18/1/2004

Test Breakdown

A >= 93% (63)

A- >= 90% (60)

B+ >= 87% (57)

B >= 83% (54)

B- >= 80% (52)

C+ >= 77% (51)

C >= 70% (45)

D >= 60% (38)

F < 60% (37 or less)

Average: 63.2% + 10%

Std Dev: 21.5%

Scores will be adjusted up by 10%

Chapter 14

Gravitation

38/1/2004

Gravity

One of the fundamental forces of Nature

Not just the reason things fall….

Why the Earth is roundWhy the moon goes around the earthWhy the earth goes around the sunWhy there are ocean tides

48/1/2004

Gravitational Force

Fg is not a constant unless you have a small object near the surface of a big sphere

Fg is an attractive force between any two masses

221

r

mmGFg

G = 6.672 x 10-11 N m2/kg2

r

m1 m2

Fg Fg

58/1/2004

Multiple Objects

Obeys principle of superposition:

...23

22

21

r

mMG

r

mMG

r

mMGFg

r1 r3

M1 M3

M2

m

r2

68/1/2004

Extended Objects

1 2Fg Fg

Each bit of #1 attracts each bit of #2

Need to integrate over whole object to get Fg

FdFg

78/1/2004

Extended Objects: Special Case

Can treat uniform spherical shells (and thus spheres) like point masses located at geometric center

No gravitational force inside uniform spherical shell (it integrates to zero)

Fg = 0

88/1/2004

Gravitational Force Examples

Force between earth and sun

Force between two people (assumed spherical)

221

r

mmGF

N.

m.

kg.kg.

kg

mN.F

22

211

2430

2

211

10523

)10491(

)10985)(10991(106726

N.m

kgkg

kg

mN.F 7

22

211 1042

)1(

)60)(60(106726

98/1/2004

Example: (Problem 14.16)

a)What will an object weigh on the Moon’s surface if it weighs 100 N on Earth’s surface?

b)How many Earth radii must this same object be from the center of Earth if it is to weigh the same as it does on the Moon?

108/1/2004

Example:

Find the mass of the object:

kgsm

N

g

Fm

Earth

Earthg 2.10/8.9

1002

,

On the Moon:

26

2211

2

)1074.1(

)2.10)(1035.7)(10672.6( 2

2

m

kgkg

r

mGMF

kgNm

Moon

Moong

NFg 5.16

118/1/2004

Example:

Distance from the earth with the same weight:

EarthEarth

kgNm

g

Earth

Rm

Rmr

N

kgkgr

F

mGMr

5.21038.6

11057.1

5.16

)2.10)(1097.5)(10672.6(

67

24112

2

2r

mGMF Earthg

128/1/2004

Objects Near Earth’s Surface

Re

As long as the distance above earth’s surface is small compared to RE, the force is approximately constant

g

E

Eg ma

R

MmGF 2 2

E

Eg

R

MGa

138/1/2004

Variation of Gravitational Force on Earth’s Surface

1)Earth is not uniformly dense

Variations in crust from region to region

2)Earth is not a sphere

Bulge at the equator

3)Apparent change from earth’s rotation

In this case, gag

148/1/2004

Variation of g from Rotation

At the earth’s pole, there is no centripetal acceleration:

N

Fg

ac=0

0

g

cnet

FN

maFmg

R

MGmN

e

e

2

At the pole, g=ag

158/1/2004

Variation of g from Rotation

At the equator:

RmFN

maF

g

cnet

2

mgRR

MGmN

e

e

2

2

NFg

ac

At the equator, g < ag!

g = 9.801 m/s2 in Pittsburgh

g = 9.786 m/s2 in Jamaica!!

Effectively lower gravity

168/1/2004

Gravitation Inside a Sphere

Recall: No gravitational force exerted on an object inside a spherical shell

As you travel further into a sphere, the layers above can be thought of as many spherical shells, which exert no gravitational force!

Only the mass of the sphere below you matters for calculating Fg!

178/1/2004

“Inner” Gravity

Movie claims that earth’s core is a trillion trillion tons…

1012 1012 tons = 1027 kgMass of entire Earth: 61024 kg !!!

Real inner core: m = 1.7% MEarth

ginner = 0.017g

They walk as if under 1 g!!!

Only the mass of the inward sphere contributes to Fg

188/1/2004

Gravitational Potential Energy

x

mmGdx

x

mmGUUW

x21

221 )(-(x)

Gravity is a conservative force – what is the associated potential energy?

ΔU = -W and

So for point masses or spheres m1 and m2

Taking U = 0 at x =

2

1

)(x

xdxxFW

x

mmGUW 21(x)

198/1/2004

Gravitational Potential Energy

Note: U = at r = 0

For point masses or spheres m1 and m2

U

r F = - dU/dr Always attractive

r

mmGU 21 (r)

208/1/2004

Gravitational Potential Energy for Astronaut between Earth and Moon

rEM

x

x

mMGxU AE

Earth )(

xr

mMGxU

EM

AMMoon

)(

xr

M

x

MGmxUxUxU

EM

MEAMoonEarth )( )()(

Ftot can be zero at some x… What about Utot?

218/1/2004

Example:

rEM

x

Find where Fnet on the astronaut equals zero.

22 )( xr

mGM

x

mGM

FF

EM

AMAE

MoonEarth

2

x

mMGF AE

Earth

2)(

xr

mMGF

EM

AMMoon

)2( 222 xxrrMxM EMEMEM

02)( 22 EMEEMEME rMxrMxMM

EMME

E rMM

Mx

Which solution is real?

228/1/2004

Path Independence

The amount of work done against a gravitational potential does not depend on the path taken (conservative force)

x

y

A

B1

23

238/1/2004

Escape Velocity

What speed does an object need to escape the Earth’s gravity?

02

1 2 UR

mMGmvUKE

e

eesciii

mphsmvesc 000,25/1012.1 4

It needs just enough KE to get to r and stop

e

eesc R

GMv

2

Escape velocity from Earth is:

248/1/2004

Example: (Problem 14.29)

The mean diameters of Mars and Earth are 6.9x103 km and 1.3x104 km, respectively. The mass of Mars is 0.11 times Earth’s mass.

a)What is the ratio of the mean density of Mars to that of Earth?

b)What is the value of the gravitational acceleration on Mars?

c)What is the escape speed on Mars?

258/1/2004

Kepler’s 1st Law

All planets move in elliptical orbits, with the Sun at one focus

a

F F’

ea

r

Eccentricities, e, of planets are small (close to circular)

268/1/2004

Kepler’s 2nd Law

The rate at which a planet sweeps out an area A is constant. (Constant areal velocity)

F F’

A

A

278/1/2004

Kepler’s 3rd Law

The square of the period of any planet is proportional to the cube of the semi-major axis of its orbit

32

2 4r

GMT

288/1/2004

Example: (Problem 14.42)

Determine the mass of Earth from the period T (27.3 days) and the radius r (3.82x105 km) of the Moon’s orbit about Earth. Assume that the Moon orbits the center of Earth rather than the center of mass of the Earth-Moon system.

298/1/2004

Example:

32

2 4r

GMT

Earth

Kepler’s 3rd law:

2611

382

2

32

)1036.2)(10672.6(

)1082.3(4

4

2

2

s

mM

TG

rM

kgNmEarth

Earth

s1036.2days3.27 6

kgM Earth24109.5

308/1/2004

Circular Orbits

Simplest case:

rFg

v

r

mvma

r

mMGF c

etot

2

2

esce v

r

GMv

318/1/2004

Geosynchronous Orbit

r

A satellite can stay over one location on earth.

2/322r

GMv

rT

e

3/2

2

T

GMr e

Period = 1 day

milesmr 000,261022.4 7

r

GMv e

328/1/2004

Why is there free fall on the orbiting space shuttle?

Both shuttle and occupants accelerating toward center of earth with same acceleration

R=Rorbit < 2Re so gravitational force is not negligible

a = GME/R2

a = GME/R2

338/1/2004

Energy in a Circular Orbit

r

MmGU

M

m

r

GMm

r

GMmmvK

22

1

2

12

2

r

GMm

r

GMm

r

GMmUKE

22

For an elliptical orbit, substitute a for r

a

GMmE

2

348/1/2004

Ocean Tides

Caused by difference in gravitational force across the extent of the earth.

r

Water closest to Moon pulled “upward”

Water farthest from Moon pulled less and thus bulges outward

Smaller effect from the sun even though it has stronger gravitational pull. Why?

358/1/2004

Tides

The effects from the sun and moon can work together to

form a spring tide…

or against each other to form a neap tide