Understanding Gravity and Orbital Motion: From Ptolemy to Newton, Slides of Physics

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8. Gravity

CassiniApparent

Ptolemaic (Geo-Centric) System

epicycle

equant deferent

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 constant ina geocentric system.

8.2. Universal Gravitation

Newton’s law of universal gravitation

:

1

2

12

12 2

m m G

r

F^

r

m^1

& m

are 2 point masses. 2

r^12

= position vector from 1 to 2. F^12

= force of 1 on 2. G^

= Constant of universal gravitation= 6.

^

10

^11

N m

2 / kg

Law also applies to spherical masses.

m 1

m 2

r^12

F^12

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

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 8.3. Geosynchronous Orbit

What altitude is required for geosynchronous orbits?

3

r

T^

G M

^

2/

1/

T 2

r^

G M

 ^

^

^

^

2/

1/

11

2

2

24

/^

s^

N m

kg

kg



^

^

^

^

^

^

^

^

^

7

m

^

Altitude =

r

^ R^ E

7

6

m

m

^

^

^

^

6

m

^

^

km

Earth circumference =

6

m

^

^

^

km

Earth not perfect sphere



orbital correction required every few weeks.

Open Orbits

Closed(circle)

Closed(ellipse)

Open(hyperbola)

Borderline(parabola)

8.4. Gravitational Energy

How much energy is required to boost a satellite to geosynchronous orbit?

(^21)

12

d

U

^

r  r^

F^

r

(^21)

12

2

r r

M m

U

G

d r

r

^

^

^

^

1

2 1

G M m

r^

r

^

^

^

^

U

12

depends only on radial positions.

U = 0on this path

… so



U 12

is

the same as ifwe start here.

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

^

To catch the satellite, the shuttle needs to lose energy.It does so by turning to fire its engine opposite its direction of motion.It drops lower,turns again , and fires its engine to achieve a circular orbit, now faster and lower than before.

Higher

K

or v



Lower

E

& orbit (r).

0 E U K K

E U

r > r K K

8.5. The Gravitational Field

Two descriptions of gravity:1.^

body attracts another body (action-at-a-distance)

2.^

Body creates gravitational field.Field acts on another body. Near Earth:

g   g^

j^2

G M

r

  g^

r

Large scale:

2

g^

m

s

^

 / N

kg

Action-at-a-distance



instantaneous messages

Field theory



finite propagation of information

Only field theory agrees with relativity.

near earth in space

Great advantage of the field approach:No need to know how the field is produced.

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