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Gravitation
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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)
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Newton’s Law of Universal Gravitation
r^2
M m
W G
E
W = mg
r^2
M
g = G E
where W is the weight of an object with mass m due to the earth’s gravitational force, G is the universal gravitational constant = 6.67x10-11^ m^2 /kg^2 , M E is the mass of the earth, r is the distance between the object and the center of mass of the earth.
Write W in the usual form,
We get
R E = 6.38x10^6 m M E = 5.97x10^24 kg G = 6.67x10-11^ m 2 kg-2^4
Newton’s Law of Universal Gravitation
1. The gravitational acceleration, g depends 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 m and M , for
which M E in eqn. (1) is replaced by M and t he
distance r represents the distance between the
centers of mass of the two objects.
3. By Newton’s 3 rd^ law, the object acts on the earth with
a force having the same magnitude but pointing in
the opposite direction.
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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,
r m
mM
g G E^
r^2
M
g = G E
where g is in units of m/s 2 and r is 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.
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Gravitational field produced by m 2 at a distance
of r 12 :
Gravitational Field
m 1 m 2
F 12 =m 1 g 2
F 21 =m 2 g 1
g 2 = Gm 2 /r 122
g 1 = Gm 1 /r 122 (< g 2 since m 1 < m 2 )
r 12
Gravitational field produced by m 1 at a distance
of r 12 :
m 1 < m 2
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Gravitational force on the earth’s surface
Solution:
Find the gravitational field g on the earth’s
surface.
( )
( )
( )
( )
( ) ( ) 2
112426 2
2 2
6 2
24 11 2 2
2
9. 80 ms
6.38m
5. 98 kg
6. 67 Nm kg
6.38 10 m
5. 98 10 kg
6. 67 10 Nm kg
= ⋅ ×
×
×
= × ⋅
−+ −×
−
E
E
R
M
g G
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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/s 2 ) =14700N pointed from the earth’s center towards the car.
The acceleration on the earth due to this force is F / M E
= 14700N/(5.98x10^24 kg) = 2.46x10 -21^ m/s 2 , 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.
g =^ Gm 2
r
⇒ r 12 g 1 / m 1 = r 22 g 2 / m 2
m 2 / m 1 = ( r 2 / r 1 ) 2 ( g 2 / g 1 ) = (1/4) 2 (1/6) = 1/
(1: earth, 2: moon)
So the mass of the moon is 1/96 times of that of the
earth.
Solution:
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Three masses on a straight line
(a) Three masses, of mass 2 M , M , and 3 M 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 (^) M 3M
R R
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(ii) What’s the magnitude of the net force experienced by mass 2M?
2M (^) M 3M
R R
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:
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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 10^4 s. 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
r
GM
From last example, v = E
But, for uniform circular motion,
r
GM
r E π
τ 2
3 / 2
4.22 x 10 7 m
So, H = r – R E = 3.58 x 10^7 m
Solution:
r
⇒ r^ = GME τ
2 π R E^ = 6.38x
(^6) m M E = 5.97x10^24 kg G = 6.67x10-11^ m 2 kg-
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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 2 nd^ 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 m and M and 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 = R E to r = R E + h , the equation above gives the same change in potential energy as mgh, where g = GM / R (^) E^2 as found before.
g
GmM U r
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Four objects in a square
Four objects of equal mass, m, are placed at the corners of a square that measures L on each side. How much gravitational potential energy is associated with this configuration of masses?
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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 mg 12 , where g 12 = − Gm / L is 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 [ g 13 + g 23 ]. (4) Finally, you bring object 4 in. The PE required is m [ g 14 + g 24 + g 34 ].
Among these, U 12 , U 23 , U 34 , and U 14 all equal - Gm^2 / L And U 13 and U 24 equal - Gm^2 /(√ 2 L )
The gravitational PE is the sum of all the six U ij’s, which is -(4+2√2) Gm^2 / L.
Convince yourself that the answer is independent of the order by which we bring the charges in.
Four objects in a square
1
(^2 )
4
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Gravitational potential energy of a distribution of masses
In general, the gravitational potential energy, U, of a
distribution of masses comprising masses m i (where i
= 1, 2, 3, …, N) is:
U = -G[m 1 m 2 /r 12 + m 1 m 3 /r 13 + … + m 1 m N /r 1N +
m 2 m 3 /r 23 + m 2 m 4 /r 24 + … + m 2 m N /r 2N +
+ m N-1m N /r N-1,N ],
where r ij is the distance between m i and mj
(i, j = 1, 2, 3, …, N.)
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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.
Ui + Ki + Wnc = Uf + Kf
Conservation of energy:
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Escape speed
Which terms can we cross out immediately?
- Assume no resistive forces, so W nc = 0.
- Assume that the object barely makes it to infinity, so both Uf and Kf are zero.
This leaves: U (^) i + Ki = 0
⇒ - GmM E / R E + ½ mv escape^2 = 0
⇒ v escape^2 = 2 GM E / R E x( R E / R E ) = 2( GM E / R E^2 ) R E
= 9.8 m/s 2 ⇒ vescape^2 = 2(9.8 m/s 2 )(6.38x10^6 m)
⇒ v escape = √(125x10^6 )m/s = 11.2 km/s
Simulation
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Escape speed
Which terms can we cross out immediately?
- Assume no resistive forces, so W nc = 0.
- Assume that the object barely makes it to infinity, so both U (^) f and Kf are zero.
This leaves: Ui + Ki = 0
⇒ - GmM E / R E + ½ mv escape^2 = 0
⇒ v escape^2 = 2 GM E / R E x( R E / R E ) = 2( GM E / R E^2 ) R E
= 9.8 m/s 2 ⇒ vescape^2 = 2(9.8 m/s 2 )(6.38x10^6 m)
⇒ v escape = √(125x10^6 )m/s = 11.2 km/s
Simulation
