topic 6.1 gravitational force and fields

8/10/2019 Topic 6.1 Gravitational Force and Fields

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6.1.1 State Newtons universal law of

gravitation.

6.1.2 Define gravitational field strength.

6.1.3 Determine the gravitational field due to

one or more point masses.

6.1.4 Derive an expression for gravitational

field strength at the surface of a planet,

assuming that all its mass is concentrated at

its centre.

6.1.5 Solve problems involving gravitationalforces and fields.

Topic 6: Fields and forces

6.1 Gravitational force and field


2/26

State Newtons universal law of gravitation.

The gravitational force is the weakest of the

four fundamental forces, as the following visual

shows:



GRAVITYSTRONG ELECTROMAGNETIC WEAK

+

+

nuclear

force

light, heat,

charge and

magnets

radioactivity freefall,

orbits

ELECTRO-WEAK

WEAKESTSTRONGEST


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In the 1680s in his groundbreaking book

PrincipiaSir Isaac Newton published not

only his works on physical motion, but what

has been called by some the greatest

scientific discovery of all time, his

universal law of gravitation.

The law states that the gravitational force

between two point masses m1and m2is

proportional to their product, and inversely

proportional to the square of their separation r.

The actual value of G, the gravitational

constant, was not known until Henry Cavendishconducted a tricky experiment in 1798 to find it.



F= Gm1m2/r2 Universal law

of gravitationwhere G= 6.671011Nm2kg

2


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Newton spent much time developing integral

calculus to prove that a spherically symmetric

shell of massMacts as if all of its mass is

located at its centre.

Thus the law works not only for point masses,

which have no radii, but for any spherical

distribution of mass at any radius like planets

and stars.



m

M

r

-Newtons shell theorem.


5/26


The earth has many

layers, kind of like

an onion:

Since each shell is

symmetric, the gravi-

tational force caused by that shell acts as

though it is all concentrated at its centre.

Thus the net force at mcaused by the shells isgiven by

F= GMim/r2+ GMom/r

2+ GMmm/r2+ GMcm/r

2

F= G(Mi+ Mo+ Mm+ Mc)m/r2

F= GMm/r2where M= Mi+ Mo+ Mm+ Mc

which is the total mass of the earth.



inner core Mi

outer core Mo

mantle Mm

crust Mc

m

r

You do not have

to recall this!


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Be very clear that ris the distance between the

centresof the masses.



EXAMPLE: The earth has a mass of M= 5.981024kg

and the moon has a mass of m= 7.361022kg. The

mean distance between the earth and the moon is

3.82108m. What is the gravitational force

between them?

SOLUTION:

F= GMm/r2

F= (6.671011)(5.981024)(7.361022)/(3.8210

8)2

F= 2.01

1020

n.

m1 m2

r

F12 F21

* The radii of each planet isimmaterial to this problem.


7/26

Define gravitational field strength.

Suppose a mass m is located a distance rfrom a

another mass M.

We define the gravitational field strengthgas

the force per unit mass acting on mdue to the

presence of M. Thus

The units are newtons per kilogram (Nkg-1).

Note that from Newtons second law, F= ma, we

see that a Nkg-1

is also a ms-2

, the units foracceleration.

Note further that weight has the formula F= mg,

and the gin this formula is none other than the

gravitational field strength!

On the earths surface, g= 9.8 Nkg-1= 9.8 ms-2.



g= F/m gravitational field strength


8/26

Derive an expression for gravitational field

strength at the surface of a planet assuming that

all its mass is concentrated at its center.

Suppose a mass mis located on the surface of a

planet of radius R. We know that its weight is

F= mg.

But from the law of universal gravitation, the

weight of mis equal to its attraction to the

planets mass Mand equals F= GMm/R2.

Thusmg= GMm/R2.

This same derivation works for any r.



g= GM/R2 gravitational field strength at the surface

of a planet of mass M and radius R

g= GM/r2 gravitational field strength at a

distance r from the center of a planet


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Determine the gravitational field due to one or

more point masses.



PRACTICE: Given that the mass of the earth is

M= 5.981024kg and the radius of the earth is

R= 6.37106m, find the gravitational field

strength at the surface of the earth, and at a

distance of one earth radii above the surface.

SOLUTION:

For r= R:

g= GM/R2

g= (6.671011)(5.981024)/(6.37106)2

g= 9.83 Nkg-1(ms

-2).

For r= 2R: Since ris squared

just divide by 22= 4. Thus

g= 9.83/4 = 2.46 ms-2.

FYI

A(N.kg-1) is

the same as a

(m.s-2)


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more point masses.



PRACTICE: A 525-kg satellite is launched from the

earths surface to a height of one earth radii

above the surface. What is its weight (a) at the

surface, and (b) at altitude?

SOLUTION:

(a) From the previous problem we found

gsurface= 9.83 ms-2. From F= mgwe get

F= (525)(9.83) = 5160 n

(b) From the previous problem we found

gsurface+R= 2.46 ms-2. From F= mgwe get

F= (525)(2.46) = 1290 n


11/26


Compare the gravitational force formula

F= GMm/r2(Force)

with the gravitational field formula

g= GM/r2 (Field)Note that the force formula has two masses, and

the force is the result of their interaction at a

distance r.

Note that the field formula has just one mass,

namely the mass that sets up the local field inthe space surrounding it.

The field view of the universe (spatial

disruption by a single mass) is currently

preferred over the force view (action at a

distance) as the next slides will try to show.




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Consider the force view (action at a distance).

In the force view, the masses know where each

other are at all times, and the force is

instantaneously felt by both masses at all times.

This requires the force signal

to be transferred between the

masses instantaneously.

As we will learn later,

Einsteins special theoryof relativity states

unequivocally that the

fastest anysignal can

travel is at the (finite)

speed of light c.



SUN


13/26


Thus the action at a distance force signal will

be slightly delayed in telling the orbital mass

when to turn.

The end result would have to be an expanding

spiral motion, as illustrated in the following

animation:

We do notobserve

planets leaving their

orbits as they travel

around the sun.

Thus action at a

distance doesnt work

if we are to believe

special relativity.



SUN


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So how does the field view take care of this

signal lag problem?

Simply put - the gravitational field distorts the

space around the mass that is causing it so that

any other mass placed at any position in the

field will know how to respond immediately.

The next slide illustrates this gravitational

curvature of the space around, for example, the

sun.




15/26


Note that each mass feels a different slope

and must travel at a particular speed to orbit.



FYI

The field view eliminates the need for long

distance signaling between two masses. Rather, it

distorts the space about one mass.


16/26

FYI

(a) The field arrow is bigger for m2than m1. Why?

(b) The field arrow always points to M. Why?

M


In the space surrounding the mass Mwhich sets up

the field we can release test masses m1and m2

as shown to determine the strength of the field.



m1

m2 g1g2

(a) Because g= GM/r2.

It varies as 1/r2.(b) Because the

gravitational force

is attractive.


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FYI

The field arrows of the inner ring are longer

than the field arrows of the outer ring and allfield arrows point to the centerline.

M


By placing a series of test masses about a

larger mass, we can map out its gravitational

field:




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If we take a top view, and

eliminate some of the field

arrows, our sketch of the

gravitational field is vastly

simplified:

In fact, we dont even have to

draw the sun-the arrows are sufficient to denote

its presence.

To simplify field drawings

even more, we take the

convention of drawing

field lines with arrows

in their center.



SUN

SUN


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In the first sketch the strength of

the field is determined by the length

of the field arrow.

Since the second sketch has lines,

rather than arrows, how do we know how

strong the field is at a particular

place in the vicinity of a mass?

We simply look at the concentrationof

the field lines. The closer together the

field lines, the stronger the field.

In the red regionthe field lines

are closer together than in the

green region.

Therefore the red fieldis stronger than the

green field.



SUN

SUN


20/26


more point masses.



PRACTICE: Sketch the gravitational field about

the earth (a) as viewed from far away, and (b) as

viewed locally (at the surface).

SOLUTION:

(a)

(b)

orFYI

Note that the closer to the

surface we are, the more uniformthe field concentration.


21/26


more point masses.



EXAMPLE: Find the gravitational field strength at

a point between the earth and the moon that is

right between their centers.

SOLUTION:

A sketch

may help.

Let r= d/2. Thus

gm= Gm/(d/2)2

gm= (6.6710

11)(7.361022)/(3.82108/2)2

gm= 1.3510-4 n.

gM= GM/(d/2)2

gM= (6.671011)(5.981024)/(3.82108/2)2

gM= 1.0910-2 n. Thus g= gMgm= 1.0810

-2 n.

M= 5.981024kg

m= 7.361022kg

d= 3.82

108m

gMgm


22/26

EXAMPLE: Two masses of 225-kg each are located at

opposite corners of a square having a side

length of 645 m. Find the gravitational

field vector at (a) the center of thesquare, and (b) one of the unoccupied

corners.

SOLUTION: Start by making a sketch.

(a) The opposing fields cancel so g= 0.

(b) The two fields are at right angles.

g1= (6.671011)(225)/(645)2= 3.6110-14 n

g2= (6.671011)(225)/(645)2= 3.6110-14 n

g2= g12+ g2

2= 2(3.6110-14)2= 2.6110-27

g= 5.1110-14 n.


more point masses.



s

s m

m

g1

g2

g1g2

sum points

to center

of square

(a)

(b)


23/26


more point masses.



PRACTICE: Determine

the gravitational

field strength at

the points A and B.

SOLUTION: Organize masses and sketch fields.

For point A:

g1= (6.671011)(125)/(225)2= 1.6510-13 n

g2= (6.671011

)(975)/(625 - 225)2

= 4.06

10-13

ng= g2g1= 4.0610

-13 - 1.6510-13 = 2.4110-13 n.

For point B:

g1= (6.671011)(125)/(625 + 225)2= 1.1510-14 n

g2= (6.671011)(975)/(225)2= 1.2810-12 n

g= g1+ g2= 1.1510-14 + 1.2810-12 = 1.2910-12 n.

975 kg125 kg

625 m

225 m 225 m

A B

1 2

g1 g2 g1g2


24/26

Solve problems involving gravitational forces and

fields.



PRACTICE:

Jupiters gravitational field strength at its

surface is 25 Nkg-1while its radius is 7.1107 m.

(a) Derive an expression for the gravitational

field strength at the surface of a planet in

terms of its mass Mand radius R and the

gravitational constant G.

SOLUTION: This is for a general planet

F= Gm1m2/r2 (law of universal gravitation)

F= GMm2/R2 (substitution)

g= F/m2 (definition of gravitational field)

g= (GMm2/R2)/m2 (substitution)

g= GM/R2


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fields.



PRACTICE:

Jupiters gravitational field strength at its

surface is 25 Nkg-1while its radius is 7.1107 m.

(b) Using the given information and the formula

you just derived deduce Jupiters mass.

(c) Find the weight of a 65-kg man on Jupiter.

SOLUTION:

(b)g= GM/R2

(just derived in (a))M= gR2/G (manipulation)

M= (25)(7.1107)2/6.671011

M= 1.91027kg.

(c) F= mg

F = 65(25) = 1600 n.


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fields.



PRACTICE: Two isolated spheres

of equal mass and different

radii are held a distance d

apart. The gravitational fieldstrength is measured on the line joining the two

masses at position xwhich varies. Which graph

shows the variation of gwith xcorrectly?

There is a point between Mand mwhere g= 0.

Since g = Gm/R2

and Rleft < Rright, gleft > gright ath f f h

topic 6.1 gravitational force and fields

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