Physics 201: Lecture 24, Pg 1 Chapter 13 The beautiful rings of Saturn consist of countless centimeter-sized ice crystals, all orbiting the planet under

Physics 201: Lecture 24, Pg 1

Chapter 13

The beautiful rings of Saturn consist of countless centimeter-sized ice

crystals, all orbiting the planet under the influence of gravity.

Goal: To use Newton’s theory of gravity to understand the motion of satellites and planets.


Who discovered the basic laws of planetary orbits?

A. NewtonB. KeplerC. FaradayD. EinsteinE. Copernicus


What is geometric shape of a planetary or satellite orbit?

A. CircleB. HyperbolaC. SphereD. ParabolaE. Ellipse


The value of g at the height of the space shuttle’s orbit is

A. 9.8 m/s2.

B. slightly less than 9.8 m/s2.

C. much less than 9.8 m/s2.

D. exactly zero.


Newton proposed that every object in the universe attracts every other object.The Law: Any pair of objects in the Universe attract each other with a force that is proportional to the products of their masses and inversely proportional to the square of their distance

Newton’s Universal “Law” of Gravity



2by1on12221

1by 2on r̂ Fr

mmGF

“Big” G is the Universal Gravitational Constant

G = 6.673 x 10-11 Nm2 / kg2

Two 1 Kg particles, 1 meter away Fg= 6.67 x 10-11 N(About 39 orders of magnitude weaker than the electromagnetic force.)

The force points along the line connecting the two objects.



2by1on12221

1by 2on r̂ Fr

mmGF

You have three 1 kg masses, 1 at the origin, 2 at (1m, 0) and 3 at (0, 1m). What would be the Net gravitational force on object 1 by objects 2 & 3?

G = 6.673 x 10-11 Nm2 / kg2

Two 1 Kg particles, 1 meter away Fg= 6.67 x 10-11 N

Each single force points along the line connecting two objects.

3by1on13221

1by 3on r̂ Fr

mmGF

Fg=9.4 x 10-11 N along diagonal


Motion Under Gravitational Force

Equation of Motion

r̂)(r

22

2

r

MmG

dttd

m

If M >> m then there are two classic solutions that depend on the initial r and v

Elliptical and hyperbolic paths


Anatomy of an ellipse

a: Semimajor axis b: Semiminor axis c: Semi-focal length a2 =b2+c2

Eccentricity e = c/a

If e = 0, then a circle

Most planetary

orbits are close to

circular.

12

2

2

2

b

y

a

x


A little history Kepler’s laws, as we call them today, state that

1. Planets move in elliptical orbits, with the sun at one focus of the ellipse.

2. A line drawn between the sun and a planet sweeps out equal areas during equal intervals of time.

3. The square of a planet’s orbital period is proportional to the cube of the semimajor-axis length.

Focus Focus


Kepler’s 2nd Law The radius vector from the sun to a planet sweeps out equal

areas in equal time intervals (consequence of angular momentum conservation).

Why?

r̂

)(r

2

2

2

r

MmG

dt

tdm


Conservation of angular momentum Area of a parallelogram = base x height

A

C

||21 rdrdA

sin|| ACCA

dtvrddtvrdA ||2

1

mdtvmr

dA|)(|2

1

||||21 Lpr

dt

dAm

L is constant (no torque)


Kepler’s 3rd Law

Square of the orbital period of any planet is proportional to the cube of semimajor axis a.

3

Sun

22 4

aGM

T

222

)2

(T

rrrv

ac

2Sun

lCentripeta r

mMGmaF c

2Sun2)

2(

rM

GT

r

23 Tr


Suppose an object of mass m is on the surface of a planet of mass M and radius R. The local gravitational force may be written as

Little g and Big G

where we have used a local constant acceleration:

On Earth, near sea level, it can be shown that gsurface ≈ 9.8 m/s2.

surfacemgFG

2surface RGM

g

mMMm R

mGMF ,2by on r̂


Cavendish’s Experiment

F = m1 g = G m1 m1 / r2

g = G m2 / r2

If we know big G, little g and r then will can find the mass of the Earth!!!


Orbiting satellites vt = (gr)½

Net Force: ma = mg = mvt

2 / r gr = vt

2

vt = (gr)½

The only difference is that g is less because you are further from the Earth’s center!


Geostationary orbit


Geostationary orbit

The radius of the Earth is ~6000 km but at 36000 km you are ~42000 km from the center of the earth.

Fgravity is proportional to 1/r2 and so little g is now ~10 m/s2 / 50 vT = (0.20 * 42000000)½ m/s = 3000 m/s

At 3000 m/s, period T = 2 r / vT = 2 42000000 / 3000 sec =

= 90000 sec = 90000 s/ 3600 s/hr = 24 hrs

Orbit affected by the moon and also the Earth’s mass is inhomogeneous (not perfectly geostationary)

Great for communication satellites

(1st pointed out by Arthur C. Clarke)


Tuesday

All of Chapter 13

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Physics 201: Lecture 24, Pg 1 Chapter 13 The beautiful rings of Saturn consist of countless centimeter-sized ice crystals, all orbiting the planet under