• Main points of today’s lecture:

Physics 231 Lecture 22

• Newton’s law of universal gravitation:

2rGMmF =

• Kepler’s laws and the relation between the orbital period and orbital radius.

r

32

2 4

π

• Gravitation potential energyGM M

32 4 rGM

T

= π

• Tensile stress and strain

1 2grav

12

GM MPEr

= −

• Bulk stress and strain:0LLY

AF Δ=Δ

VVBP

AF Δ=Δ=Δ

A uniform10 kg beam is support at one end by a hinge and at the other by a cable. The cable from the wall is attached to the beam at an angle of 30°. What is the tension in the cable?What is the tension in the cable?a) 9.8 N b) 19.6 N c) 29.4 N d) 39.2 Nd) 49 N e) 58.8 N f) 68.6 N e) 98 N

T

1. Draw the forces.2. Choose the axis for calculating the torques so as to avoid calculating torques

30otorques so as to avoid calculating torques you don’ t want to know.

T

W

hingeF

LDon’t want to know Fhinge, so put axis at hinge.

3. Set the net torque about this axis to zero and solve for the tension. barW hinge p g

0τ = ( ) bTLsin 30 m g L / 2= ° − ii

0τ =( ) barTLsin 30 m g L / 2° = ( ) barT sin 30 m g / 2° =

1T / 2

( ) barTLsin 30 m g L / 2

bar1T m g / 22

= barT m g= T 98N=

hinge ii

To get F , use F 0.=

Newton’s law of universal gravitation

• All objects (even light photons) feel a gravitational force attracting• All objects (even light photons) feel a gravitational force attracting them to other objects. This force is proportional to the two masses and inversely proportional to the square of the distance between them.

221

12mmGF = G = 6 673 x 10G = 6 673 x 10--1111 N m² /kg²N m² /kg²212r G = 6.673 x 10G = 6.673 x 10 N m /kgN m /kg

A hi i j t th Th f th th d

Example

• A spaceship is on a journey to the moon. The masses of the earth and moon are, respectively, 5.98x1024 kg and 7.36x1022 kg. The distance between the centers of the earth and the moon is 3.85x106m. At what point as measured from the center of the earth does the gravitationalpoint, as measured from the center of the earth, does the gravitational force exerted on the craft by the earth balance the gravitational force exerted by the moon? This point lies on a line between the centers of the earth and the moon. moon shipGM m GM mthe earth and the moon. moon ship

moon 2moon _ ship

Fr

=

moon ship earth ship2 2

GM m GM m=

Moonearth ship

earth 2earth _ ship

GM mF

r= =

2 2moon _ ship earth _ shipr r

moon earth2 2moon ship earth ship

M Mr r

=

moon _ shipr2earth _ ship earth2moon ship moon

r Mr M

=moon _ ship earth _ ship

earth _ ship earth

moon _ ship moon

r Mr M

=

earth shipr

moon _ ship moon

24

22

5.98x10 kg 9.017.36x10 kg

= =

earth _ ship moon _ ship moon _ ship moon _ shipr r 9.01r r+ = +5

moon _shipr 3.85x10 m=

earth _ ship

Earth

6moon _ship10.01r 3.85x10 m= =

6earth _ shipr 3.47x10 m =

Example

G i h 100 N th E th Wh t ld hi i ht b th• George weighs 100 N on the Earth. What would his weight be on the surface of another planet that has a planetary mass, which is 3 times the mass of Earth and mass and a planetary radius, which is 2 times the radius of Earth? GM mradius of Earth?– a) 75 N– b) 100 N planet GeorgeGM m

W =

Earth GeorgeEarth 2

Earth

GM mW

R=

( )earth GeorgeG 3M m=– c) 125 N

– d) 150 N

planet 2planet

WR

=( )2

earth2R=

earth George3GM m= earth GeorgeGM m3=

planet earth3W W4

=

Mp 3ME

Rp 2Re

W 100 N3100N 75N4

= =

2earth4R 2

earth4 R=

planet earth4WE 100 NWp ?

4

Hint: This is a ratio problem.

Conceptual question

A t t i ft fl ti “ i htl l ” i bit• An astronaut is on a spacecraft floating “weightlessly” in orbit. While her feet are attached securely to the spacecrafte, she shakes a large iron anvil rapidly back and forth. She reports back to Earth thatback to Earth that– a) the shaking costs her no effort because the anvil has no

inertial mass in space.b) th h ki t h ff t b t id bl l– b) the shaking costs her some effort but considerably less than on Earth.

– c) although weightless, the inertial mass of the anvil is the E thsame as on Earth.

3

Kepler’s laws• Johannes Kepler proposed three laws of planetary

motion:motion:1. All planets moved in elliptical orbits with the

Sun at one of the focal points.2 l bi l i l

2a

2. Planetary orbits sweep out equal areas in equal times. (This is a consequence of angular momentum conservation.) r

Δθ111 22 θΔ Δθ

.constr

tr21t

tr

21rr

21area

2

22

=ω

Δω≈ΔΔ

θΔ=θΔ⋅=

3 Th f th bit l i d f l t i ti l t th b

2star planet planet

grav planet c2

GM m m vF m a= = = 2

2

2

3 ω==vGMstar

/ /

3. The square of the orbital period of any planet is proportional to the cube of the average distance from the planet to the Sun. For a circular orbit:

grav planet c2r r 23 rr

TT πωπω 22 ==

2

32

=

TrGM star π

32

2 4 rGM

Tstar

=

πG star

3

star

22 a

GM4T

π=44. For an elliptical orbit:

Conceptual question

T t llit A d B f th i d E th• Two satellites A and B of the same mass are going around Earth in concentric circular orbits.The distance of satellite B from Earth’s center is twice that of satellite A.What is the ratio of the centripetal force acting on B to that acting on A?centripetal force acting on B to that acting on A?– a) 1/8– b) 1/4

E sc,A 2

A

GM mF

r= E s

c.B 2B

GM mFr

=

– c) 1/2– d)– e) 1

21

A B

( )E s

c B 2

GM mF = E s2

GM m4

=A

1 F=)RB 2RA

Fc,B/Fc,A ?

( )c.B 2A2r 2

A4r c,AF4

c,BF 1=c,AF 4

Hint: This is a ratio problem.5

Example

O t llit i i bit b t J it f di d i d f 100• One satellite is in an orbit about Jupiter of radius r1 and a period of 100 days. If a second satellite is placed in an orbit with 4 times the radius (i.e. r2=4r1), what is the period for the orbit of the second satellite?

) 200 d – a) 200 d– b) 400 d– c) 800 d

22 3

1 1Jupiter

4T rGM

π =

– d) 1600 d 22 32 2

Jupiter

4T rGM

π =

( )2

31

Jupiter

4 4rGM

π =

r2 4r1

T1 100 dT ?

23

1J it

464 rGM

π =

2

164T=T2 ?

2 2T 64THi t thi l

JupiterGM

T 8 T 8 100 d 800 d2 22 1T 64T =Hint: this can also

be solved as a ratio problem6

2 1T 8 T 8 100 days 800 days = = ⋅ =

Conceptual quiz

S E th h d t h d b ll fi d f• Suppose Earth had no atmosphere and a ball were fired from the top of Mt. Everest in a direction tangent to the ground. If the initial speed were high enough to cause the ball to travel in a circular trajectory around Earth the ball’s acceleration wouldcircular trajectory around Earth, the ball s acceleration would– a) be much less than g (because the ball doesn’t fall to the

ground).b) b i t l Note: someone in a space– b) be approximately g.

– c) depend on the ball’s speed.Note: someone in a spaceship in the same orbit wouldfeel “weightless” and manywould call this a “zero g” environment.“weightlessness” is aweightlessness is a sensation that occurs when one feels no effects of gravity.It h h 0 hIt happens when g=0 or whensomeone is falling freely.

7

Example

Th E th bit th i i l bit f di 1 5 1011• The Earth orbits the sun in an circular orbit of radius rE=1.5x1011 m. What is the mass of the sun?

22 34T π 2

2 34M T π 2 3

sun

T rGM

= 2 3

2

4 rM π =

2 3sunM T r

G =

sun 2MG T

( ) ( )32 1130

211

1.50x 10 m4 3.142.0x10 kg

⋅= =2-11 2.0x10 kg

6.673 x 10 N m² /kg² 365d 24h 3600s1yy d d

8

Example• A uniform steel beam of length 5.00 m has a weight of 4.5x103 N. One end of the beam is

bolted to a vertical wall. The beam is held in a horizontal position by a cable attached p ybetween the other end of the beam and a point on the wall. The cable makes an angle of 25o

above the horizontal. A load whose weight is 12.0x103 N is hung from the beam at a point that is 3.5 m from the wall. Find (a) the magnitude of the tension in the supporting cable and (b) the magnitude of the force exerted on the end of the beam by the bolt that attaches to the(b) the magnitude of the force exerted on the end of the beam by the bolt that attaches to the wall.

( ) ( ) ( )oi load beam

i

Choose pivot to be at hinge:

0 T(5m)sin 25 W 3.5m W 2.5mτ = = − −( ) ( )i

T

( )( )

( )( )load beamo o

3.5m 2.5mT W W

(5m)sin 25 (5m)sin 25= ⋅ + ⋅

4T 2.54x10 N =25o

load

TR

beamW

( )oi x

iHinge forces: F 0; x-comp: R -Tcos 25 0= =

( )o 4xR Tcos 25 2.29x10 N= =

oy comp: R W W Tsin(25 ) 0+load

loadW y load beamy-comp: R W W Tsin(25 ) 0− − + =

o 3 3 3y load beamR W W Tsin(25 ) 12.0x10 N 4.5x10 N 10.7x10 N= + − = + −

3 2 2 3y x yR 5.8x10 N R R R 23.6x10 N= = + =y x y

30

R R4

5.8x10 Ntan( ) .25 14 above horizontal2.3x10 N

θ θ= = =