other satellites orbit as low as a few hundred km), the energy and costs required for launching a satellite into geostationary orbit are high.

## Worked examples: Orbital motion

These examples look at orbital motion under gravity. They can be tackled independently by most students, but solutions are provided if you wish to use them as worked examples.

Episode 403-1: Worked examples – orbital motion (Word, 34 KB)

## Student questions

Episode 403-2: Using Kepler's third law (Word, 43 KB)

Episode 403-3: Changing orbits (Word, 53 KB)

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### Up next

### Gravitational potential energy

## Episode 404: Energy and gravitational fields

Lesson for 16-19

- Activity time 110 minutes
- Level Advanced

In this episode, students will appreciate the changes to the ways that energy is stored as a body moves in a gravitational field. They have seen this concept before, for a uniform gravitational field, in the form

change in energy stored gravitationally, or change in gravitational potiential energy = m x g x Δh

where Δh = change in height, but this will be generalised to non-uniform fields around point or spherical masses. They will then be introduced to the concept of gravitational potential and its uses, before finally making the link between that rate at which gravitational potential changes from place to place and the field strength.

Note that the mention of infinity often gets students’ minds racing, and puzzling over seeming paradoxes such as “If the gravitational field is infinite in extent, what does it mean to be at infinity where the field is zero?” It’s best to approach this pragmatically; by infinity, we really mean as far away from all masses in the universe as we need to be to make the gravitational forces negligible in whatever context we are looking at.

This is quite a long episode, and worth spending the time on to give a good grounding in these ideas. They will be important again with electric fields.

Lesson Summary

- Discussion: Work and energy (5 minutes)
- Worked examples: Energy in a constant gravitational field (10 minutes)
- Discussion: Energy in a non-uniform field (10 minutes)
- Discussion: Potential (5 minutes)
- Worked examples: Energy and potential (15 minutes)
- Student questions (15 minutes)
- Spreadsheet exercise (20 minutes)
- Discussion: Equipotentials, potential gradient and field strength (10 minutes)
- Spreadsheet exercise: Potential gradient and field strength (20 minutes)

## Discussion: Work and Energy

This draws on what students should already know about work and energy.

What is work? (Work is done when a force moves its point of application through some distance in the direction of the force. In the absence of frictional forces, this work done is stored as the energy of the body on which the force was acting.)

If I lift an object up from the floor to above my head (demonstrate it!), have I done work on it? (Yes. I applied an upwards force on the object which itself moved upward too.)

What happened to the work I did on the object? (It is stored as energy of the object. We say that the energy is stored gravitationally. 'Gravitationally' because we have to do work against gravity to lift the object. If you have something breakable (but safe and cheap!) they will really see this energy being realised.)

## Worked examples: Energy in a constant field

Now, students have come across these ideas before at pre-16 level in the form:

Change in *E*_{G} = *m**g**h*.

It is as well to revise this with a few worked examples now.

Episode 404-1: Energy in a constant field (Word, 26 KB)

## Discussion: Energy in a non-uniform field

Now you can extend these ideas to energy in a non-uniform field.

All the previous examples involved changes to the energy stored gravitationally near the surface of the Earth. What would be the problem if we wanted to use the same equation to work out changes to the energy stored gravitationally for, say, a rocket launched to the moon? (The gravitational field strength is not constant – the value of *g* changes.)

We therefore need another way of calculating changes to the energy stored gravitationally in non-uniform fields. The full treatment of how we arrive at this formula requires off-syllabus calculus that would actually be accessible to more able students. We find that we can calculate energy of a mass *m*
at a point distance *r* from a (point or spherical) mass *M*
by:

*E*_{G} = − *G**M**m**r*

There are several very important points to note about this equation:

- We know that the further you get from an object, the greater the energy stored in the gravitational field. (As something must have done more work against gravity to get you there). Thus when you are infinitely far away, you have as high an energy relative to it as possible. We choose (arbitrarily) to make the value of energy of all bodies at infinity zero. Then since this is the highest value of energy, all real values of energy stored gravitationally (closer than infinity) must be negative. Therefore the minus sign in the equation is
*not*optional; it must always be included and all values of energy stored in a gravitational field are negative. (This is not the case when we come on to electric fields, because they can be repulsive too). - Note that we have written energy here, and not
Change in energy

. By defining a point relative to which all energy is measured, we can now talk about*absolute*values of energy rather than just changes. This point is at infinity (see note 1 above). - Note that energy follows an inverse proportion law (1r) and not an inverse square law (1
*r*^{ 2})

## Discussion: Potential

The weight of an object in the Earth’s gravitational field depends upon the mass of the object (as well as the mass of the Earth). However, as we have already seen, the field strength at a point is independent of the object placed there (because it is defined as force per unit mass of the object). Thus we can think of field strength as a property of the field at a point, and not the particular object placed there.

Similarly, the energy stored in the gravitational field by moving a body to a place relative to the Earth depends upon the mass of the object (as well as the mass of the Earth) and the distance between them. How do you think we can get a quantity related to energy in the field at a point, which does not depend upon the object placed there? (By looking at the energy stored per unit mass of the object, thus removing the dependence on the mass of the object just as we did with field strength.)

We define the potential at a point in a field as the energy *per unit mass* stored by placing 1kg at that point in the field. We can get equations for potential using this definition. For a field due to a (point or spherical) mass *M*, we have:

*E*_{G} = − *G**M**m**r*

And so the potential, *V* , is given by:

*V* = − *G**M**r*

A few points to make:

- This only relates to the field due to a (point or spherical) mass M.
*V*is measured in J kg^{-1}. It follows an inverse proportion law (1*r*) not inverse square (1*r*^{ 2}).- Just as with the equation for energy, the minus sign is not optional. All real potentials are negative, and the zero of potential is at infinity (since all objects store no energy gravitationally at infinity).
- Potential, like field strength is a property of the field at a point, and is independent of the object placed there. Two objects with different masses at the same point in the field are subject to the same potential, but have different energies.
- For uniform fields (e.g. close to the surface of the Earth), we can use

change in potential = *g**h*

(since change in *E*_{G} = *m**g**h*
and
potential = *E*_{G}*m*).

(Potential will be returned to again when we study electric fields. There, differences in potential (or potential differences, pds) are what we often call voltages.)

## Worked examples:Energy and potential

Episode 404-2: Energy and potential (Word, 30 KB)

## Student questions

These questions relate to a spacecraft travelling from the Moon to the Earth.

You may wish to omit questions 3–5 inclusive if your students have not covered the topic of momentum.

Episode 404-3: Gravitational field between Earth and moon (Word, 74 KB)

## Spreadsheet exercise

You can extend the spreadsheet activity from Episode 402 by including the idea of gravitational potential.

Episode 402-3: Data from the Apollo 11 mission (Word, 85 KB)

Episode 404-4: Analysing Apollo 11 data, gravitational potential (Word, 29 KB)

## Discussion: Equipotentials, potential gradient and field strength

Equipotentials join points of equal potential. They are very simple in the cases of uniform fields (close to the surface of the Earth) and radial fields (for point and spherical masses):

How far apart are the equipotential lines in the first diagram? (Since change in potential for a constant field is simply *g**h* (where *h* is change in height), we have *g**h* = 9.8 J kg^{-1}
}. Therefore *h* = 1 m – the lines are 1 m apart.)

What shapes are equipotentials in the real world? (Equipotentials are surfaces rather than lines in the real 3-dimensional world (i.e. horizontal planes rather than lines close to the surface of the Earth, and concentric spheres rather than concentric circles about a point/spherical mass), but we can only capture a slice of them on paper.)

Equipotentials are always perpendicular to field lines. Diagrams of equipotential lines give us information about the gravitational field in much the same way as contour maps give us information about geographical heights.

What does it mean on a contour map if the contours are very close together? (On a contour map, the contours may be marked off at, say, 5 m intervals. Therefore, if they are close together, it means that the land on which they lie must be very steep.)

What do you think it therefore means if equipotential surfaces are close together? (An educated guess would be that it means that the gravitational field is very strong there. That would be correct – if the equipotentials are close together, a lot of work must be done over a relatively short distance to move a mass from one point to another against the field – i.e. the field is very strong. Hence on the drawing above of the equipotentials around point or spherical masses, the equipotential surfaces get further and further apart as the field strength decreases with distance.)

In fact, the field strength is given by the negative of the gradient of the potential:

*g* = − Δ *V* Δ *r*

For students that might struggle with a derivative as above, it could be introduced as:

*g* = − change in potentialchange in distance.

## Spreadsheet exercise: Potential gradient and field strength

In this exercise, students will manipulate (calculated) raw data on the variation of potential with height above the Earth’s surface. By the end, they will calculate the change in potential per metre (i.e. the potential gradient) and compare it to field strength and find that the two are equal. When they calculate the change in potential between say 200 km and 300 km, and divide it by the 100 km distance to get a field strength, they are actually getting an *average* field strength between 200 km and 300 km, which is likely to be very close to the field at 250 km (though not exactly because the variation in potential is non-linear). Hence the field strengths on sheet 2 being given at 50 km, 150 km, etc.

They are expected to have used spreadsheets before and are asked to do some fairly simple spreadsheet manipulations.

Episode 404-5: Potential gradient and field strength (Word, 28 KB)

Episode 404-6: Excel file potential gradient and field strength

(Word, 53 KB)