Gravity - UCSBclas.sa.ucsb.edu/staff/vince/Physics 6B/12.1 Physics 6B Gravity.pdfGRAVITY One more...

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!



GRAVITY


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.

GRAVITY


Newton’s Law of Universal Gravitation gives us a formula to calculate the attractive force between 2 objects:

221

gravr

mmGF

⋅=





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



Example:




1012 m 3 x 1012 m

FDP on HFApes on H




Example:




1012 m 3 x 1012 m

FDP on HFApes on H


Our formula will find the forces (we supply the direction from looking at the diagram): 2

21grav

r

mmGF

⋅=



Example:




1012 m 3 x 1012 m

FDP on HFApes on H



21grav

r

mmGF

⋅=

( )( )( )212

2024

kgNm11

HonApesm10

kg106kg1061067.6F 2

2 ⋅⋅

⋅−= − This is negative because

the force points to the left



Example:




1012 m 3 x 1012 m

FDP on HFApes on H



21grav

r

mmGF

⋅=

( )( )( )

N104.2m10

kg106kg1061067.6F 11

212

2024

kgNm11

HonApes 2

2⋅−=

⋅⋅





Example:




1012 m 3 x 1012 m

FDP on HFApes on H



21grav

r

mmGF

⋅=

( )( )( )

N104.2m10

kg106kg1061067.6F 11

212

2024

kgNm11

HonApes 2

2⋅−=

⋅⋅



( )( )( )

N103.1m103

kg106kg1031067.6F 11

212

2025

kgNm11

HonDP 2

2⋅+=

⋅

⋅⋅

⋅+= − This is positive because the

force points to the right



Example:




1012 m 3 x 1012 m

FDP on HFApes on H



21grav

r

mmGF

⋅=

( )( )( )

N104.2m10

kg106kg1061067.6F 11

212

2024

kgNm11

HonApes 2

2⋅−=

⋅⋅



( )( )( )

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


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






( )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

⋅=






( )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



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.


2. As a planet moves in its orbit, it sweeps out an equal amount of area in an equal amount of time.


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.


What should you use for the constant?


What should you use for the constant?

Let’s derive the formula to find out.


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

221

gravr

mmGF

⋅=



221

gravr

mmGF

⋅=

Notice that the gravity force points toward the center: i.e. it is the centripetal force.centripetal force.



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⋅

=



rvm

r

mMG

2

2⋅

=⋅

Set these equal to get our formula:

M is the mass of the central object.



rv

r

MG

2

2=


The mass of the orbiting object will cancel out:

rrG

2=

How do we get the period (T) involved in our formula?



rv

r

MG

2

2=


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

=



r

)(MG

2Tr2

2

π

=


Plug in our expression for v:

rrG

2=

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



r

)(MG

2Tr2

2

π

=


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.



Example:





Answer: Deimos is farther from Mars. Inspection of the formula in Kepler’s 3rd Law should make this clear.



Example:





We will use the formula in Kepler’s 3rd Law for this one, but we will need to figure out the constant.



Example:






Previously we derived an expression for the constant. Our formula is:



23

rGM2

T ⋅π

=What do we use for M in this formula?

Example:









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:









23

rGM2

T ⋅π


Rearrange this formula to solve for radius, then plug in the numbers:

Example:









23

rGM2

T ⋅π



32

T4

GMr 3

2⋅

π=

Example:









23

rGM2

T ⋅π



m1036.2)s1010.1(4

)kg1045.6)(1067.6(T

4

GMr 753

2

23

kgNm11

32

322

2

32

⋅=⋅⋅π

⋅⋅=⋅

π=

−

Gravity - UCSBclas.sa.ucsb.edu/staff/vince/Physics 6B/12.1 Physics 6B Gravity.pdfGRAVITY One more...

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