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Transcript of Gravity - 6B/12.1 Physics 6B GRAVITY One more useful detail about gravity: The acceleration

• 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:

2 21

grav r

mm GF

⋅ =

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:

2 21

grav r

mm GF

⋅ =

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

21 grav

r

mm GF

⋅ =

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

21 grav

r

mm GF

⋅ =

( )( ) ( )212

2024

kg Nm11

HonApes m10

kg106kg106 1067.6F 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

21 grav

r

mm GF

⋅ =

( )( ) ( )

N104.2 m10

kg106kg106 1067.6F 11

212

2024

kg Nm11

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

21 grav

r

mm GF

⋅ =

( )( ) ( )

N104.2 m10

kg106kg106 1067.6F 11

212

2024

kg Nm11

HonApes 2 2

⋅−= ⋅⋅

 

  

 ⋅−= − This is negative because the force points to the left

( )( ) ( )

N103.1 m103

kg106kg103 1067.6F 11

212

2025

kg Nm11

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

21 grav

r

mm GF

⋅ =

( )( ) ( )

N104.2 m10

kg106kg106 1067.6F 11

212

2024

kg Nm11

HonApes 2 2

⋅−= ⋅⋅

 

  

 ⋅−= − This is negative because the force points to the left

( )( ) ( )

N103.1 m103

kg106kg103 1067.6F 11

212

2025

kg Nm11

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 11

net ⋅−= 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 planet

grav R

mm GF

⋅ =

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 planet

grav R

mm GF

⋅ =

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

mm Gmg

⋅ =

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