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Newton’s Law of
Universal Gravitation
©2009 by Goodman & Zavorotniy
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Table of Contents: UG and UCM
Click on the topic to go to that section
· Universal Gravitation
· Gravitational Field
· Surface Gravity
· Gravitational Field in Space
· Orbital Motion
· Kepler's 3rd Law
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Universal Gravitation
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Newton’s Law of Universal Gravitation
It was well known, from
ancient times, that Earth
is a sphere. Which
means that all objects
are attracted to the
center of Earth.
FG
FG
FG
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Newton’s Law of Universal Gravitation
Newton connected the
idea that objects, like
apples, fall towards the
center of Earth with the
idea that the moon orbits
around Earth...it's also
falling towards the center
of Earth.
It just stays in circular
motion since it has a
velocity perpendicular to
its acceleration.
FG
FG
FG
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Newton’s Law of Universal Gravitation
In fact, this view of Earth
is as good as any...just
not one that we're used
to seeing in the USA.
FG
FG
FG
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Newton’s Law of Universal Gravitation
While the magnitude of the force is given by:
FG = G
m 1m 2
r2
G = 6.67 x 10-11 N-m2/kg2
r is the distance between the centers of the masses
The direction of the force is along the line connecting the
centers of the two masses.
r
Each mass feels a force of attraction towards the other
mass...along that line.
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Newton’s Law of Universal Gravitation
While the magnitude of the force is given by:
FG = G
m 1m 2
r2
G = 6.67 x 10-11 N-m2/kg2
r is the distance between the centers of the masses
The direction of the force is along the line connecting
the centers of the two masses.
Each mass feels a force of attraction towards the other
mass...along that line.
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Newton’s Law of Universal Gravitation
Newton's third law tells us that the force on each mass is
equal.
That means that if I drop a pen, the force of Earth pulling
the pen down is equal to the force of the pen pulling
Earth up.
However, since the mass of Earth is so much larger, that
force causes the pen to accelerate down, while the
movement of Earth up is completely unmeasurable.
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1
The gravitational force between two objects is F.
What is the force F' between those objects when
the distance between them is halved?
A
1/2 F
B
1/4 F
C
F
D
2F
E
4F
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2
The gravitational force between two objects is F.
What is the force F' between those objects when
the mass of one object is doubled?
A
1/4 F
B
1/2 F
C
F
D
2F
E
4F
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3
The gravitational force between two objects is F.
What is the force F' between those objects when
the distance between them is doubled?
A 1/4 F
B 1/2 F
C F
D 2F
E
4F
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Gravitational Field
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Gravitational Field
While the force between two objects can always be computed by
using the formula for FG; it's sometimes convenient to consider
one mass as creating a gravitational field and the other mass
responding to that field.
m 1m 2
FG = G 2
r
FG = G
FG =
m1 m
2
r2
GM
r2
m
FG = g m
FG = mg
M
g = G2
r
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Gravitational Field
The magnitude of the gravitational field created by an object
varies from location to location in space; it depends on the
distance from the object and the object's mass.
M
g = G2
r
Gravitational field, g, is a vector. It's direction is always
towards the object creating the field.
That's the direction of the force that a test mass would
experience if placed at that location. In fact, g is the
acceleration that a mass would experience if placed at that
location in space.
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Gravitational Field
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Surface Gravity
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Surface Gravity
Planets, stars, moons, all have a gravitational field...since they
all have mass.
That field is largest at the object's surface, where the distance
from the center of the object is the smallest...when "r" is the
radius of the object.
BTW, only the mass of the
planet that's closer to the
center of the planet than
you are contributes to its
gravitational field. So the
field actually gets smaller if
you tunnel down below the
surface.
R
M
M
g= G 2
R
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4
Compute g for the surface of a planet whose
radius is double Earth's and whose mass is triple
Earth's.
A
1/4 g on Earth
B
1/2 g on Earth
C
3/4 g on Earth
D
g on Earth
E
2 g on Earth
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5
Compute g for the surface of a planet whose
radius and mass are half that of Earth.
A
1/4 g on Earth
B
1/2 g on Earth
C
3/4 g on Earth
D
g on Earth
E
2 g on Earth
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6
Compute g for the surface of a planet whose
radius and mass are double that of Earth.
A
1/4 g on Earth
B
1/2 g on Earth
C
3/4 g on Earth
D
g on Earth
E
2 g on Earth
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Gravitational Field in Space
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Gravitational field in space
While gravity gets weaker as you get farther
from a planet, it never becomes zero. There
is always some gravitational field present due
to every planet, star and moon in the
universe.
However, the local field is usually dominated
by nearby masses since gravity gets weaker
as the inverse square of the distance.
The contribution of a planet to the local
gravitational field can be calculated using the
same equation we've been using. You just
have to be careful about "r".
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Gravitational field in space
The contribution of a planet to the local gravitational field can
be calculated using the same equation we've been using. You
just have to be careful about "r".
If a location, "A", is a height "h" above a planet of radius "R",
it is a distance "r" from the planet's center, where r = R + h.
M
gA = G 2
r
R
gA =
M
r
h
A
GM
(R+h)2
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7
What is g at a height of 2 REarth above the surface
of Earth?
A
1/9 g on Earth
B
1/4 g on Earth
C
1/3 g on Earth
D
1/2 g on Earth
E
0.95 g on Earth
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8
What is g at a height of 1 REarth above the surface
of Earth?
A
1/9 g on Earth
B
1/4 g on Earth
C
1/3 g on Earth
D
1/2 g on Earth
E
0.95 g on Earth
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The International Space Station (ISS) is a research facility currently being
assembled in outer space, the on-orbit construction of which began in 1998.
The space station is in a
Low Earth Orbit and can
be seen from Earth with
the naked eye; it orbits at
an altitude of
approximately 350 km
(190 mi) above the surface
of the Earth, and travels
at an average speed of
27,700 kilometres (17,210
mi) per hour, completing
15.7 orbits per day.
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9
What is g at the height of the space station, 3.5 x
105 m. (REarth = 6.4 x 106m)?
A
1/9 g on Earth
B
1/4 g on Earth
C
1/3 g on Earth
D
1/2 g on Earth
E
0.90 g on Earth
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10 How does the gravitational field acting on the
occupants of the space station compare to that
acting on you now?
A
there is no gravity acting on them
B
it's about 1/3 as much
C
it's about 1/2 as much
D
it's close to the same
E
it's exactly the same
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Orbital Motion
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Orbital Motion
We've already determined that the gravitational field acting on
the occupants of the space station, and on the space station
itself, is not very different than the force acting on us.
How come they don't
fall to Earth?
R
Let's start with the
diagram we used
earlier.
M
r
h
A
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Orbital Motion
The gravitational field will be pointed towards the center of
Earth and represents the acceleration that a mass would
experience at that location (regardless of the mass).
In this case any object would
simply fall to Earth. How could
that be avoided?
a
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Orbital Motion
If the object has a tangential velocity perpendicular to its
acceleration, it will go in a circle. It will keep falling to Earth, but
never strike Earth.
v
a
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Orbital Motion
If the object has a
tangential velocity
perpendicular to its
acceleration, it will go
in a circle. It will keep
falling to Earth, but
never strike Earth.
v
a
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Here is Newton's own
drawing of a thought
experiment where a
cannon on a very high
mountain (above the
atmosphere) shoots a
shell with increasing
speed, shown by
trajectories for the shell of
D, E, F, and G and finally
so fast that it never falls to
earth, but goes into orbit.
(From M. Fowler,
http://galileoandeinstein.physics.vi
rginia.edu/lectures/newton.html)
http://ircamera.as.arizona.edu/NatS
ci102/lectures/newton.htm
Orbital Motion
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Orbital Motion
We can calculate the velocity
necessary to maintain a stable
orbit at a distance "r" from the
center of a planet of mass "M".
ΣF = ma
v
a
GMm/r2 = mv2/r
GM/r2 = v2/r
GM/r = v2
v = (GM/r)1/2
[g = v2/r]
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Orbital Motion
From that, we can calculate the
period, T, of any object's orbit.
v = 2πr/T
T = 2πr/v
OR
g = v2/r
v = (gr)1/2
T = 2πr/(gr)1/2
T = 2π(r/g)1/2
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11 By what factor is g at a height of 59 REarth lower
than g at Earth's surface?
A
1/59
B
1/3600
C
1/60
D
1/978
E
1/5432
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12 Use your previous answer to determine the
magnitude of velocity that would allow an object to
have a circular orbit at a height of 59REarth above
Earth's surface.
A
7894.48 m/s
B
1019.17 m/s
C
3678.23 m/s
D
2848.46 m/s
E
6705.56 m/s
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Orbital Motion
Now, we can find the relationship
between the period, T, and the orbital
radius, r, for any orbit.
v = 2πr/T
v2 = 4π2r2 / T2
T2 = 4π2r2 / v2
T2 = 4π2r2 / (GM/r)
T2 = (4π2r2)(r / GM)
T2 = (4π2r3) / (GM)
T2 4π2
=
r3 GM
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Kepler's 3rd Law
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Kepler's Third Law
T2 4π2
=
r3 GM
Kepler had noted that the ratio of T2 / r3 yields the same
result for all the planets. That is, the square of the period
of any planet's orbit divided by the cube of its distance
from the sun always yields the same number.
We have now shown why: (4π2) / (GM) is a constant; its
the same for all orbiting objects, where M is the mass of
the object being orbited; it is independent of the object
that is orbiting.
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** Kepler's Third Law
If you know the period (T) of
a planet's orbit, you can
determine its distance (r)
from the sun.
T2 4π2
=
r3 GM
Since all planets orbiting
the sun have the same
period to distance ratio, the
following is true:
T(red)2
Click object to see
orbital motion
r(red)3
=
T(green)2
r(green)3
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13 The moon orbits Earth in about 28 days. What
would be the orbital period of an object whose
orbital radius is half that of the moon?
A
56 days
B
40 days
C
24 days
D
14 days
E
10 days
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14 The moon orbits Earth in about 28 days. What
would be the orbital period of an object whose
orbital radius is double that of the moon?
A
224 days
B
79 days
C
56 days
D
40 days
E
10 days
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15 Earth orbits the sun in about 365 days. What would be the
orbital period of an object whose orbital radius is double
that of Earth?
A
2920 days
B
1460 days
C
1032 days
D
730 days
E
129 days
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