Topic 6.1 Gravitational Force and Fields
Transcript of 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
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State Newtons universal law of gravitation.
The gravitational force is the weakest of the
four fundamental forces, as the following visual
shows:
Topic 6: Fields and forces
6.1 Gravitational force and field
GRAVITYSTRONG ELECTROMAGNETIC WEAK
+
+
nuclear
force
light, heat,
charge and
magnets
radioactivity freefall,
orbits
ELECTRO-WEAK
WEAKESTSTRONGEST
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State Newtons universal law of gravitation.
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.
Topic 6: Fields and forces
6.1 Gravitational force and field
F= Gm1m2/r2 Universal law
of gravitationwhere G= 6.671011Nm2kg
2
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State Newtons universal law of gravitation.
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.
Topic 6: Fields and forces
6.1 Gravitational force and field
m
M
r
-Newtons shell theorem.
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State Newtons universal law of gravitation.
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.
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6.1 Gravitational force and field
inner core Mi
outer core Mo
mantle Mm
crust Mc
m
r
You do not have
to recall this!
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State Newtons universal law of gravitation.
Be very clear that ris the distance between the
centresof the masses.
Topic 6: Fields and forces
6.1 Gravitational force and field
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.
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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.
Topic 6: Fields and forces
6.1 Gravitational force and field
g= F/m gravitational field strength
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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.
Topic 6: Fields and forces
6.1 Gravitational force and field
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.
Topic 6: Fields and forces
6.1 Gravitational force and field
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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Determine the gravitational field due to one or
more point masses.
Topic 6: Fields and forces
6.1 Gravitational force and field
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
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Define gravitational field strength.
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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Define gravitational field strength.
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.
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6.1 Gravitational force and field
SUN
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Define gravitational field strength.
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.
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6.1 Gravitational force and field
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Define gravitational field strength.
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.
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Define gravitational field strength.
Note that each mass feels a different slope
and must travel at a particular speed to orbit.
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6.1 Gravitational force and field
FYI
The field view eliminates the need for long
distance signaling between two masses. Rather, it
distorts the space about one mass.
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FYI
(a) The field arrow is bigger for m2than m1. Why?
(b) The field arrow always points to M. Why?
M
Define gravitational field strength.
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.
Topic 6: Fields and forces
6.1 Gravitational force and 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
Define gravitational field strength.
By placing a series of test masses about a
larger mass, we can map out its gravitational
field:
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6.1 Gravitational force and field
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Define gravitational field strength.
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.
Topic 6: Fields and forces
6.1 Gravitational force and field
SUN
SUN
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Define gravitational field strength.
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.
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6.1 Gravitational force and field
SUN
SUN
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Determine the gravitational field due to one or
more point masses.
Topic 6: Fields and forces
6.1 Gravitational force and field
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.
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Determine the gravitational field due to one or
more point masses.
Topic 6: Fields and forces
6.1 Gravitational force and field
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
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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.
Determine the gravitational field due to one or
more point masses.
Topic 6: Fields and forces
6.1 Gravitational force and field
s
s m
m
g1
g2
g1g2
sum points
to center
of square
(a)
(b)
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Determine the gravitational field due to one or
more point masses.
Topic 6: Fields and forces
6.1 Gravitational force and field
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
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Solve problems involving gravitational forces and
fields.
Topic 6: Fields and forces
6.1 Gravitational force and field
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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Solve problems involving gravitational forces and
fields.
Topic 6: Fields and forces
6.1 Gravitational force and field
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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Solve problems involving gravitational forces and
fields.
Topic 6: Fields and forces
6.1 Gravitational force and field
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