PHY GRA

1
1
Gravitation
2
Definition of Weight Revisited
The weight of an object on or above the earth is the
gravitational force that the earth exerts on the object.
The weight always points toward the center of mass of the
earth.
On or above another astronomical body, the weight is the
gravitational force exerted on the object by that body. The
direction of the weight (or gravitational force) points towards
the center of mass of that body.
SI Unit of Weight: Newton (N)
3
Newton’s Law of Universal Gravitation
2
r
mM
GW E
=
mgW =
2
r
M
Gg E
=
where Wis the weight of an object with
mass mdue to the earth’s gravitational
force, Gis the universal gravitational
constant = 6.67x10-11 m2/kg2, MEis the
mass of the earth, ris the distance
between the object and the center of
mass of the earth.
(1)
Write Win the usual form,
We get
RE= 6.38x106m
ME= 5.97x1024 kg
G = 6.67x10-11 m2kg-2 4
Newton’s Law of Universal Gravitation
1.The gravitational acceleration, gdepends on the
distance, r, between the object and the earth’s
center of mass.
2.Equation (1) can be generalized for the gravitational
force between two objects with masses mand M, for
which MEin eqn. (1) is replaced by M and the
distance rrepresents the distance between the
centers of mass of the two objects.
3.By Newton’s 3rd law, the object acts on the earth with
a force having the same magnitude but pointing in
the opposite direction.
5
Gravitational Field
The gravitational field, g, at a point is the gravitation
force an object experiences when placed at that point
divided by the object’s mass. For gravitational field
coming from the earth,
m
r
mM
Gg E1
2⋅=
2
r
M
Gg E
=
where g is in units of m/s2and ris the distance the
point is from the center of mass of the earth. This result
shows that the gravitational field is the same as the
gravitational acceleration. 6
Gravitational Field
Since the gravitational field is essential the
gravitational force experienced by a unit mass at the
point of interest, it should have a direction. The
direction of the gravitational field is pointed towards
the body that produces the field. In other words,
gravitational force always attracts the object towards
the body producing the field.
Note that gravitational force is a kind of
interaction forces. So both the object and the body
involved experience the same magnitude of
attractive force from each other.
11/15/2010 (Mon)

2
7
Gravitational field produced by m2at a distance
of r12:
Gravitational Field
m1m2
F12=m1g2
F21=m2g1
g2= Gm2/r122
g1= Gm1/r122(< g2since m1< m2)
r12
Gravitational field produced by m1at a distance
of r12:
m1< m2
8
Gravitational force on the earth’s surface
Solution:
Find the gravitational field gon the earth’s
surface.
()
()
()
()
()
()
2
622411
2
22
2
6
24
2211
2
sm 80.9
10
m 6.38
kg98.5
kgmN67.6
m 106.38
kg 1098.5
kgmN1067.6
=
×⋅=
×
×
⋅×=
=
×−+−
−
E
E
R
M
Gg
9
Gravitational force acting on the earth
by a car
Find the gravitational acceleration acting on the earth by
a car with mass 1500 kg running on its surface.
− mg
+mg
Solution:
The gravitational force, F, acting on the
earth by the car is (1500kg)(9.8m/s2)
=14700N pointed from the earth’s
center towards the car.
The acceleration on the earth due to
this force is F/ME
= 14700N/(5.98x1024kg)
= 2.46x10-21m/s2, which would be too small to be detected. 10
Earth and Moon
Using the fact that the gravitational field at the surface of
the Earth is about six times larger than that at the
surface of the Moon, and the fact that the Earth’s radius
is about four times the Moon’s radius, determine how the
mass of the Earth compares to the mass of the Moon.
=2
Gm
gr
⇒r12g1/m1= r22g2/m2
m2/m1= (r2/r1)2(g2/g1) = (1/4)2(1/6) = 1/96
(1: earth, 2: moon)
So the mass of the moon is 1/96 times of that of the
earth.
Solution:
11
Three masses on a straight line
(a) Three masses, of mass
2M, M, and 3M are equally
spaced along a line, as
shown. The only forces
each mass experiences are the forces of gravity from the
other two masses.
(i) Which mass experiences the largest magnitude net force?
1.2M
2.M
3.3M
4.Equal for all three
5.Net force magnitude on 2M is equal to that
on 3M but bigger than that on M
2M M3M
RR
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(ii) What’s the magnitude of the net force experienced by
mass 2M?
2M M3M
RR
Three masses on a straight line
Solution
We can just add the forces from the other two objects. The
net force on the 2M object is directed right with a magnitude
of:

3
13
A triangle of masses
Three point objects, 1 through
3 with identical mass, are
placed at the corners
of an equilateral triangle.
In what direction is the net gravitational field at point A,
halfway between objects 2 and 3?
23
1
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Net gravitational field at point A
The net gravitational field at point A comes from three
sources, objects 1, 2, and 3. The diagram below shows the
corresponding three gravitational fields, g1, g2and g3.
Obviously, g2and g3cancel.
The net field at A is due only to g1, which points up.
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1
g3
g2g1
gi= Gm/riA2
∝1/riA2
where riA is the distance
between point A and
object i, m is the mass of
the objects.
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Net gravitational force at point A
Find the net gravitational force experienced by an object of
mass 2M at point A in terms of the mass of the three source
objects m and the length of each side of the equilateral
triangle, L.
From the above discussion, the net gravitational field is g1,
which is Gm/r1A2. But r1A = Lsin60o= (√3/2)L. So,
g1= 2Gm/(3L2).
The net gravitational force, F, experienced by an object with
mass 2M is 2M times the field at that point. So,
F = (2M)g1= 4GmM/(3L2).
Solution
16
There is only one speed, v, that a satellite can have if the
satellite is to remain in a circular orbit with radius, r. What is
the relation between vand r ? What is vat r= 10RE?
2
2
r
mM
Gmg
r
v
mE
==
Velocity of Orbiting Satellites
r
GM
vE
=⇒
Solution:
Fnet = mg
For the satellite to remain in a
circular orbit with radius r, the
acceleration of the satellite must
equal ac= v2/rfor otherwise the
unbalanced force, Fnet –macwill
cause the satellite to move away
from the (circular) orbit.
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2M M3M
RR
Three masses on a straight line
(iii) Which of the following changes would cause the
magnitude of the force experienced by the 2M object to
increase by a factor of 4? Select all that apply.
[ ] double the mass of all three objects
[ ] change the mass of the 2M object to 8M, without
changing the mass of the other objects
[ ] double R
[ ] Move the system to a parallel universe where the value of
the universal gravitational
constant is four times larger than its value in our universe
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Velocity of Orbiting Satellites
r
GM
vE
=
Substitute r= 10REto find v in the orbit:
m1038.610 kg1097.5kgm1067.6 6
242-211
xx xxx −
=
= 2.50 km/s
This is faster than the speed of any plane ever flown on
earth (which is ~2 km/s)!
() 2/1612411
2-2 10
m38.6 kg97.5kgm67.6 ×−−+−
×
×
=

4
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In many applications of satellites such as digital satellite
system television, it is desirable that the motion of the satellite
follows a circular orbit and be synchronized with the
The Orbital Radius for Synchronous
satellites
of the satellite be exactly one day,
i.e., 8.64 x 104s. What is the height,
H, of the satellite above the earth’s
surface?
earth’s self rotation (so that the
satellite is always at the same
location above the earth’s surface).
This requires that the period, τ,
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The Orbital Radius for Synchronous
satellites
τ
π
r
v 2 =
r
GM
vE
=
From last example,
But, for uniform circular motion,
=⇒
=⇒
r
GM
rE
π
τ
2
2/3
4.22 x 107m
So, H= r–RE= 3.58 x 107m
Solution:
(2)
r
GM
rE
=⇒
τ
π
2RE= 6.38x106m
ME= 5.97x1024 kg
G = 6.67x10-11 m2kg-2
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General Orbital Motions of satellites
1.They often trace out an ellipse. Therefore, the
gravitational force is not always perpendicular to
the satellite’s velocity.
2.K + U is conserved throughout the orbit
3.Angular momentum is conserved throughout the
orbit.
4.Linear momentum is not conserved since there
is gravitational force.
5.The orbit period does not depend on the mass
of the satellite.
6.They obey the Kepler’s 2nd Law. That is, equal
areas of the orbit are swept out in equal time
intervals. 22
Gravitational potential energy
The gravitational interaction or potential energy of two objects
with masses mand Mand separation r is:
The negative sign tells us that the interaction is attractive.
Note that with this equation the potential energy is defined to
be zero when r= infinity.
What matters is the change in gravitational potential energy.
For small changes in height at the Earth’s surface, i.e., from r
= REto r= RE+h, the equation above gives the same change
in potential energy as mgh, where g= GM/RE2as found
before.
gGmM
Ur =−
23
Four objects in a square
Four objects of equal mass, m, are placed at the corners
of a square that measures Lon each side. How much
gravitational potential energy is associated with this
configuration of masses?
24
Label the four objects by 1, 2, 3, 4 as shown. Imagine that
you start with no object on the square. (1) You first bring
object 1 in place. That causes no energy change since there
is no other masses in the space. (2) Then you bring object 2
in place. The potential energy (PE) involved is mg12, where
g12 = −Gm/Lis the gravitational field produced by object 1 at
where object 2 is placed. (3) Then you bring object 3 in. The
PE involved is m[g13 + g23]. (4) Finally, you bring object 4 in.
The PE required is m[g14 + g24 + g34].
Among these, U12, U23, U34, and U14 all equal -Gm2/L
And U13 and U24 equal -Gm2/(√2L)
The gravitational PE is the sum of all the six
Uij’s, which is -(4+2√2)Gm2/L.
Convince yourself that the answer is independent
of the order by which we bring the charges in.
Four objects in a square
1
23
4

5
25
Gravitational potential energy of a
distribution of masses
In general, the gravitational potential energy, U, of a
distribution of masses comprising masses mi(where i
= 1, 2, 3, …, N) is:
U = -G[m1m2/r12 + m1m3/r13 + … + m1mN/r1N +
m2m3/r23 + m2m4/r24 + … + m2mN/r2N +
……
+ mN-1mN/rN-1,N],
where rij is the distance between miand mj
(i, j = 1, 2, 3, …, N.) 26
Escape speed
How fast would you have to throw an object so it never
came back down? Ignore air resistance. Find the escape
speed - the minimum speed required to escape from a
planet's gravitational pull.
Solution:
Approach to use: Forces are hard to work with here
because the size of the force changes as the object gets
farther away. Energy is easier to work with in this case.
+
+=+
ii ncff
UKW UK
Conservation of energy:
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Escape speed
Which terms can we cross out immediately?
1. Assume no resistive forces, so Wnc = 0.
2. Assume that the object barely makes it to infinity, so both
Ufand Kfare zero.
This leaves: +=0
ii
UK
⇒-GmME/RE+ ½ mvescape2 = 0
⇒vescape2 = 2GME/REx(RE/RE)
= 2(GME/RE2)RE
= 9.8 m/s2
⇒vescape2= 2(9.8 m/s2)(6.38x106m)
⇒vescape = √(125x106)m/s = 11.2 km/s
Simulation
28
Escape speed
Which terms can we cross out immediately?
1. Assume no resistive forces, so Wnc = 0.
2. Assume that the object barely makes it to infinity, so both
Ufand Kfare zero.
This leaves:
+
=0
ii
UK
⇒-GmME/RE+ ½ mvescape2 = 0
⇒vescape2 = 2GME/REx(RE/RE)
= 2(GME/RE2)RE
= 9.8 m/s2
⇒vescape2= 2(9.8 m/s2)(6.38x106m)
⇒vescape = √(125x106)m/s = 11.2 km/s
Simulation