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.
Physics 201: Lecture 24, Pg 1

Who discovered the basic laws of planetary orbits?
A. Newton
B. Kepler
C. Faraday
D. Einstein
E. Copernicus
Physics 201: Lecture 24, Pg 2

What is geometric shape of a planetary or satellite orbit?
A. Circle
B. Hyperbola
C. Sphere
D. Parabola
E. Ellipse
Physics 201: Lecture 24, Pg 3

The value of g at the height of the space shuttle’s orbit is
A. 9.8 m/s 2 .
B. slightly less than 9.8 m/s 2 .
C. much less than 9.8 m/s 2 .
D. exactly zero.
Physics 201: Lecture 24, Pg 4

Newton’s Universal “Law” of Gravity
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
Physics 201: Lecture 24, Pg 5


F on 2 by 1
 
G m
1 m
2 r 2 rˆ
12
 

F on 1 by 2
“Big” G is the Universal Gravitational Constant
G = 6.673 x 10 -11 Nm 2 / kg 2
Two 1 Kg particles, 1 meter away  F g
= 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.
Physics 201: Lecture 24, Pg 6

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 Nm 2 / kg 2
Two 1 Kg particles, 1 meter away  F g
= 6.67 x 10 -11 N
Each single force points along the line connecting two objects.

F on 2 by 1
 
G m
1 m
2
12
 

F on 1 by 2 r 2

F on 3 by 1
 
G m
1 r m
2
2
13
 

F on 1 by 3
F g
=9.4 x 10 -11 N along diagonal
Physics 201: Lecture 24, Pg 7

Motion Under Gravitational Force
 Equation of Motion m d
2
 r ( t ) dt 2
 
G
Mm r 2 rˆ
 If M >> m then there are two classic solutions that depend on the initial r and v
 Elliptical and hyperbolic paths
Physics 201: Lecture 24, Pg 8

 a: Semimajor axis
 b: Semiminor axis
 c: Semi-focal length
 a 2 =b 2 +c 2
 Eccentricity e = c/a
Anatomy of an ellipse x a 2
2
 y b 2
2

1
If e = 0, then a circle
Most planetary orbits are close to circular.
Physics 201: Lecture 24, Pg 9


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
Physics 201: Lecture 24, Pg 10

Kepler’s 2 nd Law


The radius vector from the sun to a planet sweeps out equal areas in equal time intervals (consequence of angular momentum conservation).
Why?
m d 2
 r ( t ) dt 2


G
Mm r 2 rˆ
Physics 201: Lecture 24, Pg 11

Conservation of angular momentum
 dA dA dt
 dA dA

1
2


|
1
2 r


| A

C |

 d r
1
2
|
 d r r


 v

|
|
 v dt dt
AC sin

| r


(
 m v ) | dt m
2
1 m
| r


 p |

|

L |
 L is constant (no torque)


A


C
Physics 201: Lecture 24, Pg 12

rd
 Square of the orbital period of any planet is proportional to the cube of semimajor axis a.
F
Centripeta l
 ma c

G
M
Sun m r
2
T 2
4

2

GM
Sun a c
 v 2 r
 r

2
 r (
2

T
) 2 a 3 r (
2

T
)
2

G
M
Sun r 2 r 3

T 2
Physics 201: Lecture 24, Pg 13

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
F
G
 mg surface where we have used a local constant acceleration:

F on m by M
  mGM
R 2 rˆ
M , m g
 surface
GM
R
2
On Earth, near sea level, it can be shown that g surface
≈ 9.8 m/s 2 .
Physics 201: Lecture 24, Pg 14

Cavendish’s Experiment
F = m
1 g = G m
1 m
1
/ r 2 g = G m
2
/ r 2
If we know big G , little g and r then will can find the mass of the Earth!!!
Physics 201: Lecture 24, Pg 15

Orbiting satellites v t
= (gr)
½
Net Force: ma = mg = mv t
2 gr = v t
2
/ r v t
= (gr)
½
The only difference is that g is less because you are further from the
Earth’s center!
Physics 201: Lecture 24, Pg 16

Physics 201: Lecture 24, Pg 17

 The radius of the Earth is ~6000 km but at 36000 km you are
~42000 km from the center of the earth.


F gravity is proportional to 1/r 2 and so little g is now ~10 m/s 2 / 50
 v
T
= (0.20 * 42000000)
At 3000 m/s, period T = 2
 r / v
T
½
= 2
 m/s = 3000 m/s
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
(1 st pointed out by Arthur C. Clarke)
Physics 201: Lecture 24, Pg 18

 All of Chapter 13
Physics 201: Lecture 24, Pg 19