Understanding Gravity and Orbital Motion: Universal Gravitation and Kepler's Laws, Slides of Physics

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8. Gravity

Toward a Law of Gravity

Universal Gravitation

Orbital Motion

Gravitational Energy

The Gravitational Field

This TV dish points at a satellite in a fixed position in the sky.How does the satellite manage to stay at that position?

period = 24 h

8.1. Toward a Law of Gravity

1543: Copernicus – Helio-centric theory.1593: Tycho Brahe – Planetary obs.1592-1610: Galileo – Jupiter’s moons,

sunspots, phases of Venus.

1609-19: Kepler’s Laws1687: Newton – Universal gravitation.

Phases of Venus:Size would be constantin a geocentric system.

Kepler’s Laws

d A

const

dt

2

3

T^

a 

Explains retrograde motion

Example 8.1. Acceleration of Gravity

Use the law of gravitation to find the acceleration of gravity(a) at Earth’s surface.(b) at the 380-km altitude of the International Space Station.(c) on the surface of Mars.

E 2 m m

F^

m g

G

r

^

^



E 2 m

g^

G

r 

(a)

^

^

24

11

2

2

2 6

kg

g^

N m

kg

m

^

^

2

m

s

(b)

^

^

^

^

24

11

2

2

2

6

3

kg^10

g^

N m

kg

m

m

^

^

^

^

2

m

s

(c)

^

^

24

11

2

2

2 6

kg

g^

N m

kg

m

^

^

2

m

s

see App.E

TACTICS 8.1. Understanding “Inverse Square”

Given Moon’s orbital period T & distance R from Earth,Newton calculated its orbital speed v and hence acceleration a = v

2 / R.

He found a ~ g / 3600. 

Moon-Earth distance is about 60 times Earth’s radius.

8.3. Orbital Motion

Orbital motion:

Motion of object due to gravity from another larger body.

E.g. Sun orbits the center of our galaxy with a period of ~200 million yrs.Newton’s “thought experiment”

2

2 M

m

v

G

m

r^

r  G M

v^

r



Condition for circular orbitSpeed for circular orbit

r

T^

 v

Orbital period

3

r G M

 Kepler’s 3

rd^ law

g = 0

orbit

projectiles

Example 8.2. The Space Station

The ISS is in a circular orbit at an altitude of 380 km.What are its orbital speed & period?

G M

v^

r



3

r

T^

G M

Orbital speed:

^

^ 

11

2

2

24

6

3

/^

N m

kg

kg

m

m

 ^

^

^

^

^

km

s

Orbital period:

^

^

3

6

3

11

2

2

24

/^

m

m

N m

kg

kg

^

^

^

^

3

s

^

^

90 min 

Near-Earth orbit T ~ 90 min.Moon orbit T ~ 27 d.Geosynchronous orbit T = 24 h.

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Elliptical Orbits

Orbits of most known comets, are highly elliptical.

Perihelion: closest point to sun.Aphelion: furthest point from sun.

Projectile trajectory is parabolic onlyif curvature of Earth is neglected.

ellipse

Open Orbits

Closed(circle)

Closed(ellipse)

Open(hyperbola)

Borderline(parabola)

Example 8.4. Steps to the Moon

Materials to construct an 11,000-kg lunar observatory are boosted from Earth to geosyn orbit.There they are assembled & launched to the Moon, 385,000 km from Earth.Compare the work done against Earth’s gravity on the 2 legs of the trip.

12

1

2 1

E

W

U

G M

m

r^

r

^

^

^

^

st 1 leg:

^

^





11

2

2

24

6

7

1

1

10

/^

10

11, 000

10

10

W

N m

kg

kg

kg

m^

m

^

^



^

^

^



^



^



^



^







11

2

2

24

7

8

1

1

10

/^

10

11, 000

10

10

W

N m

kg

kg

kg

m^

m

^

^



^

^

^



^



^



^



11

10

J

^



nd 2 leg:

10

10

J

^



Zero of Potential Energy^  

G M m

U

r

r

 

12

1

2 1

U

G M m

r^

r

^

^

^

^

^

^

^

U



Gravitational potential energy

E > 0, open orbit

Open Closed

E < 0, closed orbit

Bounded motion

Turning point

Escape Velocity

Body with

E

 0 can escape to



2

1

E G M m

m v

R

^



esc

E G M

v^

R

^

11

2

2

24

6

/^

esc

N m

kg

kg

v^

m

 ^

^

^

km

s

Escape velocity

km

h

Moon trips have

v <

v^ esc

.

Open Closed

Energy in Circular Orbits

Circular orbits:

2

G M

v^

a r

r

^

^

2

1 2 K

m v 



G M m

r



G M m

U

r

G M m

E^

K^

U

r

^

^



K

1 U 2

^

Higher

K

or v



Lower

E

& orbit (r).

0 E U

K K