How we know what
we know
An introduction into orbital
mechanics
Matt Hamill
History
• Isaac Newton (1643 –
1727)
• Did he discover
gravity?
– NO!
• What did he do?
– He used physics to
connect the force that
causes an apple to fall
to the force that
causes the moon to
orbit the Earth
Newton’s Thought Experiment
• If you launch a
cannonball it will
follow a curved path
due to gravitational
influences
• If the cannonball is
launched at a high
enough velocity it will
“fall” around the
Earth.
Newton’s Hypothesis
• Every particle in the Universe attracts every
other particle with a force (gravity) that is
directly proportional to the product of their
masses and inversely proportional to the
square of the distance between them.
m1  m2
FG 
2
d
The legendary apple.
• Newton knew a falling
apple accelerated
toward the Earth
around 32 ft/s2 or 10
m/s2
The mysterious Moon
• Newton also knew
– the Moon orbited the
Earth in approximately
28 days
– The moon’s distance
from the Earth was
about 60 times the
Earth’s radius
In the 17th Century how can you
prove Newton’s hypothesis?
• Newton used his inversesquare prediction to
reason that the Moon’s
acceleration toward the
Earth (centripetal
acceleration) should be
proportional to:
1
aM  2
rM
rM  3.84 10 m
8
Inverse square continued…
• Using the same logic,
the acceleration of the
apple toward the
Earth should be
proportional to:
1
g 2
RE
RE  6.37 10 m
6
Newton’s hypothesis predicts
• Newton predicted the ratio of the Moon’s
acceleration (aM) to the apple’s acceleration
(g) would be
2
2
aM (1 r )  RE   6.37 10 
4
  


 

2
.
75

10
g
(1 R )  rM   3.84 108 
2
M
2
E
6
• Therefore the centripetal acceleration of the
Moon should be around
aM  (2.75 104 )(9.80m / s 2 )  2.70 103 m / s 2
What is the Moon’s actual
centripetal acceleration?
• We know centripetal acceleration can be
calculated with the following formula
2
v
ac 
r
• If we assume the orbit of the Moon is circular it
travels a distance that is equal to the
circumference of a circle.
C  2r
What is the Moon’s actual
centripetal acceleration?
• The Moon completes its orbital period in a time
interval T = 27.32 days or 2.36 x 106 s.
v
(2rM T )
4 rM 4 (3.84 10 m)
aM 



2
r
rM
T
(2.36 106 s ) 2
2
2
2
3
2
aM  2.72 10 m / s
2
8
Predicted vs. Actual
• Predicted centripetal
acceleration of the
Moon
• Actual centripetal
acceleration of the
Moon
• Less than 1%
difference
3
aM  2.70 10 m / s
2
aM  2.72 103 m / s 2
Newton’s Law of Universal
Gravitation
Gm1m2
FG 
2
r
• G—gravitational
constant
G = 6.67 x 10-11 N·m2/kg2
• m1, m2—mass
• r—distance from their
centers of mass
Three ways to calculate
gravitational force
• Equation1:
• Equation 2:
• Equation 3
F  mg
v
F m
r
2
Gm1m2
F
2
r
Resources
• Serway, Raymond, and John Jewett. Physics for
Scientists and Engineers. 6th ed.. USA:
Brooks/Cole, 2004. Print.
• http://en.wikipedia.org/wiki/File:GodfreyKnellerIsaacNewton-1689.jpg
• http://scienceiq.com/Images/FactsImages/apple
_falling.gif
• http://www.chemheritage.org/women_chemistry/
univ/images/clark_moon.jpg
• http://www.astronautix.com/lvs/newannon.htm