10/10/2013PHY 113 C Fall 2013-- Lecture 131 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin 101 Plan for Lecture 13: Chapter 13 – Fundamental gravitational

PHY 113 C Fall 2013-- Lecture 13 110/10/2013

PHY 113 C General Physics I11 AM - 12:15 PM MWF Olin 101

Plan for Lecture 13:

Chapter 13 – Fundamental gravitational force law and planetary motion

1. Gravitational force law; relationship with g near Earth’s surface

2. Orbital motion of planets, satellites, etc.

3. Energy associated with gravitation and planetary motion



Question from Webassign Assignment #11


X

t1

t3

t2

Frτ


X

t1

t3

t2

Frτ



X

vmvmL

m

21 22

:caseour In

vrL



2

2

2

12

1

fff

iii

fffiii

IK

IK

ILIL

iclicker question: A. Ki=Kf ? B. Ki=Kf ?


Universal law of gravitation Newton (with help from Galileo, Kepler, etc.) 1687

212

122112

ˆ

r

mGm rF

2

211

kg

mN 10674.6

G


Newton’s law of gravitation: m2 attracts m1 according to:

x

y

m1

m2

r2

r1

r2r1

212

122112

ˆ

r

mGm rF

G=6.67 x 10-11 N m2/kg2

NNF

rmm

82

11

1221

1017.82

70701067.6

:m 2 kg; 70 :Example

PHY 113 C Fall 2013-- Lecture 13 10

F12

10/10/2013

Vector nature of Gravitational law:

m1 m2

m3

x

y

d

d

jiF1 3221 mm

d

mG

F13


Gravitational force of the Earth

RE m

2226

2411

2

2

m/s8.9m/s)1037.6(

1098.51067.6

E

E

E

E

R

GMg

R

mGMF

Note: Earth’s gravity acts as a point mass located at the Earth’s center.


Question:

Suppose you are flying in an airplane at an altitude of 35000ft~11km above the Earth’s surface. What is the acceleration due to Earth’s gravity?

2226

2411

2

2

m/s796.9m/s)10)011.037.6((

1098.51067.6

)(

)(

hR

GMa

mahR

mGMF

E

E

E

E

a/g = 0.997


Attraction of moon to the Earth:

N1099.1N)1084.3(

1036.71098.51067.6 2028

222411

2

EM

ME

R

MGMF

Acceleration of moon toward the Earth:

F = MM a a = 1.99x2020 N/7.36x1022kg =0.0027 m/s2

REM



Gravity on the surface of the moon

EM

M

MM

M

gg

g

mgmR

GmMF

165.0

m/s 62.1)1074.1(

1035.71067.6

N)1074.1(

1035.71067.6

226

2211

26

2211

2

Gravity on the surface of mars

EMars

Mars

MarsMars

gg

R

GMg

380.0

m/s 73.3)1039.3(

1042.61067.6 2

26

2311

2


iclicker question:In estimating the gravitational acceleration near the surfaces of the Earth, Moon, or Mars, we used the full mass of the planet or moon, ignoring the shape of its distribution. This is a reasonable approximation because:

A. The special form of the gravitational force law makes this mathematically correct.

B. Most of the mass of the planets/moon is actually concentrated near the center of the planet/moon.

C. It is a very crude approximation.


days 27.4 s 32367353.951098.51067.6

)1084.3(π2

π2

π2ω

:Earth todueMoon for law sNewton'

2411

38

3

2

2

E

EM

EMEM

EM

E

EM

MMM

GM

RT

RT

Rv

R

GM

R

va

M aF

Stable circular orbit of two gravitationally attracted objects (such as the moon and the Earth)

REM

F

a

v


iclicker question:

In the previous discussion, we saw how the moon orbits the Earth in a stable circular orbit because of the radial gravitational attraction of the moon and Newton’s second law: F=ma, where a is the centripetal acceleration of the moon in its circular orbit. Is this the same mechanism which stabilizes airplane travel? Assume that a typical cruising altitude of an airplane is 11 km above the Earth’s surface and that the Earth’s radius is 6370 km.

(a) Yes (b) No


!mi/h! 18000km/s 7.9v

hours 1.4 s 07051098.51067.6

)1038.6(π2

π2

π2ω

2411

36

3

2

2

E

Ea

EaEa

Ea

E

Ea

GMR

T

RT

Rv

RGM

Rv

a

Stable (??) circular orbit of two gravitationally attracted objects (such as the airplane and the Earth)

REa

F

a

v


2226

2411

2

2

m/s8.9m/s)1037.6(

1098.51067.6

E

E

E

E

R

GMg

R

mGMF

212

122112

ˆ

r

mGm rF Newton’s law of gravitation:

Earth’s gravity:

Stable circular orbits of gravitational attracted objects:

RE m

REMF

a

vMM

Summary:


More details

If we examine the circular orbit more carefully, we find that the correct analysis is that the stable circular orbit of two gravitationally attracted masses is about their center of mass.

m1

R2

R1

m2

2211

:mass ofCenter

RmRm


m1

R2

R1

m2

Radial forces on m1:

2

2211

1

1

11

1

21

111221

211

π2

π2

GmRRR

T

TR

v

Rv

mamRR

mGmF rr

T2 ?


iclicker question:

What is the relationship between the periods T1 and T2 of the two gravitationally attracted objects rotating about their center of mass? (Assume that m1 < m2.)

(A) T1=T2 (B) T1<T2 (C) T1>T2

m1

R2

R1

m2


21

321

21

21

22222

21

212111

1

21

1

2211

2

mmG

RRTT

RmRR

mGmRm

R

vm

RmRm

10/10/2013

m1

R2

R1

m2

1

2212

2

2211

21

321

21

22

2

: thatNote

Gm

RRR

Gm

RRR

mmG

RRTT


iclicker questions:How is it possible that all of these relations are equal?

A. Magic.B. Trick.C. Algebra.

2211

2

2211

2

2211

21

321

since 11 because

21

122

2

1

1

2

2

1

1

2

RmRm

Gm

RRR

Gm

RRR

mmG

RRT

mm

RR

mm

RR

1

2212

2

2211

21

321 222

Gm

RRR

Gm

RRR

mmG

RRT


21

321

21

21

22222

21

212111

1

21

1

2211

2

mmG

RRTT

RmRR

mGmRm

R

vm

RmRm

10/10/2013

m1

R2

R1

m2

For m2>>m1 R2 << R1:

m2

m1

R12

31

1 2Gm

RT

Larger mass


What is the physical basis for stable circular orbits?

1. Newton’s second law?

2. Conservation of angular momentum? L = (const)

Note: Gravitational forces exert no torque

0

0

ˆ

121212

212

122112

dt

d

r

mGm

L

Frτ

rF


(const)

0

L

Lτ

dtd

m1

R2

R1

m2

v1

v2

L1=m1v1R1

L2=m2v2R2

L = L1 + L2

Question:

How are the magnitudes of L1 and L2 related?

2

2

1

1

R

L

R

L

Note: More generally, stable orbits can be elliptical.


Satellites orbiting earth (approximately circular orbits):

RE ~ 6370 km

Examples:

sRhRhGM

RT EE

E

E 23

23

/15058/1π23

Satellite h (km) T (hours) v (mi/h)

Geosynchronous 35790 ~24 6900

NOAA polar orbitor 800 ~1.7 16700

Hubble 600 ~1.6 16900

Inter. space station* 390 ~1.5 17200


Planets in our solar system – orbiting the sun

Planet Mass (kg) Distance to Sun (m)

Period of orbit (years)

Mercury 3.30x1023 5.79x1010 0.24

Venus 4.87x1024 1.08x1011 0.61

Earth 5.97x1024 1.496x1011 1.00

Mars 6.42x1023 2.28x1011 1.88

Jupter 1.90x1027 7.78x1011 11.85

Saturn 5.68x1026 1.43x1012 29.43


m1

R2

R1

m2

v1

v2

Review: Circular orbital motion about center of mass

CM

21

321

21

2

111

1

2

1

11

1

21

1

2211

2

22

2221

21

1

21

1

2

212

mmG

RRTT

TRm

RT

Rm

R

vm

RmRm

R

vm

RR

mGm

R

vm

2

31

21

321

21

1212

22

then if that Note

Gm

R

mmG

RRTT

RRmm


m1

R1

m2

v1

Review: Circular orbital motion about center of mass

2

31

21

321

21

1212

22

then if that Note

Gm

R

mmG

RRTT

RRmm


Example: Satellite in circular Earth orbit

kgm

kgM

GM

hRT

S

E

E

E

3

24

3

10

1097.5 :Note

2

onous)(geosynchr

11053.8

1083.35 If4

6

daysT

mh


Work of gravity:

ri

rf

f

i

f

i

f

i

r

r

r

r

fi

r

r

fi

r

mGm dr

r

mGmW

r

mGmdW

212

21

221 ˆ

r

FrF

ifif

fi rUrUr

mGm

r

mGmW

2121


Gravitational potential energy

r

mGm dr

r

mGmrU

r

mGmdrU

r

gravity

r

r

gravity

ref

212

21

221

''

)(

ˆ )(

rFrF

hR

mGMhRrU

E

SEEgravity

)(

Example:


hU=mgh

iclicker exercise:We previously have said that the gravitational potential of an object of mass m at a height h isU=mgh. How is this related to

A. No relationB. They are approximately equalC. They are exactly equal

? hR

mGMU

E

SEgravity


h

hR

mGM

R

mGM

hR

mGMRUhRU

R

h

RRh

RRhRhR

E

E

E

E

E

EEgravityEgravity

EEE

EEEE

2

2

)()(

1/1

1

/1

111

gNear the surface of the Earth U=mgh is a good approximation to the gravitation potential.


iclicker exercise:How much energy kinetic energy must be provided to an object of mass m=1000kg, initially on the Earth’s surface to outer space?

A. This is rocket science and not a fair question.B. It is not possible to escape the Earth’s

gravitational field.C. We can estimate the energy by simple

conservation of energy concepts.

J

JR

mGMK

KR

mGMU

UKUKE

E

Ei

iE

Ei

ffii

10

6

2411

1025.6

1037.6

10001097.510674.6

0


Total energy of a satellite in a circular Earth orbit

hR

mGMhRrU

E

SEEgravity

)(

hR

mGMvmK

hR

mGM

hR

vm

vmKUKE

E

SES

E

SE

ES

Sgravity

22

1

:orbitcircular aFor

2

1

2

2

2

2


Total energy of a satellite in a circular Earth orbit

hR

mGME

hR

mGMU

hR

mGMK

UKE

E

SE

E

SEgravity

E

SE

gravity

2

2

iclicker question:Compared to the energy needed to escape the Earth’s gravitational field does it take

A. moreB. less

energy to launch a satellite to orbit the Earth?


JJ

JJK

kmh

R

mGM

hR

mGM

R

mGMK

R

mGMU

hR

mGMUKUKE

i

E

E

E

E

E

Ei

E

Ei

E

Effii

1010

6

2411

6

2411

1038.393.62

37.611025.6

1093.62

10001097.510674.6

1037.6

10001097.510674.6

Telescope) Hubble assuch orbit earth (low 560 example,For

2

2


Energy involved with changing orbits

h’

J

E

kmhkmh

hR

mGM

hR

mGMEEE

E

E

E

E

8

6

2411

1065.1

97.6

1

93.6

1

102

10001097.510674.6

600' and 560 example,For

2'2'


Stable Elliptical Orbits

r

MGM

rM

LvME

vv

M

pS

prp

r

p

2

22

22

1

ˆˆ θrv

vrL

Documents

10/10/2013PHY 113 C Fall 2013-- Lecture 131 PHY 113 C General Physics I 11 AM - 12:15 PM MWF Olin 101 Plan for Lecture 13: Chapter 13 – Fundamental gravitational