Gravity
Physics 6A
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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!
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
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 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.
m2
m1
r
F2 on 1
F1 on 2
Prepared by Vince Zaccone
For Campus Learning
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
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
Prepared by Vince Zaccone
For Campus Learning
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
Planet of the Apes:
Planet Hollywood:
Daily Planet:
mass=6 x 1024 kg
mass=6 x 1020 kg
mass=3 x 1025 kg
We should start by defining our coordinate system.
Let’s put the origin at planet H and say positive is to the right.
Prepared by Vince Zaccone
For Campus Learning
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
We can also draw the forces on planet H in our diagram.
Prepared by Vince Zaccone
For Campus Learning
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
Prepared by Vince Zaccone
For Campus Learning
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
m m
Fgrav  G 1 2
r2
Our formula will find the forces (we supply the
direction from looking at the diagram):


24
20

11 Nm2  6  10 kg 6  10 kg
FApes on H   6.67  10

2
kg2 

1012m



Prepared by Vince Zaccone
For Campus Learning
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
m m
Fgrav  G 1 2
r2
Our formula will find the forces (we supply the
direction from looking at the diagram):



24
20

11 Nm2  6  10 kg 6  10 kg
FApes on H   6.67  10
 2.4  1011N

2 
2
kg 

1012m


This is negative because the
force points to the left
Prepared by Vince Zaccone
For Campus Learning
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
m m
Fgrav  G 1 2
r2
Our formula will find the forces (we supply the
direction from looking at the diagram):



24
20

11 Nm2  6  10 kg 6  10 kg
FApes on H   6.67  10
 2.4  1011N

2 
2
kg 

1012m





25
20

11 Nm2  3  10 kg 6  10 kg
FDP on H   6.67  10
 1.3  1011N

2 
2
kg 

3  1012m


This is negative because the
force points to the left
This is positive because the
force points to the right
Prepared by Vince Zaccone
For Campus Learning
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
m m
Fgrav  G 1 2
r2
Our formula will find the forces (we supply the
direction from looking at the diagram):



24
20

11 Nm2  6  10 kg 6  10 kg
FApes on H   6.67  10
 2.4  1011N

2 
2
kg 

1012m





25
20

11 Nm2  3  10 kg 6  10 kg
FDP on H   6.67  10
 1.3  1011N

2 
2
kg 

3  1012m

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
to the left
Prepared by Vince Zaccone
For Campus Learning
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
m
Rplanet
Prepared by Vince Zaccone
For Campus Learning
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)
m
Rplanet
Prepared by Vince Zaccone
For Campus Learning
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
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Kepler’s Laws
1st Law: The orbits of the planets are elliptical,
with the sun at one focus of the ellipse.
a=Semi-major axis
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Kepler’s Laws
2nd Law: A line from the sun to a given planet
sweeps out equal areas in equal times.
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Kepler’s Laws
3rd Law: The periods of the planets are proportional to the 3/2 power
of the major axis length.
Here is a formula:
 2
 3
r 2
T
 GM

cen 

Mass of central object –
orbiting mass does not matter
Radius of orbit (assuming
circular instead of elliptical)
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example: Find the period of a satellite of mass m that orbits the earth
at an altitude of 1500km. How fast is the satellite moving?
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB
Example: Find the period of a satellite of mass m that orbits the earth
at an altitude of 1500km. How fast is the satellite moving?
We can use Kepler’s 3rd Law.
Look up data for the Earth:
Mearth=5.97x1024 kg
Rearth=6.38x106 m
Prepared by Vince Zaccone
For Campus Learning
Assistance Services at UCSB