Gravity
Physics 6B
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 Universal Gravitation gives us a formula to calculate the attractive force between 2 objects:
221
gravr
mmGF
⋅=
m1 and m2 are the masses, and r is the center-to-center distance between them
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
m1
m2
r
F1 on 2
F2 on 1
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.
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:
221
gravr
mmGF
⋅=
m1 and m2 are the masses, and r is the center-to-center distance between them
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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
m1
m2
r
F1 on 2
F2 on 1
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.
*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.
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).
Planet Hollywood:Planet of the Apes: Daily Planet:
1012 m 3 x 1012 m
mass=6 x 1020 kgmass=6 x 1024 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).
Planet Hollywood:Planet of the Apes: Daily Planet:
1012 m 3 x 1012 m
FDP on HFApes on H
mass=6 x 1020 kgmass=6 x 1024 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).
Planet Hollywood:Planet of the Apes: Daily Planet:
1012 m 3 x 1012 m
FDP on HFApes on H
mass=6 x 1020 kgmass=6 x 1024 kg mass=3 x 1025 kg
Our formula will find the forces (we supply the direction from looking at the diagram): 2
21grav
r
mmGF
⋅=
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).
Planet Hollywood:Planet of the Apes: Daily Planet:
1012 m 3 x 1012 m
FDP on HFApes on H
mass=6 x 1020 kgmass=6 x 1024 kg mass=3 x 1025 kg
Our formula will find the forces (we supply the direction from looking at the diagram): 2
21grav
r
mmGF
⋅=
( )( )( )212
2024
kgNm11
HonApesm10
kg106kg1061067.6F 2
2 ⋅⋅
⋅−= − 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).
Planet Hollywood:Planet of the Apes: Daily Planet:
1012 m 3 x 1012 m
FDP on HFApes on H
mass=6 x 1020 kgmass=6 x 1024 kg mass=3 x 1025 kg
Our formula will find the forces (we supply the direction from looking at the diagram): 2
21grav
r
mmGF
⋅=
( )( )( )
N104.2m10
kg106kg1061067.6F 11
212
2024
kgNm11
HonApes 2
2⋅−=
⋅⋅
⋅−= − 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).
Planet Hollywood:Planet of the Apes: Daily Planet:
1012 m 3 x 1012 m
FDP on HFApes on H
mass=6 x 1020 kgmass=6 x 1024 kg mass=3 x 1025 kg
Our formula will find the forces (we supply the direction from looking at the diagram): 2
21grav
r
mmGF
⋅=
( )( )( )
N104.2m10
kg106kg1061067.6F 11
212
2024
kgNm11
HonApes 2
2⋅−=
⋅⋅
⋅−= − This is negative because
the force points to the left
( )( )( )
N103.1m103
kg106kg1031067.6F 11
212
2025
kgNm11
HonDP 2
2⋅+=
⋅
⋅⋅
⋅+= − 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).
Planet Hollywood:Planet of the Apes: Daily Planet:
1012 m 3 x 1012 m
FDP on HFApes on H
mass=6 x 1020 kgmass=6 x 1024 kg mass=3 x 1025 kg
Our formula will find the forces (we supply the direction from looking at the diagram): 2
21grav
r
mmGF
⋅=
( )( )( )
N104.2m10
kg106kg1061067.6F 11
212
2024
kgNm11
HonApes 2
2⋅−=
⋅⋅
⋅−= − This is negative because
the force points to the left
( )( )( )
N103.1m103
kg106kg1031067.6F 11
212
2025
kgNm11
HonDP 2
2⋅+=
⋅
⋅⋅
⋅+= − This is positive because the
force points to the right
Add the forces to get the net force on H: N101.1F 11net ⋅−=
Net force is to the left Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
GRAVITYOne 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.
( )2planet
planetgrav
R
mmGF
⋅=
Rplanet
m
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
GRAVITYOne 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.
( )2planet
planetgrav
R
mmGF
⋅=
We already know that Fgrav is the weight of the
Rplanet
m
We already know that Fgrav is the weight of the object, and that should just be mg (if the planet is the Earth)
( )2planet
planet
R
mmGmg
⋅=
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
GRAVITYOne 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.
( )2planet
planetgrav
R
mmGF
⋅=
We already know that Fgrav is the weight of the
Rplanet
m
We already know that Fgrav is the weight of the object, and that should just be mg (if the planet is the Earth)
( )2planet
planet
R
mmGmg
⋅=
This part is g
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
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.
221
gravr
mmGF
⋅=
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
221
gravr
mmGF
⋅=
Notice that the gravity force points toward the center: i.e. it is the centripetal force.centripetal force.
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
221
gravr
mmGF
⋅=
Notice that the gravity force points toward the center: i.e. it is the centripetal force.centripetal force.
We have a formula for centripetal force:
rvm
F2
cent⋅
=
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
rvm
r
mMG
2
2⋅
=⋅
Set these equal to get our formula:
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.
rv
r
MG
2
2=
Set these equal to get our formula:
The mass of the orbiting object will cancel out:
rrG
2=
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.
rv
r
MG
2
2=
Set these equal to get our formula:
The mass of the orbiting object will cancel out:
rrG
2=
Each orbit is one trip around the circle, so the distance traveled is 2∏r.
Speed is distance/time, so we get:
Tr2
vπ
=
12-3 Kepler’s Laws of Orbital Motion
Start with Newton’s Law of Gravity applied to a circular orbit.
r
)(MG
2Tr2
2
π
=
Set these equal to get our formula:
Plug in our expression for v:
rrG
2=
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.
r
)(MG
2Tr2
2
π
=
Set these equal to get our formula:
Plug in our expression for v:
rrG
2=
Now we can rearrange this to solve for T. Do this now.
23
rGM2
T ⋅π
=
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.
Prepared by Vince Zaccone
For Campus Learning 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.
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.
Prepared by Vince Zaccone
For Campus Learning 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.
Prepared by Vince Zaccone
For Campus Learning 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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
23
rGM2
T ⋅π
=What do we use for M in this formula?
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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
23
rGM2
T ⋅π
=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
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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
23
rGM2
T ⋅π
=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:
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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
23
rGM2
T ⋅π
=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:
32
T4
GMr 3
2⋅
π=
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:
Prepared by Vince Zaccone
For Campus Learning Assistance Services at UCSB
23
rGM2
T ⋅π
=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:
m1036.2)s1010.1(4
)kg1045.6)(1067.6(T
4
GMr 753
2
23
kgNm11
32
322
2
32
⋅=⋅⋅π
⋅⋅=⋅
π=
−