gravity physics 6b prepared by vince zaccone for campus learning assistance services at ucsb

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

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

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.

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

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:

mass=6 x 1020 kg

Planet of the Apes:

mass=6 x 1024 kg

Daily Planet:

mass=3 x 1025 kg

1012 m 3 x 1012 m

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:

mass=6 x 1020 kg

Planet of the Apes:

mass=6 x 1024 kg

Daily Planet:

mass=3 x 1025 kg

1012 m 3 x 1012 m

FDP on HFApes on H

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:

mass=6 x 1020 kg

Planet of the Apes:

mass=6 x 1024 kg

Daily Planet:

mass=3 x 1025 kg

1012 m 3 x 1012 m

Our formula will find the forces (we supply the direction from looking at the diagram):

FDP on HFApes on H

221

gravr

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:

mass=6 x 1020 kg

Planet of the Apes:

mass=6 x 1024 kg

Daily Planet:

mass=3 x 1025 kg

1012 m 3 x 1012 m

Our formula will find the forces (we supply the direction from looking at the diagram):

FDP on HFApes on H

221

gravr

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:

mass=6 x 1020 kg

Planet of the Apes:

mass=6 x 1024 kg

Daily Planet:

mass=3 x 1025 kg

1012 m 3 x 1012 m

Our formula will find the forces (we supply the direction from looking at the diagram):

FDP on HFApes on H

221

gravr

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:

mass=6 x 1020 kg

Planet of the Apes:

mass=6 x 1024 kg

Daily Planet:

mass=3 x 1025 kg

1012 m 3 x 1012 m

Our formula will find the forces (we supply the direction from looking at the diagram):

FDP on HFApes on H

221

gravr

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:

mass=6 x 1020 kg

Planet of the Apes:

mass=6 x 1024 kg

Daily Planet:

mass=3 x 1025 kg

1012 m 3 x 1012 m

Our formula will find the forces (we supply the direction from looking at the diagram):

FDP on HFApes on H

221

gravr

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

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.

2planet

planetgrav

R

mmGF

Rplanet

m

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.

2planet

planetgrav

R

mmGF

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

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.

2planet

planetgrav

R

mmGF

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.

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.

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:

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:

Each orbit is one trip around the circle, so the distance traveled is 2∏r.

Speed is distance/time, so we get:

Tr2

v

12-3 Kepler’s Laws of Orbital Motion

Start with Newton’s Law of Gravity applied to a circular orbit.

r

)(

r

MG

2T

r2

2

Set these equal to get our formula:Plug in our expression for v:

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

)(

r

MG

2T

r2

2

Set these equal to get our formula:Plug in our expression for v:

Now we can rearrange this to solve for T. Do this now.

23

rGM

2T

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.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

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 3 rd Law should make this clear.

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

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.

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

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:

23

rGM

2T

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.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

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:

23

rGM

2T

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.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

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:

23

rGM

2T

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.

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB

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:

23

rGM

2T

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

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

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:

23

rGM

2T

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

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.