Download (T) involved in our formula? 12-3 Kepler`s Laws of

Gravity
Physics 6B
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GRAVITY
Any pair of objects, anywhere in the universe, feel a mutual attraction due to gravity.
There are no exceptions – if you have mass, every other mass is attracted to you, and you are attracted to
every other mass. Look around the room – everybody here is attracted to you!
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GRAVITY
Any pair of objects, anywhere in the universe, feel a mutual attraction
due to gravity.
There are no exceptions – if you have mass, every other mass is attracted to you, and you are
attracted to every other mass. Look around the room – everybody here is attracted to you!
Newton’s Law of Universal Gravitation gives us a formula to calculate the attractive force
between 2 objects:
Fgrav  G
m1  m2
r2
m1 and m2 are the masses, and r is the center-to-center distance between them
G is the gravitational constant – it’s tiny: G≈6.674*10-11 Nm2/kg2
Use this formula to find the magnitude of the gravity
force.
Use a diagram or common sense to find the direction.
The force will always be toward the other mass.
m1
r
m2
F2 on 1
F1 on 2
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GRAVITY
Any pair of objects, anywhere in the universe, feel a mutual attraction
due to gravity.
There are no exceptions – if you have mass, every other mass is attracted to you, and you are
attracted to every other mass. Look around the room – everybody here is attracted to you!
Newton’s Law of Universal Gravitation gives us a formula to calculate the attractive force
between 2 objects:
Fgrav  G
m1  m2
r2
m1 and m2 are the masses, and r is the center-to-center distance between them
G is the gravitational constant – it’s tiny: G≈6.674*10-11 Nm2/kg2
Use this formula to find the magnitude of the gravity
force.
Use a diagram or common sense to find the direction.
The force will always be toward the other mass.
m1
r
*Note: If you are dealing with spherical objects with
uniform density (our typical assumption) then you can
pretend all the mass is concentrated at the center.
m2
F2 on 1
F1 on 2
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Example:
Three planets are aligned as shown. The masses and distances are given in the diagram.
Find the net gravitational force on planet H (the middle one).
1012 m
3 x 1012 m
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
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Example:
Three planets are aligned as shown. The masses and distances are given in the diagram.
Find the net gravitational force on planet H (the middle one).
1012 m
3 x 1012 m
FApes on H
FDP on H
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
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Assistance Services at UCSB
Example:
Three planets are aligned as shown. The masses and distances are given in the diagram.
Find the net gravitational force on planet H (the middle one).
1012 m
3 x 1012 m
FApes on H
FDP on H
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
Our formula will find the forces (we supply
the direction from looking at the diagram):
m m
Fgrav  G 1 2
r2
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Example:
Three planets are aligned as shown. The masses and distances are given in the diagram.
Find the net gravitational force on planet H (the middle one).
1012 m
3 x 1012 m
FApes on H
FDP on H
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
m m
Fgrav  G 1 2
r2
24
20
11 Nm2  6  10 kg 6  10 kg

FApes on H   6.67  10

2
kg2 

1012 m
Our formula will find the forces (we supply
the direction from looking at the diagram):





This is negative because
the force points to the left
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Example:
Three planets are aligned as shown. The masses and distances are given in the diagram.
Find the net gravitational force on planet H (the middle one).
1012 m
3 x 1012 m
FApes on H
FDP on H
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
Our formula will find the forces (we supply
the direction from looking at the diagram):


m m
Fgrav  G 1 2
r2

24
20

11 Nm 2  6  10 kg 6  10 kg
FApes on H   6.67  10
 2.4  1011N

2 
2
kg 

1012 m


This is negative because
the force points to the left
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Example:
Three planets are aligned as shown. The masses and distances are given in the diagram.
Find the net gravitational force on planet H (the middle one).
1012 m
3 x 1012 m
FApes on H
FDP on H
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
Our formula will find the forces (we supply
the direction from looking at the diagram):


m m
Fgrav  G 1 2
r2

24
20

11 Nm 2  6  10 kg 6  10 kg
FApes on H   6.67  10
 2.4  1011N

2 
2
kg 

1012 m





25
20

11 Nm 2  3  10 kg 6  10 kg
FDP on H   6.67  10
 1.3  1011N

2 
2
kg 

3  1012 m


This is negative because
the force points to the left
This is positive because the
force points to the right
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Assistance Services at UCSB
Example:
Three planets are aligned as shown. The masses and distances are given in the diagram.
Find the net gravitational force on planet H (the middle one).
1012 m
3 x 1012 m
FApes on H
FDP on H
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
Our formula will find the forces (we supply
the direction from looking at the diagram):


m m
Fgrav  G 1 2
r2

24
20

11 Nm 2  6  10 kg 6  10 kg
FApes on H   6.67  10
 2.4  1011N

2 
2
kg 

1012 m





25
20

11 Nm 2  3  10 kg 6  10 kg
FDP on H   6.67  10
 1.3  1011N

2 
2
kg 

3  1012 m

Add the forces to get the net force on H:

11
Fnet  1.1 10 N
This is negative because
the force points to the left
This is positive because the
force points to the right
Net force is
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to the left
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GRAVITY
One more useful detail about gravity:
The acceleration due to gravity on the surface of a planet is right there in the formula.
Here is the gravity formula, modified for the case where m is the mass of an object on the
surface of a planet.
Fgrav  G
mplanet  m
Rplanet 2
m
Rplanet
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GRAVITY
One more useful detail about gravity:
The acceleration due to gravity on the surface of a planet is right there in the formula.
Here is the gravity formula, modified for the case where m is the mass of an object on the
surface of a planet.
Fgrav  G
mplanet  m
Rplanet 2
We already know that Fgrav is the weight of the
object, and that should just be mg (if the planet
is the Earth)
mg  G
mplanet  m
Rplanet 
2
m
Rplanet
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Assistance Services at UCSB
GRAVITY
One more useful detail about gravity:
The acceleration due to gravity on the surface of a planet is right there in the formula.
Here is the gravity formula, modified for the case where m is the mass of an object on the
surface of a planet.
Fgrav  G
mplanet  m
Rplanet 2
We already know that Fgrav is the weight of the
object, and that should just be mg (if the planet
is the Earth)
mg  G
mplanet  m
Rplanet 
2
m
Rplanet
This part is g
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12-3 Kepler’s Laws of Orbital Motion
Johannes Kepler made detailed studies of the
apparent motions of the planets over many years,
and was able to formulate three empirical laws:
1. Planets follow elliptical orbits, with the Sun at
one focus of the ellipse.
12-3 Kepler’s Laws of Orbital Motion
2. As a planet moves in its orbit, it sweeps out an
equal amount of area in an equal amount of time.
12-3 Kepler’s Laws of Orbital Motion
3. The period, T, of a planet increases as its
mean distance from the Sun, r, raised to the 3/2
power.
This can be shown to be a consequence of the
inverse square form of the gravitational force.
12-3 Kepler’s Laws of Orbital Motion
What should you use for the constant?
12-3 Kepler’s Laws of Orbital Motion
What should you use for the constant?
Let’s derive the formula to find out.
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Fgrav  G
m1  m2
r2
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Fgrav  G
m1  m2
r2
Notice that the gravity force points
toward the center: i.e. it is the
centripetal force.
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Fgrav  G
m1  m2
r2
Notice that the gravity force points
toward the center: i.e. it is the
centripetal force.
We have a formula for centripetal
force:
Fcent
m  v2

r
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Set these equal to get our formula:
M  m m  v2
G 2 
r
r
M is the mass of
the central object.
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Set these equal to get our formula:
The mass of the orbiting object
will cancel out:
M v2
G 2 
r
r
How do we get the period (T)
involved in our formula?
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Set these equal to get our formula:
The mass of the orbiting object
will cancel out:
M v2
G 2 
r
r
Each orbit is one trip around the
circle, so the distance traveled is 2∏r.
Speed is distance/time, so we get:
2r
v
T
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Set these equal to get our formula:
Plug in our expression for v:
2
M (2Tr )
G 2 
r
r
Now we can rearrange this to
solve for T. Do this now.
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
Set these equal to get our formula:
Plug in our expression for v:
2
M (2Tr )
G 2 
r
r
Now we can rearrange this to
solve for T. Do this now.
3
2
T
 r2
GM
Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
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Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
Answer: Deimos is farther from Mars. Inspection of the formula in Kepler’s 3rd Law should make this clear.
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Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant.
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Assistance Services at UCSB
Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant.
Previously we derived an expression for the constant. Our formula is:
3
2
T
 r2
GM
What do we use for M in this formula?
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Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant.
Previously we derived an expression for the constant. Our formula is:
3
2
T
 r2
GM
Here M is the mass of the object being orbited, in this case Mars.
We can look this up in a table: MMars = 6.45x1023 kg
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Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant.
Previously we derived an expression for the constant. Our formula is:
3
2
T
 r2
GM
Here M is the mass of the object being orbited, in this case Mars.
We can look this up in a table: MMars = 6.45x1023 kg
Rearrange this formula to solve for radius, then plug in the numbers:
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Assistance Services at UCSB
Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant.
Previously we derived an expression for the constant. Our formula is:
3
2
T
 r2
GM
Here M is the mass of the object being orbited, in this case Mars.
We can look this up in a table: MMars = 6.45x1023 kg
Rearrange this formula to solve for radius, then plug in the numbers:
r3
2
GM
3

T
42
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Example:
The Martian moon Deimos has an orbital period that is greater than the other Martian moon,
Phobos. Both moons have approximately circular orbits.
a) Is Deimos closer to or farther from Mars than Phobos? Explain.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
b) Calculate the distance from the center of Mars to Deimos given that its period is 1.1x105 s.
We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant.
Previously we derived an expression for the constant. Our formula is:
3
2
T
 r2
GM
Here M is the mass of the object being orbited, in this case Mars.
We can look this up in a table: MMars = 6.45x1023 kg
Rearrange this formula to solve for radius, then plug in the numbers:
2
r
3
2
GM
3

T

2
4
3
(6.67  1011 Nm2 )(6.45  1023 kg)
kg
42
5
2
3
 (1.10  10 s)  2.36  107 m
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12-4 Gravitational Potential Energy
Gravitational potential energy of an object of
mass m a distance r from the Earth’s center:
12-5 Energy Conservation
Total mechanical energy of an object of mass m
a distance r from the center of the Earth:
This confirms what we already know – as an
object approaches the Earth, it moves faster
and faster.