physics ch 09_nopw

8/9/2019 Physics Ch 09_noPW

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MOTION IN FIELDS9.1 Projectile motion

9.2 Gravitational eld, potential and energy

9.3 Electric eld, potential and energy

9

9.1.1 State the independence of the vertical andthe horizontal components of velocity for aprojectile in a uniform eld.

9.1.2 Describe and sketch the trajectory ofprojectile motion as parabolic in theabsence of air resistance.

9.1.3 Describe qualitatively the effect of airresistance on the trajectory of a projectile.

9.1.4 Solve problems on projectile motion. IBO 2007

9.1.1 9.1.3 P ROJECTILE MOTION

Projectile launched horizontally

In this section we shall look at the motion of a projectilethat is launched horizontally from a point above thesurface of the Earth.

In the Figure 901a projectile is red horizontally from acliff of height h with an initial horizontal velocity v h

Our problem is effectively to nd where it will land, withwhat velocity and the time of ight. We shall assume thatwe can ignore air resistance and that the acceleration dueto gravity g is constant.

v h

g ms2

cliff h

dFigure 901 The path of a horizontal projectile

Since there is no force acting in the horizontal directionthe horizontal velocity will remain unchanged throughoutthe ight of the particle. However, the vertical accelerationof the projectile will be equal to g .

We can nd the time of ight t by nding the time it takesthe particle to fall a height h.

To start with, we consider only the vertical motion of theobject:

u = 0

s = h

t = ?

ha = g

v vv=

Figure 902 The vertical motion of the object

9.1 PROJECTILE MOTION


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Time of ight

is is calculated from the denition of acceleration

i.e., using v = u + at , we have that

v v 0 g t t +v v g -----= =

where v v is the vertical velocity with which the objectstrikes the ground.

To nd v v we use the equation v 2 u2 2as+= , so that

as the initial vertical velocity ( u) is zero, the accelerationa = g and s = h.

en v v 2 02 2 g h v v + 2 gh= =

From which we have that

a

t 2 g h g

------------- 2 h g ------= =

nd so, the time of ight is given by

t 2 h g ------=

Since the horizontal velocity is constant, the horizontaldistance d that the particle travels before striking theground is v h t . (i.e., using s = ut + at

2 = ut, wherein the horizontal direction we have that a = 0 andu = v h = constant)

is gives

d v v 2h g ------=

is is the general solution to the problem and it is notexpected that you should remember the formula for thisgeneral result. You should always work from rst principleswith such problems.

An interesting point to note is that, since there is nohorizontal acceleration, then if you were to drop aprojectile from the top of the cliff vertically down, at themoment that the other projectile is red horizontally,then both would reach the ground at the same time. isis illustrated by the copy of a multiash photograph, asshown in Figure 903.

Figure 903 Using multi-ash photography

is is irrespective of the speed with which the particleis red horizontally. e greater the horizontal speed, thefurther this projectile will travel from the base of the cliff.It is also possible to show that the path of the particle isparabolic.

To nd the velocity with which the particle strikes theground we must remember that velocity is a vectorquantity. So, using Pythagoras theorem at the point ofimpact (to take into account both the vertical componentof velocity and the horizontal component of velocity) wehave that the velocity has a magnitude of

vv

vh

V

V v v 2 v h

2+=

and the direction will be given by nding

ar cv v v h---- tan=

where the angle is quoted relative to the horizontal. If theangle is to be given relative to the vertical then we evaluate

90 ( )


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or

ar cv hv v ----- tan=

Notice that at impact the velocity vector is tangential tothe path of motion. As a matter of fact, the velocity vector

is always tangential to the path of motion and is made upof the horizontal and vertical components of the velocitiesof the object.

Projectiles launched at an angle

to the horizontal

Consider the problem of a projectile that is launched fromthe surface of the Earth and at an angle to the surface of theEarth. Ignore air resistance and assume that g is constant.In Figure 905 the particle is launched with velocity v atangle to the surface.

v

v v

v h

Range

maximum height (H )

v v 0 a, g = =+ ve

T 2v v g --------=

g time to impact

y

x

Figure 905 Projectile launched at an angle

e vertical component of the velocity, v v , is

v v v sin=

e horizontal component of the velocity, v h, is

v h v cos=

As in the case of the projectile launched horizontally,

there is no acceleration in the horizontal direction and theacceleration in the vertical direction is g .

If we refer the motion of the projectile to a Cartesian co-ordinate system, then a er a time t, the horizontal distancetravelled will be given by

x v ht v cos( )t = =

and the vertical distance can be found by using theequation

s ut 1

2--- at

2+=

so that

y v v t 1

2--- g ( )t 2+= y v sin( )t 1

2--- g t

2=

If we now substitute for

t x v cos---------------=

into this equation we get

y v sin( ) x v cos-------------- - 1

2--- g x

v cos--------------- 2=

x sin cos------------

1

2--- g x

v -- 2 1

cos------------ 2=

x 12--- g x

v -- 2 2sectan=

is is the general equation of the motion of the projectilethat relates the vertical and horizontal distances. isequation is plotted below for a projectile that is launchedwith an initial speed of 20 m s 1 at 60 to the horizontal.e path followed by the projectile is a parabola.

60

20

20

20 60 17.32sin

20 60 cos 10=

10

10

10

10

17.32

v v 0= y

x

Figure 906 The parabolic path

e maximum height H that the projectile reaches can befound from the equation

v 2

u2

2as+=

where u is the initial vertical component of the velocityand v the nal (vertical component) of the velocity at thehighest point, where at this point, the vertical componentis zero. So that,

02 v sin( )2 2 g H += H v 2 2sin

2 g -------------------=

If we use the gures in the example above ( v = 20 m s1, = 60)

with g = 10 m s2 then we see that H 20 60 sin( )

2 10------------------------------ 15= =


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i.e., the object reaches a maximum height of 15 m.

e time T to reach the maximum height is found usingv= u + a t , such v = 0,

u v sin= and a = g , to give

0 v g T g T sin v sin= =

Hence,

T v sin g

--------------=

For the example above the value of T is 1.73 s. is means(using symmetry) that the projectile will strike the ground3.46 s a er the launch. e horizontal range R is given byR v cos( ) 2 T = which for the example gives R = 34.6 m.

(We could also nd the time for the projectile to strike theground by putting y = 0 in the equation

y v sin( )t 12--- g t

2=

Although we have established a general solution,essentially solving projectile problems, remember thatthe horizontal velocity does not change and that whenusing the equations of uniform motion you must use thecomponent values of the respective velocities. Do not tryto remember the formulae.

The effect of air resistance onprojectile motion

We have seen that in the absence of air resistance, the pathfollowed by a projectile is a parabola and that the pathdepends only on the initial speed and angle of projection.Of course, in the real world all projectiles are subject to airresistance. Fig 907 shows the free body force diagram for aprojectile subject to air resistance.

vertical drag

horizontal drag

weight

Figure 907 The effect of air resistance

Experiment shows that both the horizontal and verticaldrag forces depend on the speed of the projectile. eeffect of the horizontal drag will be to foreshorten therange of the projectile and the effect of the vertical dragwill be to reduce the maximum height reached by theprojectile. However, the presence or air resistance alsomeans that the mass of the projectile will now affect

the path followed by the projectile. In the absence ofair resistance there is no acceleration in the horizontaldirection and the acceleration in the vertical direction is g ,the acceleration of free fall. With air resistance present, tond the horizontal ( aH)and vertical ( aV) accelerations wehave to apply Newtons second law to both the directions.If we let the horizontal drag equal kv H and the verticaldrag equal Kv V where k and K are constants and v H and v V are the horizontal and vertical speeds respectively at anyinstant, then we can write

H H kv ma= and V V mg Kv ma =

From this we can now see why the mass affects the pathsince both aH and aV depend on height. ( For those of youdoing HL maths, you will realise that the above equacan be written as differential equations but nding solution is no easy matter! ) We have here, another exampleof the Newton method for solving the general mechanicsproblem- know the forces acting at a particular instantand you can in principle predict the future behaviour ofthe system.

9.1.4 S OLVE PROJECTILE PROBLEMSNotice that at impact the velocity vector is tangential tothe path of motion. As a matter of fact, the velocity vectoris always tangential to the path of motion and is made upof the horizontal and vertical components of the velocitiesof the object.

Example

A particle is red horizontally with a speed of 25 ms -1 fromthe top of a vertical cliff of height 80 m. Determine

(a) the time of ight

(b) e distance from the base of the cliff where itstrikes the ground

(c) the velocity with which it strikes the ground


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Solution

cliff

+veVertical: u = 0

a = g

Horizontal: u = 25 m s-1

a = 0

80 m

(a) e vertical velocity with which it strikes the groundcan be found using the equation

v 2 u2 2as+= , with u = 0, a = g and s = 80 (= h).

is then givesv v 2 g h 2 10 80= =

= 40

at is, the vertical velocity at impact is 40 m s1.

e time to strike the ground can be found usingv = u + at, with u = 0, a = g and v = v v . So that,

t v v g -----

4010------= = = 4.

at is, 4 seconds.

(b) e distance travelled from the base of the cliff using

s ut 12--- at 2+= , with u = 25,

a = 0 and t = 4 is given by

s 25 4= = 100.

at is, the range is 100 m.

(c) e velocity with which it strikes the ground is given by the resultant of the vertical and horizontalvelocities as shown.

40

Ground level

v v =

v h 25=

V

e magnitude of this velocity is

402

252

+ 47= m s1

and it makes an angle to the horizontal of

ar c 4025------ tan=

= 68 (or to the vertical of 32 ).

Conservation of energy and

projectile problems

In some situations the use of conservation of energy canbe a much simpler method than using the kinematicsequations. Solving projectile motion problems makesuse of the fact that Ek E p+ constant= at every pointin the objects ight (assuming no loss of energy due tofriction).

In Figure 909, using the conservation of energy principlewe have that the

Total energy at A = Total energy at B = Total energy at C

i.e.,1

2--- mv A

2 1

2--- mv B

2mg H +

1

2--- mv C

2mg h+= =

Notice that at A, the potential energy is set at zero ( h = 0).

v A

v v

v h

H

12---mv B

2 mg H +v B

h

12---mv C

2mgh+

A

B

C

v C

12---mv A

20+

Figure 909 Energy problem


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Example

A ball is projected at 50 ms -1 at an angle of 40 o above thehorizontal. e ball is released 2.00 m above ground level.Taking g = 10 m s -2, determine

(a) the maximum height reached by the ball

(b) the speed of the ball as it hits the ground

Solution

2 m

A

B

C

H

R = range

50

40

a. e total energy at A is given by

Ek E p+ 1

2---m 50.0( )2 mg 2.00+=

1250 m 20 m+=1270 m=

Next, to nd the total energy at B we need to rst determinethe speed at B, which is given by the horizontal componentof the speed at A.

Horizontal component:50.0 40 cos 38.3= m s1.

erefore, we have that

Ek E p+

1

2---

m 38.3

( )2

mg H +=

y 15 t 5 t 2=

733.53 m 10 mH +=

Equating, we have1270 m 733.53 m 10 mH +=

1270 733.53 10 H +=

H 53.6=

at is, the maximum height reached is 53.6 m.

b. At C, the total energy is given by

Ek E p+ 1

2--- mv C

2mg 0+ 1

2--- mv C

2= =

Using the total energy at A,Ek E p+ 1270 m=

Equating, we have that

1270 m 1

2--- mv C

2v C

2 2540= =

v C 2540 50.4= =

at is, the ball hits the ground with a speed of 50.4 m s1.

Exercise

1. A projectile is red from the edge of a vertical cliffwith a speed of 30 m s1 at an angle of 30 to thehorizontal. e height of the cliff above the surfaceof the sea is 100 m.

(a) If g = 10 m s2 and air resistance is ignoredshow that at any time t a er the launch the vertical displacement y of the projectile asmeasured from the top of the cliff is givenby:

y = 15t - 5t 2

Hence show that the projectile will hit the surfaceof the sea about 6 s a er it is launched.

(b) Suggest the signicance of the negative value of t that can be obtained in solvingthe equation?

(c) Determine the maximum height reached bythe projectile and the horizontal distance towhere it strikes the sea as measured fromthe base of the cliff.


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9.2.1 Dene gravitational potential and

gravitational potential energy .

9.2.2 State and apply the expression forgravitational potential due to a pointmass.

9.2.3 State and apply the formula relatinggravitational eld strength togravitational potential gradient.

IBO 2007

9.2.1 G RAVITATIONAL POTENTIALWe have seen that if we li an object of mass m to a height h above the surface of the Earth then its gain in gravitationalpotential energy is mgh. However, this is by no means thefull story. For a start we now know that g varies with h andalso the expression really gives a difference in potentialenergy between the value that the object has at the Earthssurface and the value that it has at height h. So what wereally need is a zero point. Can we nd a point where thepotential energy is zero and use this point from which tomeasure changes in potential energy?

Well, the point that is chosen is in fact innity. At innitythe gravitational eld strength of any object will in factbe zero. So let us see if we can deduce an expression forthe gain in potential energy of an object when it is li edfrom the surface of the Earth to innity. is in effectmeans nding the work necessary to perform this task.

r

r r r +

r =

m

g

A B

M e

Figure 911 Gravitational forces

In the diagram we consider the work necessary to movethe particle of mass m a distance r in the gravitationaleld of the Earth.

e force on the particle at A is F G M em

r 2----------------=

If the particle is moved to B, then since r is very small,we can assume that the eld remains constant over thedistance AB. e work W done against the gravitationaleld of the Earth in moving the distance AB is

W GM em

r 2---------------- r =

(remember that work done against a force is negative)

To nd the total work done, W , in going from the surfaceof the Earth to innity we have to add all these little bitsof work. is is done mathematically by using integralcalculus.

W GM e m

r 2----------------

r d

R

GM em 1r 2----- r d

R

GM em 1r --- R

= = =

GM em 0 1

R---

=

G M emR

----------------=

Hence we have, where R is the radius of the Earth, that thework done by the gravitational eld in moving an object ofmass m from R (surface of the Earth) to innity, is given by

W GM em

R----------------=

We can generalise the result by calculating the worknecessary per unit mass to take a small mass from thesurface of the Earth to innity. is we call the gravitationalpotential, V , i.e.,

We would get exactly the same result if we calculated thework done to bring the point mass from innity to thesurface of Earth. In this respect the formal denition ofgravitational potential at a point in a gravitational eldis therefore dened as the work done per unit mass inbringing a point mass from innity to that point.

Clearly then, the gravitational potential at any point inthe Earths eld distance r from the centre of the Earth(providing r > R) is

V GM e

r ----------- -=

9.2 GRAVITATIONAL FIELD, POTENTIAL ANDENERGY


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e potential is therefore a measure of the amount of workthat has to be done to move particles between points in agravitational eld and its unit is the J kg 1. We also notethat the potential is negative so that the potential energyas we move away from the Earths surface increases until itreaches the value of zero at innity.

If the gravitational eld is due to a point mass of mass m,then we have the same expression as above except that M e is replaced by m and must also exclude the value of thepotential at the point mass itself i.e. at r = 0.

We can express the gravitational potential due to the Earth(or due to any spherical mass) in terms of the gravitationaleld strength at its surface.

At the surface of the Earth we have

g 0 ReGM

Re---------=

So that,

g 0Re2 GM =

Hence at a distance r from the centre of the Earth thegravitational potential V can be written as

V GM e

r ----------- -

g 0 Re2

r ------------= =

e potential at the surface of the Earth

(r = Re) is therefore - g 0Re

It is interesting to see how the expression for thegravitational potential ties in with the expression mgh.e potential at the surface of the Earth is - g 0Re (see theexample above) and at a height h will be g 0 Re h+( ) ifwe assume that g 0 does not change over the distance h.e difference in potential between the surface and theheight h is therefore g 0h. So the work needed to raise anobject of mass m to a height h is mgh , i.e., m difference

in gravitational potential

is we have referred to as the gain in gravitationalpotential energy (see 2.3.5).

However, this expression can be extended to any twopoints in any gravitational eld such that if an object ofmass m moves between two points whose potentials areV 1 and V 2 respectively, then the change in gravitationalpotential energy of the object is m(V 1 V 2).

9.2.3 G RAVITATIONAL POTENTIAL GRADIENT

Let us consider now a region in space where thegravitational eld is constant. In Figure 912 the two points

A and B are separated by the distance x .

A B

direction of uniform gravitationalfield of strength I

x

Figure 912 The gravitational potential gradient

e gravitational eld is of strength I and is in the directionshown. e gravitational potential at A is V and at B is V+ V .

e work done is taking a point mass m from A to B isF x = mI x .

However, by denition this work is also equal to - mV .

erefore mI x = -mV

orV

I x

=

Effectively this says that the magnitude of the gravitationaleld strength is equal to the negative gradient of thepotential. If I constant then V is a linear function of x and I is equal to the negative gradient of the straight linegraph formed by plotted V against x . If I is not constant(as usually the case), then the magnitude of I at any pointin the eld can be found by nd the gradient of the V-x graph at that point. An example of such a calculation canbe found in Section 9.2.9.

For those of you who do HL maths the relationship be eld and potential is seen to follow from the expressthe potential of a point mass viz:

mV G r =

2

ddV mG I r r

= + =


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9.2.4 Determine the potential due to one ormore point masses.

9.2.5 Describe and sketch the pattern ofequipotential surfaces due to one and twopoint masses.

9.2.6 State the relation between equipotentialsurfaces and gravitational eld lines.

9.2.7 Explain the concept of escape speed from a

planet.

9.2.8 Derive an expression for the escape speedof an object from the surface of a planet.

IBO 2007

9.2.4 P OTENTIAL DUE TO ONE OR MORE POINT MASSES

Gravitational potential is a scalar quantity so calculatingthe potential due to more than one point mass is a matterof simple addition. So for example, the potential Vdue tothe Moon and Earth and a distance x from the centre ofEarth is given by the expression

E M M M V Gx r x

= +

where M E = mass of Earth, M M= mass of Moon and r =distance between centre of Earth and Moon.

9.2.5-6 E QUIPOTENTIALS AND FIELD LINES

If the gravitational potential has the same value at all pointson a surface, the surface is said to be an equipotentialsurface. So for example, if we imagine a spherical shell

about Earth whose centre coincides with the centre ofEarth, this shell will be an equipotential surface. Clearly,if we represent the gravitational eld strength by eldlines, since the lines radiate out from the centre of Earth,then these lines will be at right angles to the surface If theeld lines were not normal to the equipotential surfacethen there would be a component of the eld parallel tothe surface. is would mean that points on the surfacewould be at different potentials and so it would no longerbe an equipotential surface! is of course holds true forany equipotential surface.

Figure 913 shows the eld lines and equipotentials for twopoint masses m.

m m

Figure 913 Equipotentials for two point masses

It is worth noting that we would get exactly the samepattern if we were to replace the point masses with twoequal point charges. (See 9.3.5)

9.2.7-8 E SCAPE SPEEDe potential at the surface of Earth is

whichmeans that the energy required to take a particle of massm from the surface to innity is equal to

But what does it actually mean to take something toinnity? When the particle is on the surface of the Earthwe can think of it as sitting at the bottom of a potentialwell as in gure 914.

satellite

surface of E arth

innity

GM

R---------

FIgure 914 A potential well

e depth of the well is and if the satellite gains anamount of kinetic energy equal to where m is itsmass then it will have just enough energy to li it out ofthe well.

In reality it doesnt actually go to innity it just means thatthe rocket is effectively free of the gravitational attractionof the Earth. We say that it has escaped the Earthsgravitational pull. We meet this idea in connection withmolecular forces. Two molecules in a solid will sit at their


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equilibrium position, the separation where the repulsiveforce is equal to the attractive force. If we supply just enoughenergy to increase the separation of the molecules suchthat they are an innite distance apart then the moleculesare no longer affected by intermolecular forces and thesolid will have become a liquid. ere is no increase in thekinetic energy of the molecules and so the solid melts at

constant temperature.

We can calculate the escape speed of a satellite very easilyby equating the kinetic energy to the potential energy suchthat

12---mv escape

2 GM emRe

----------------=

v escape 2 GM e

Re--------------- 2 g 0 Re= =

Substituting for g 0 and Re gives a value for v escape of about11 km s1 for the Earth.

You will note that the escape speed does not depend onthe mass of the satellite since both kinetic energy andpotential energy are proportional to the mass.

In theory if you want to get a rocket to the moon it canbe done without reaching the escape speed. However thiswould necessitate an enormous amount of fuel and it islikely that the rocket plus fuel would be so heavy that itwould never get off the ground. It is much more practicalto accelerate the rocket to the escape speed and then intheory just point it at the Moon to where it will now coastat constant speed.

9.2.9 S OLVE PROBLEMS INVOLVING GRAVITATIONAL POTENTIAL ENERGY AND GRAVITATIONAL POTENTIAL

Example

Use the following data to determine the potential at thesurface of Mars and the magnitude of the acceleration offree fall

mass of Mars = 6.4 1023 kg

radius of Mars = 3.4 10 6 m

Determine also the gravitational eld strength at a distanceof 6.8 106 m above the surface of Mars.

Solution

2311 7

6

6.4 106.7 10 1.3 103.4 10

M V G R

= = = N kg

-1

But V = -g 0R

erefore7

0 6

1.3 103.8

3.4 10V

g R

= = = m s

-2

To determine the eld strength g h at 6.8 106 m above thesurface, we use the fact that0 2

M g G

R= such that GM = g 0R2

erefore2 2

0h 2 2 2

h h

3.8 (3.4)0.42

(10.2) g RGM g

R R= = = = m s-2

(the distance from the centre is 3.4 106 + 6.8 106 = 10.2 106 m)

Exercise

1. e graph below shows how the gravitationalpotential outside of the Earth varies with distancefrom the centre.

0 10 20 30 40 50 60 70

1

2

3

4

5

6

7

r /m 106

V / J k g

1

1 0 7

(a) Use the graph to determine the gain ingravitational potential energy of a satelliteof mass 200 kg as it moves from the surfaceof the Earth to a height of 3.0 10 7 m abovethe Earths surface.

Answer: 1010 J

(b) Calculate the energy required to take it toinnity?

Answer; 1.3 1010 J


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(c) Determine the slope of the graph at thesurface of the Earth, m, above the surface ofthe Earth? Comment on your answers.

Answer: 10, and equals the gravitational eld strength atthe surface of Earth (acceleration of free fall)

9.3 ELECTRIC FIELD,POTENTIAL ANDENERGY

9.3.1 Dene electric potential and electric potential energy .

9.3.2 State and apply the expression for electricpotential due to a point charge.

9.3.3 State and apply the formula relatingelectric eld strength to electrical potentialgradient.

9.3.4 Determine the potential due to one ormore point charges.

9.3.5 Describe and sketch the pattern ofequipotential surfaces due to one and twopoint charges.

9.3.6 State the relation between equipotentialsurfaces and electric eld lines.

9.3.7 Solve problems involving electric potentialenergy and electric potential.

IBO 2007

9.3.1 D EFINING ELECTRIC POTENTIAL & ELECTRIC POTENTIAL ENERGY

Electric potential energy

e concept of electric potential energy was developedwith gravitational potential energy in mind. Just as anobject near the surface of the Earth has potential energybecause of its gravitational interaction with the Earth,so too there is electrical potential energy associated withinteracting charges.

Let us rst look at a case of two positive point charges eachof 1C that are initially bound together by a thread in a vacuum in space with a distance between them of 10 cmas shown in Figure 916. When the thread is cut, the pointcharges, initially at rest would move in opposite directions,moving with velocities v 1 and v 2 along the direction of theelectrostatic force of repulsion.

BEFORE AFTER

v v2

Figure 916 Interaction of two positive particles

e electric potential energy between two point chargescan be found by simply adding up the energy associatedwith each pair of point charges. For a pair of interactingcharges, the electric potential energy is given by:

U = Ep + E k = W = Fr = kqQ / r 2 x r = kqQ

Because no external force is acting on the system, theenergy and momentum must be conserved. Initially,Ek = 0 and Ep = k qQ / r = 9 109 1 10-12 / 0.1 m = 0.09 J.When they are a great distance from each other, E p willbe negligible. e nal energy will be equal to m v 1

2 + mv 2

2 = 0.09 J. Momentum is also conserved and the velocities would be the same magnitude but in oppositedirections.

Electric potential energy is more o en dened in terms ofa point charge moving in an electric eld as:

the electric potential energy between any two points inan electric eld is dened as negative of the work done byan electric eld in moving a point electric charge betweentwo locations in the electric eld.

U = Ep = - W = -Fd = qEx

where x is the distance moved along (or opposite to) thedirection of the electric eld.

Electric potential energy is measured in joule (J). Just aswork is a scalar quantity, so too electrical potential energyis a scalar quantity . e negative of the work done by anelectric eld in moving a unit electric charge betweentwo points is independent of the path taken. In physics,we say the electric eld is a conservative eld.

Suppose an external force such as your hand moves a smallpositive point test charge in the direction of a uniformelectric eld. As it is moving it must be gaining kinetic


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energy. If this occurs, then the electric potential energy ofthe unit charge is changing.

In Figure 917 a point charge + q is moved between pointsA and B through a distance x in a uniform electric eld.

B

A+ q

Figure 917 Movement of a positive point charge in a uniform eld

In order to move a positive point charge from point A topoint B, an external force must be applied to the chargeequal to qE (F = qE ).

Since the force is applied through a distance x , thennegative work has to be done to move the charge becauseenergy is gained, meaning there is an increase electricpotential energy between the two points. Rememberthat the work done is equivalent to the energy gained orlost in moving the charge through the electric eld. econcept of electric potential energy is only meaningful ifthe electric eld which generates the force in question isconservative.

W F x E q x = =

cos

Figure 918 Charge moved at an angle to the eld

If a charge moves at an angle to an electric eld, thecomponent of the displacement parallel to the electric

eld is used as shown in Figure 918

W F x Eq x cos= =

e electric potential energy is stored in the electric eld,and the electric eld will return the energy to the pointcharge when required so as not to violate the Law ofconservation of energy.

Electric potential

e electric potential at a point in an electric eldis dened as being the work done per unit charge inbringing a small positive point charge from innity tothat point.

V = V V f= -W /q

If we designate the potential energy to be zero at innitythen it follows that electric potential must also be zero atinnity and the electric potential at any point in an elctriceld will be:

V = -W / q

Now suppose we apply an external force to a small positivetest charge as it is moved towards an isolated positivecharge. e external force must do work on the positivetest charge to move it towards the isolated positive chargeand the work must be positive while the work done by theelectric eld must therefore be negative. So the electricpotential at that point must be positive according to theabove equation. If a negative isolated charge is used, theelectric potential at a point on the positive test chargewould be negative. Positive point charges of their ownaccord, move from a place of high electric potential toa place of low electric potential. Negative point chargesmove the other way, from low potential to high potential.In moving from point A to point B in the diagram, thepositive charge +q is moving from a low electric potentialto a high electric potential. e electric potential istherefore different at both points.

In the denition given, the term work per unit chargehas signicance. If the test charge is +1.6 10 -19C wherethe charge has a potential energy of 3.2 10 -17 J, then thepotential energy would be 3.2 10 -17J / +1.6 10-19 C =200 JC-1. Now if the charge was doubled, the potentialenergy would become 6.4 10 -17 J. However, the potentialenergy per unit charge would be the same.

Electric potential is a scalar quantity and it has units JC-1

or volts where 1 volt equals one joule per coloumb. e volt allows us to adopt a unit for the electric eld in termsof the volt.

Previously, the unit for the electric eld was NC -1.

W = qV and F = qE ,

so W / V =F / E

E = F V / W= NV / Nm = Vm -1.


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e work done per unit charge in moving a pointcharge between two points in an electric eld is againindependant of the path taken.

9.3.2 E LECTRIC POTENTIAL DUE TO A

POINT CHARGELet us take a point r metres from a charged object.e potential at this point can be calculated using thefollowing:

W = -Fr = -qV and F = q1 q2 4 0 r 2

erefore,

q 1 q 2

4 0 r 2

---------------- r q 1 q 2

4 0 r

--------------- q1q 2

4 0 r

-------------- - q 1= - =- =- = -

at is

V q4 0r -------------- -=

Or, simply

V kqr

----- -=

Example

Determine how much work is done by the electric eld ofpoint charge 15.0 C when a charge of 2.00 C is movedfrom innity to a point 0.400 m from the point charge.(Assume no acceleration of the charges).

Solution

The work done by the electric eld is W = -qV= -1/40 q (Q /r - Q / r 0.400 )

W = (- 2.00 10-6 C 9.00 109 NmC -2 15.0 10-6 C) 0.400 m = - 0.675 J

An external force would have to do +0.675 J of work .

9.3.3 E LECTRIC FIELD STRENGTH AND ELECTRIC POTENTIAL GRADIENT

Let us look back at Figure 917. Suppose again charge +q is moved a small distance by a force F fro

so that the force can be considered constant. e woris given by:W F x =

e force F and the electric eld E are oppositely dand we know that:

F = -qE and W = q V

erefore, the work done can be given as:

q V = -qE x

erefore

E ------- =

e rate of change of potential V at a point with resdistance x in the direction in which the change is mis called the potential gradient . We say that the electric = - the potential gradient and the units are Vm-1. From theequation we can see that in a graph of electric pversus distance, the gradient of the straight line eqelectric eld strength.

In reality, if a charged particle enters a uniform elecit will be accelerated uniformly by the eld and ienergy will increase. is is why we had to assuacceleration in the last worked example.

Ek 1

2--- mv 2 q E x q V

x --- x q V = = = =

Example

Determine how far apart two parallel plates must besituated so that a potential difference of 1.50 x 10 2 Vproduces an electric eld strength of 1.00 x 10 3 NC-1.


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Solution

Using E V x

------- x V E

------- 1.5 10 2 V1.00 10 3 N C 1------------------------------------------= = =

= 1.50 10-1

e plates are 1.50 10-1 m apart.

e electric eld and the electric potential at a point dueto an evenly distributed charge + q on a sphere can berepresented graphically as in Figure 919.

r

r

V

E

0

r 0

r 0

over surface

E 140------------

Q

r 2----- r r 0

>,=

V 14 0------------

Qr ---- r r 0>,=

Charge of +Q evenly distributedr

Figure 919 Electric eld and potential due to a charged sphere

When the sphere becomes charged, we know that thecharge distributes itself evenly over the surface. ereforeevery part of the material of the conductor is at the samepotential. As the electric potential at a point is dened asbeing numerically equal to the work done in bringing a unitpositive charge from innity to that point, it has a constant

value in every part of the material of the conductor.

Since the potential is the same at all points on the conductingsurface, then V / x is zero. But E = V / x. erefore,the electric eld inside the conductor is zero. ere is noelectric eld inside the conductor.

Some further observations of the graphs in Figure 915 are:

Outside the sphere, the graphs obey therelationships given as E 1 / r 2 and V 1 / r

At the surface, r = r 0. erefore, the electric eldand potential have the minimum value for r atthis point and this infers a maximum eld and

potential. Inside the sphere, the electric eld is zero. Inside the sphere, no work is done to move

a charge from a point inside to the surface.erefore, there is no potential difference and thepotential is the same as it is when r = r 0.

Similar graphs can be drawn for the electric eld intensityand the electric potential as a function of distance fromconducting parallel plates and surfaces, and these aregiven in Figure 920.

Potential plot E eld plotE eld:

++++

x x x

+

x x

x x

+ x

x

+ +

x

x

Figure 920 Electric eld and electric potential at a distance from a charged surface


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isolated positive sphere. e lines of force and someequipotential lines for an isolated positive sphere areshown in Figure 922.

Lines of equipotential

Figure 922 Equipotentials aroundan isolated positive sphere

In summary, we can conclude that

No work is done to move a charge along anequipotential.

Equipotentials are always perpendicular to theelectric lines of force.

Figure 923 and 924 show some equipotential lines fortwo oppositely charged and indentically positive spheresseparated by a distance.

+ve ve

equipotential lines

Figure 923 Equipotential linesbetween two opposite charges

equipotential lines

Figure 924 Equipotential lines betweentwo charges which are the same

+

+ + +

40 V

30 V

20 V

10 V

50 V

Figure 925 Fquipotential lines between charged parallel plates gravitational elds and electric elds

roughout this chapter the similarities and differencesbetween gravitational elds and electric elds havebeen discussed. e relationships that exists betweengravitational and electric quantities and the effects of pointmasses and charges is summarised in Table 926

Gravitational quantity Electrical quantity

Q u a n t i t i e s

g V x

-------= E V x

-------=

P o i n t m a s s e s a n

d c h a r g e s

V Gmr ----= V 14 0

------------ qr ---=

g Gm

r 2

---- -= E 1

4 0------------ q

r 2

-----=

F Gm1 m2

r 2

--------------= F 1

4 0------------

q1 q2

r 2

-----------=

Figure 926 Formulas (table)

9.3.7 S OLVING PROBLEMSere are a number of worked examples that have beengiven in section 9.3. Here are two more examples.


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Example 1

Deduce the electric potential on the surface of a goldnucleus that has a radius of 6.2 fm.

Solution

Using the formula

V = kq / r , and knowing the atomic number of gold is 79.We will assume the nucleus is spherical and it behaves as if itwere a point charge at its centre (relative to outside points).

V = 9.0 109 Nm2C -2 79 1.6 10-19 C 6.2 10-15 m= 1.8 107 V

e potential at the point is18 MV .

Example 2

Deduce the ionisation energy in electron-volts of theelectron in the hydrogen atom if the electron is in itsground state and it is in a circular orbit at a distance of 5.3x 10-11 m from the proton.

Solution

is problem is an energy, coulombic, circular motionquestion based on Bohrs model of the atom (not the acceptedquantum mechanics model). e ionisation energy is theenergy required to remove the electron from the groundstate to innity. e electron travels in a circular orbit andtherefore has a centripetal acceleration. e ionisation energywill counteract the coulombic force and the movement of the

electron will be in the opposite direction to the centripetal force

Total energy = Ek electron + E p due to the proton-electroninteraction

F = kqQ / r 2 = m v 2 / r and as such m v 2 = = kqQ / r.

erefore, Ek electron = kqQ / r.

E p due to the proton-electron interaction = - kqQ / r.

Total energy = kqQ / r + - kqQ / r = - kqQ / r

= - 9.0 109 Nm2C -2 (1.6 10-19 C)2 5.3 10-11 m =-2.17 10-18 J

= -2.17 10-18 J 1.6 10-19 = -13.6 eV.

e ionisation energy is13.6 eV .

Exercise 9.3

1. A point charge P is placed midway between twoidentical negative charges. Which one of thefollowing is correct with regards to electric eldand electric potential at point P?

Electric eld Electric potentialA non-zero zeroB zero non-zeroC non-zero non-zeroD zero zero

2. Two positive charged spheres are tied together ina vacuum somewhere in space where there are noexternal forces. A has a mass of 25 g and a chargeof 2.0 C and B has a mass of 15 g and a charge of3.0 C. e distance between them is 4.0 cm. eyare then released as shown in the diagram.

BEFORE AFTER

v1 v2

A B

(a) Determine their initial electric potentialenergy in the before situation.

(b) Determine the speed of sphere B a errelease.

3. e diagram below represents two equipotentiallines in separated by a distance of 5 cm in auniform electric eld.

+ + + + + + + +

40 V

20 V5 cm

Determine the strength of the electric eld.


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4. is question is about the electric eld due to acharged sphere and the motion of electrons in thateld. e diagram below shows an isolated, metalsphere in a vacuum that carries a negative electriccharge of 6.0 C.

(a) On the diagram draw the conventional wayto represent the electric eld pattern due tothe charged sphere and lines to representthree equipotential surfaces in the regionoutside the sphere.

(b) Explain how the lines representing theequipotential surfaces that you havesketched indicate that the strength of theelectric eld is decreasing with distancefrom the centre of the sphere.

(c) e electric eld strength at the surface ofthe sphere and at points outside the spherecan be determined by assuming that thesphere acts as a point charge of magnitude6.0 C at its centre. e radius of the sphereis 2.5 102 m. Deduce that the magnitudeof the eld strength at the surface of thesphere is 8.6 107 Vm1.

An electron is initially at rest on the surface of thesphere.

(d) (i) Describe the path followed by theelectron as it leaves the surface of thesphere.

(ii) Calculate the initial acceleration ofthe electron.

5. Determine the amount of work that is done inmoving a charge of 10.0 nC through a potentialdifference of 1.50 102 V.

6. ree identical 2.0 C conducting spheres areplaced at the corners of an equilateral triangle ofsides 25 cm. e triangle has one apex C pointingup the page and 2 base angles A and B. Determinethe absolute potential at B .

7. Determine how far apart two parallel platesmust be situated so that a potential difference of2.50 102 V produces an electric eld strength of2.00 103 NC-1.

8. e gap between two parallel plates is 1.0 10 -3 m,and there is a potential difference of 1.0 10 4 Vbetween the plates. Calculate

i. the work done by an electron in movingfrom one plate to the other

ii. the speed with which the electron reaches

the second plate if released from rest.iii. the electric eld intensity between theplates.

9. An electron gun in a picture tube is acceleratedby a potential 2.5 10 3 V. Determine the kineticenergy gained by the electron in electron-volts.

10. Determine the electric potential 2.0 x10 -2 m from acharge of -1.0 10-5 C.

11. Determine the electric potential at a point mid-way between a charge of 20 pC and anotherof + 5 pC on the line joining their centres if thecharges are 10 cm apart.

12. During a thunderstorm the electric potentialdifference between a cloud and the ground is1.0 109 V. Determine the magnitude of thechange in electric potential energy of an electronthat moves between these points in electron-volts.

13. A charge of 1.5 C is placed in a uniform electriceld of two oppositely charged parallel plates witha magnitude of 1.4 10 3 NC-1.

(a) Determine the work that must be doneagainst the eld to move the point charge adistance of 5.5 cm.

(b) Calculate the potential difference betweenthe nal and initial positions of the charge.

(c) Determine the potential difference betweenthe plates if their separation distance is 15 cm.

14. During a ash of lightning, the potential difference

between a cloud and the ground was 1.2 109

Vand the amount of transferred charge was 32 C.

(a) Determine the change in energy of thetransferred charge.

(b) If the energy released was all used toaccelerate a 1 tonne car, deduce its nalspeed.

(c) If the energy released could be used to meltice at 0 C, deduce the amount of ice thatcould be melted.


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15. Suppose that when an electron moved from A to Bin the diagram along an electric eld line that theelectric eld does 3.6 10-19 J of work on it.

Determine the differences in electric potential:

(a) V B V A(b) V C V A(c) V C V B

16. Determine the potential at point P that is locatedat the centre of the square as shown in the diagrambelow.

- 6 C

+3 C +2 C

1m

1m

5 C

P

9.4 ORBITAL MOTIONAlthough orbital motion may be circular, elliptical orparabolic, this sub-topic only deals with circular orbits.is sub-topic is not fundamentally new physics, butan application that synthesizes ideas from gravitation,

circular motion, dynamics and energy.

9.4.1 State that gravitation provides thecentripetal force for circular orbital motion.

9.4.2 Derive Keplers third law.

9.4.3 Derive expressions for the kinetic energy,potential energy and total energy of anorbiting satellite.

9.4.4 Sketch graphs showing the variationwith orbital radius of the kinetic energy,gravitational potential energy and totalenergy of a satellite.

9.4.5 Discuss the concept of weightlessnessin orbital motion, in free fall and in deepspace.

9.4.6 Solve problems involving orbital motion. IBO 2007

9.4.1 S ATELLITESe Moon orbits the Earth and in this sense it is o enreferred to as a satellite of the Earth. Before 1957 it was theonly Earth satellite! However, in 1957 the Russians launchedthe rst man made satellite, Sputnik 1. Since this date manymore satellites have been launched and there are nowliterally thousands of them orbiting the Earth. Some areused to monitor the weather, some used to enable peopleto nd accurately their position on the surface of the Earth,many are used in communications, and no doubt some are

used to spy on other countries. Figure 932 shows how, inprinciple, a satellite can be put into orbit.

e person (whose size is greatly exaggerated with respectto Earth) standing on the surface on the Earth throwssome stones. e greater the speed with which a stoneis thrown the further it will land from her. e pathsfollowed by the thrown stones are parabolas. By a stretchof the imagination we can visualise a situation in whicha stone is thrown with such a speed that, because of thecurvature of the Earth, it will not land on the surface of theEarth but go into orbit. (Path 4 on gure 932).

A

B C


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12

3

4Earth

Figure 932 Throwing a stone into orbit

e force that causes the stones to follow a parabolic pathand to fall to Earth is gravity and similarly the force thatkeeps the stone in orbit is gravity. For circular motionto occur we have seen that a force must act at rightangles to the velocity of an object, that is there must be acentripetal force. Hence in the situation we describe herethe centripetal force for circular orbital motion about theEarth is provided by gravitational attraction of the Earth.

We can calculate the speed with which a stone must bethrown in order to put it into orbit just above the surfaceof the Earth.

If the stone has mass m and speed v then we have fromNewtons 2nd law

mv 2

RE--------- - G

M Em

RE2------------=

where RE is the radius of the Earth and M E is the mass ofthe Earth.

Bearing in mind that g 0 G M ERE

2--------= , then

v g RE 10 6.4 106 8 10 3= = = .

at is, the stone must be thrown at 8 10 3m s1.

Clearly we are not going to get a satellite into orbit soclose to the surface of the Earth. Moving at this speed the

friction due to air resistance would melt the satellite beforeit had travelled a couple of kilometres! In reality thereforea satellite is put into orbit about the Earth by sending it,attached to a rocket, way beyond the Earths atmosphereand then giving it a component of velocity perpendicularto a radial vector from the Earth. See Figure 933.

Earth Satellite carriedby rocket to hereSatellite orbit

Tangential component of velocity

Figure 933 Getting a satellite into orbit

9.4.2 K EPLERS THIRD LAW( is work of Kepler and Newtons synthesis of the woan excellent example of the scientic method and maa good TOK discussion)

In 1627 Johannes Kepler (1571-1630) published his lawsof planetary motion. e laws are empirical in natureand were deduced from the observations of the Danishastronomer Tycho de Brahe (1546-1601). e third lawgives a relationship between the radius of orbit R of aplanet and its period T of revolution about the Sun. elaw is expressed mathematically as

2

3 constantT R

=

We shall now use Newtons Law of Gravitation to showhow it is that the planets move in accordance with Keplersthird law.

In essence Newton was able to use his law of gravity topredict the motion of the planets since all he had to dowas factor the F given by this law into his second law,F = ma , to nd their accelerations and hence their futurepositions.

In Figure 934 the Earth is shown orbiting the Sun and thedistance between their centres is R.

RSun

Earth

F es

F se

Figure 934 Planets move according to Keplers third law

F es is the force that the Earth exerts on the Sun and F se is theforce that the Sun exerts on the Earth. e forces are equaland opposite and the Sun and the Earth will actually orbitabout a common centre. However since the Sun is so verymuch more massive than the Earth this common centrewill be close to the centre of the Sun and so we can regardthe Earth as orbiting about the centre of the Sun. e other


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thing that we shall assume is that we can ignore the forcesthat the other planets exert on the Earth. ( is would notbe a wise thing to do if you were planning to send a spaceship to the Moon for example!). We shall also assume thatwe have followed Newtons example and indeed provedthat a sphere will act as a point mass situated at the centreof the sphere.

Kepler had postulated that the orbits of the planets areelliptical but since the eccentricity of the Earths orbit issmall we shall assume a circular orbit.

e acceleration of the Earth towards the Sun is a R2

=

where 2

T ------=

Hence,

a R 2T

------

2 42 R

T2-------------= =

But the acceleration is given by Newtons Second Law,F = ma, where F is now given by the Law of Gravitation.So in this situation

F maGM s M e

R2-------------------= = , but, we also have that

a 42R

T 2

-------------= and m = M e so that

G M s M e

R2------------------- M

e

4 2 RT 2

------------- GM s

4 2----------- R

3

T 2------= =

But the quantity

is a constant that has the same value for each of the planetsso we have for all the planets, not just Earth, that

R3

T 2

------ k=

where k is a constant. Which is of course Keplers thirdlaw.

is is indeed an amazing breakthrough. It is difficultto refute the idea that all particles attract each other inaccordance with the Law of Gravitation when. the lawis able to account for the observed motion of the planetsabout the Sun.

e gravitational effects of the planets upon each othershould produce perturbations in their orbits. Such is thepredictive power of the Universal Gravitational Law thatit enabled physicists to compute these perturbations. etelescope had been invented in 1608 and by the middle of

the 18th Century had reached a degree of perfection indesign that enabled astronomers to actually measure theorbital perturbations of the planets. eir measurementswere always in agreement with the predictions made byNewtons law. However, in 1781 a new planet, Uranus wasdiscovered and the orbit of this planet did not t withthe orbit predicted by Universal Gravitation. Such was

the physicists faith in the Newtonian method that theysuspected that the discrepancy was due to the presence ofa yet undetected planet. Using the Law of Gravitation theFrench astronomer J.Leverrier and the English astronomer.J. C. Adams were able to calculate just how massive thisnew planet must be and also where it should be. In 1846the planet Neptune was discovered just where they hadpredicted. In a similar way, discrepancies in the orbit ofNeptune led to the prediction and subsequent discoveryin 1930 of the planet Pluto. Newtons Law of Gravitationhad passed the ultimate test of any theory; it is not onlyable to explain existing data but also to make predictions.

9.4.3-4 S ATELLITE ENERGYWhen a satellite is in orbit about a planet it will have bothkinetic energy and gravitational potential energy. Supposewe consider a satellite of mass m that is in an orbit of radiusr about a planet of mass M .

e gravitational potential due to the planet at distance r from its centre is

.

e gravitational potential energy of the satellite V sat

is thereforeGM em

r ---------------- .

at is, V satGM em

r ----------------= .

e gravitational eld strength at the surface of the planetis given by

g 0GM eRe

2------------=

Hence we can write

V sat g 0 Re

2

r ------------=

e kinetic energy of the satellite K sat is equal to mv 2 ,

where v is its orbital speed.


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By equating the gravitational force acting on the satelliteto its centripetal acceleration we have

G M em

r 2---------------- mv

2

r ---------- mv 2

GM emr

----------------= = .

From which12---mv 2 1

2---

GM emr

----------------=

= 0 e2

2 g R m

r Which is actually quite interesting since it shows that,irrespective of the orbital radius the KE is numericallyequal to half the PE, Also the total energy Etot of thesatellite is always negative since

Etot K sat V sat+ 1

2---

G M emr

---------------- G M em

r ----------------

+ 12---

GM emr

----------------= = =

e energies of an orbiting satellite as a function of radialdistance from the centre of a planet are shown plotted ingure 935.

1.2

1.0

0.8

0.6

0.4

0.2

0

0.2

0.4

0.6

0.81.0

1.22 4 6 8 10 1 2

n o r m i l i s e d e n e r g y

distance / R

kinetic energytotal ener gy

potential ener gy

Y should be energy-arbitrary units

Figure 935 Energy of an orbiting satellite asa function of distance from the centre of a planet

9.4.5 W EIGHTLESSNESS

Suppose that you are in an elevator (li ) which isdescending at constant speed and you let go of a book thatyou are holding in your hand. e book will fall to theoor with acceleration equal to the acceleration due togravity. If the cable that supports the elevator were to snap(a situation that I trust will never happen to any of you)and you now let go the book that you are holding in yourother hand, this book will not fall to the oor - it will stayexactly in line with your hand! is is because the book isnow falling with the same acceleration as the elevator andas such the book cannot catch up with the oor of theelevator. Furthermore, if you happened to be standing on

a set of bathroom scales, the scales would now read zero- you would be weightless! It is this idea of free fall thatexplains the weightlessness of astronauts in an orbitingsatellite. ese astronauts are in free fall in the sense thatthey are accelerating towards the centre of the Earth.

It is actually possible to dene the weight of a body in

several different ways. We can dene it for example asthe gravitational force exerted on the body by a speciedobject such as the Earth. is we have seen that we doin lots of situations where we dene the weight as beingequal to mg . If we use this denition, then an object infree fall cannot by denition be weightless since it is stillin a gravitational eld. However, if we dene the weightof an object in terms of a weighing process such asthe reading on a set of bathroom scales, which in effectmeasures the contact force between the object and thescales, then clearly objects in free fall are weightless. Onenow has to ask the question whether or not it is possible.For example, to measure the gravitational force acting onan astronaut in orbit about the Earth. We shall return tothis idea of bodies in free fall when we look at EinsteinsGeneral eory of Relativity in Chapter 19

We can also dene weight in terms of the net gravitationalforce acting on a body due to several different objects. Forexample for an object out in space, its weight could bedened in terms of the resultant of the forces exerted on itby the Sun, the Moon, the Earth and all the other planetsin the Solar System. If this resultant is zero at a particularpoint then the body is weightless at this point.

In view of the various denitions of weight that areavailable to us it is important that when we use the wordweight we are aware of the context in which it is beingused.

9.4.6 S OLVE PROBLEMS INVOLVING ORBITAL MOTION

Example 1

Calculate the height above the surface of the Earth atwhich a geo-stationary satellite orbits.


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Solution

A geo-stationary satellite is one that orbits the Earth in sucha way that it is stationary with respect to a point on thesurface of the Earth. is means that its orbital period must

be the same as the time for the Earth to spin once on its axisi.e. 24 hours.

From Keplers third law we haveG M s

42----------- R

3

T 2

------= .

at is,

R M

e

m

h

using the fact that the force of attraction between the satelliteand the Earth is given by

F G M em

R2

----------------=

and that F = ma

where m is the mass of the satellite anda 42R

T 2

-------------=

we have,

GM em

R2

---------------- m 4 2 R

T 2

------------- GM e

4 2----------- - R

3

T 2

------= =

Now, the mass of the Earth is 6.0 1024 kg and the period, T,measured in seconds is given by T = 86,400 s.

So substitution gives R = 42 106 m

e radius of the Earth is 6.4 106

m so that the orbitalheight, h, is about 3.6 107 m.

Example 2

Calculate the minimum energy required to put a satelliteof mass 500 kg into an orbit that is as a height equal to theEarths radius above the surface of the Earth.

Solution

We have seen that when dealing with gravitationand potential it is useful to remember that

g 0G M

Re2---------= or, g 0 Re

2

GM =

e gravitational potential at the surface of the Earth

g 0 ReGM Re

---------= .

e gravitational potential at a distance R

from the centre of the Earth is

=

e difference in potential between the surface of thand a point distance R from the centre is therefore

V g 0Re 1ReR------

=

IfR 2Re= then V g 0 Re

2------------=

is means that the work required to li the satellitorbit is g 0Rm where m is the mass of the satellite.equal to

10 3.2 10 6 500 = 16000 MJ.

However, the satellite must also have kinetic energyto orbit Earth. is will be equal to

g 0 mRe2

2R-----------------

g 0mRe2

4---------------- - 8000 MJ = =

e minimum energy required is therefore

24000 MJ.


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Exercises

1. e speed needed to put a satellite in orbit doesnot depend on

A. the radius of the orbit.B. the shape of the orbit.C. the value of g at the orbit.D. the mass of the satellite.

Answer: D

2. Estimate the speed of an Earth satellite whoseorbit is 400 km above the Earths surface. Alsodetermine the period of the orbit.

Answer: 2.6 103 m s-1, 1.6 104 s

3. Calculate the speed of a 200 kg satellite, orbitingthe Earth at a height of 7.0 10 6 m.

Assume that g = 8.2 m s 2 for this orbit.

Answer: 1.0 104 m s-1

4. e radii of two satellites, X and Y, orbitingthe Earth are 2 r and 8r where r is the radius ofthe Earth. Calculate the ratio of the periods ofrevolution of X to Y.

Answer: 2

5. A satellite of mass m kg is sent from Earths surfaceinto an orbit of radius 5 R, where R is the radius ofthe Earth.Write down an expression for

(a) the potential energy of the satellite in orbit.

Answer:

(b) the kinetic energy of the satellite in orbit.

Answer:

(c) the minimum work required to send thesatellite from rest at the Earths surface intoits orbit.

Answer:

6. A satellite in an orbit of 10 r , falls back to Earth(radius r ) a er a malfunction. Determine thespeed with which it will hit the Earths surface?

Answer: 9.6 103 m s-1

7. e radius of the moon is that of the Earth

Assuming Earth and the Moon to have the samedensity, compare the accelerations of free fall atthe surface of Earth to that at the surface of theMoon.

Answer: g Moon = 4 g Earth

9. Use the following data to determine thegravitational eld strength at the surface of theMoon and hence determine the escape speed fromthe surface of the Moon.

Mass of the Moon = 7.3 10 22 kg, Radius of theMoon = 1.7 10 6 m

Answer: 2.5 N kg-1, 3.0 103 m s-1

physics ch 09_nopw

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