Kepler’s Laws of Planetary Motion
Based on the very accurate observations of Tycho Brahe, Kepler’s teacher, Kepler
developed the following laws of planetary motion. Know that these laws are
based on observations only...they do not offer reasons or try to explain what they
describe.
Kepler’s 1st Law : Planetary orbits are ellipses, with the sun at one focus.
An ellipse is a figure such that for any point P on the figure, the sum of the
distances PF1 + PF2 is always the same. (F1 and F2 are the focal points.)
The eccentricity of an ellipse is found by the ratio c / L, where c and L are as
shown in the figure below. The smaller the ratio, the more circular the ellipse.
Kepler first studied the orbit of Mars, which has the most elliptical orbit of the
known planets at that time.
P
F1
F2
c
L
Kepler’s 2nd Law : The Law of Equal Areas
An imaginary line drawn between the sun and the planet “sweeps out” equal
areas in equal time intervals. As illustrated below, this means that planets travel
fastest when they are closest to the sun. We now know (thanks to Newton) that
this is because of the greater pull of gravity when the planet is nearer to the sun.
∆t
∆t
The time period for each of the
‘triangular’ regions, whose areas
are equal, is the same. Equal areas
are "swept out" in equal time
periods! The closer to the sun that
a planet is, therefore, the faster it
must be moving!!!
Kepler’s 3rd Law : The Law of Periods
The ratio of the average radius of orbit cubed to the planet’s period squared is the
same for all the planets:
ravg3 / T2 = k
interesting twilight zone note: k = 1 when T is in years and ravg is in AU’s (1 AU
= 93,000,000 miles or the earth - sun distance).
planet
ravg
Sun
Newton’s Law of Universal Gravitation
Newton improved upon Galileo’s ideas of falling bodies, inertia and projectile
motion. Newton found that the cause for the motions Galileo studied extended
beyond the earthly realm, into the heavens. In his time, it was thought that
different rules applied for the rest of the universe than applied on earth. This
seems appropriate since Newton was born on Christmas of the year that Galileo
died.
Newton postulated that the same ‘force of gravity’ that caused objects to
accelerate towards the ground near the earth’s surface (ummm….like apples
perhaps?) was THE SAME FORCE that kept the moon in orbit around the earth.
Of course by now we too all know that an object moving in a circle must have a
net force on it.
This is where the apples came in. He wasn’t ‘hit on the head’ by an apple. See
how many people will tell you that’s what happened. And he didn’t ‘discover’
gravity. What he did do was work out the initial mathematical justification to
calculate the actual force.
He figured it like this (sorta):
If the force of gravity was dependent on the mass of a falling object (remember
W=mg) then the force was also dependent on the mass of the earth (It was his
very own deceptively simple Third Law F12 = - F21 that allowed him to see this).
Thus the gravitational force between two masses depended on the mass of each
object.
F
α m1 and m2
this symbol
means
‘proportional to’
From Kepler’s Laws and the fact that the planets moved in ellipses he could
mathematically determine that
F
α 1 / r2
He put these two ideas together to state:
F α m1 m2
/ r2
r is the distance
between the
centers
of the two objects
100 years later Cavendish did the detailed experiments with his Torsion Balance
to solidify this proportionality into a mathematical expression (don’t say the e
word).
F = G m1m2 / r2
Where G = 6.673 x 10-11 N m2 / kg2
This expression makes you scratch your head and wonder a bit. You say
“HOLD ON A SECOND MR. EISNER…I know Newton’s Second
Law pretty well by now. It says a = Fnet / m. So if the force on a
freely falling rock is smaller as the distance from the earth’s center
increases, then wouldn’t its freefall acceleration be smaller too?”
I say “why yes, that’s an insightful supposition.”
You say “But
if that’s true, then g, the freefall acceleration due
to gravity is NOT constant, but really depends on your r, or
distance from the center of the earth!”
I say “well, yes that does seem to naturally follow.”
You say “Then
what was all that ‘g is constant’ stuff
about???”
I say “well sometimes it makes more sense to take these ideas in small steps, you know how it is. I mean if we consider
how big the radius of the earth is, then any small changes in height near the earth’s surface don’t really impact the overall
r that much so we can ummmm, ignore those little differences and say that r is well, ahhh, constant when things are close
to the earth. No, really, you can put that big stick down right about now…
ACTIVITY:
See if you can use Newton’s Law of Universal gravitation to make an
estimate of the mass of the earth! The radius of the earth was known
then (thank Eratosthenes) to be about 6.38 x 106 meters. Use F = G
m1m2 / r2. Use any falling object as m1 and the earth as m2. Remember,
you know the value of the force on an object at or very near the earth’s
surface. Go.
Combining Circles and Gravity
Okay. Let’s pretend for a minute that there was this large object moving in a
constant speed circle (almost) around the earth. I know, its crazy, but humor me.
Newton figured Fc = Fg. In words, he figured the gravitational force was the
centripetal force. We know Fc isn’t its own force, so it had to be due to
something.
ACTIVITY:
You be Newton. Go. Ha ha. No really. Write a mathematical expression
(in variable form) for:
1) ac of the moon.
2) v of the moon
3) T of the moon
4) Once you have completed step 3, square both sides of the expression
for T, and rearranging the terms to show T2/r3 = 4 π2/GM
Warning! Be Careful about r !!!
1.
At the surface of a planet,
r is the radius of the planet
r
2.
For a satellite in orbit around a planet,
h
rplanet
r = rplanet + h
Gravity Problems
1. What is the acceleration due to gravity at the surface of the earth? At the top
of Mt. Everest? The summit is about 8.8 km above sea level. Report your
answers to three decimal places.
(9.803, 9.776 m/s2 )
2. The space shuttle orbits about 200 km above the earth’s surface. Find its
orbital speed and period.
(7787 m/s ; 5323 sec)
3. A 500 kg satellite orbits 300 km above the surface of Venus. Find the satellite’s
weight and its orbital period.
(4027 N ; 5717 sec)
4. Telecommunication satellites have a period of 24 hours. How far are they
from the surface of the earth? Think about this some more. What does this say
about ‘geosynchronous’ orbits?
(3.56 x 107 m)
5. Find the force of gravity between the earth and the moon.
(2.03 x 1020 N)
6. Suppose a planet is 4 times as far from the sun as Mars is. What would be the
period of that planet’s orbit?
(4.75 x 108 sec)
7. My weight, mg, is the force of gravity, Fg = Gm1m2/r2, between me and the
earth. I weigh (the earth pulls on me with a force of) 850 N when I am one earth
radius away from its center.
(a) What is the gravitational attraction between myself and a lovely math or
history teacher who weighs 50 kg and is standing 1 meter away from me. (Now I
can say without a doubt, despite what former girlfriends have said, that I truly am attractive!)
(2.8 x 10-7 N)
8 A rocket ship is launched directly at the moon. At what point will the lunar
gravitational force become equal to that of the earth? Are the astronauts on
board truly ‘weightless’ at this point?
(3.41 x 10 8 AND 4.37 x 10 8 m
explain if both answers are valid)
Universal Gravitation Thought Questions
1.
Since the earth is attracted to the sun, why don’t we just fall into it?
2.
If gravity is a force of attraction between any two masses, why is it that we
never observe people drifting toward large buildings, or your pencils and
books moving towards you?
3.
If the sun suddenly vanished, what would happen to the motion of the
planets?
4.
Why do all masses fall at the same rate near the earth’s surface (if we
neglect air resistance)?
Table of Astronomical Data
Mean Radius (m)
Mass (kg)
Earth
6.38 x 10 6
5.98 x 10 24
Moon
1.74 x 10 6
7.36 x 10 22
Sun
6.96 x 10 8
1.99 x 10 30
Venus
6.052 x 10 6
4.87 x 10 24
Mars
3.4 x 10 6
6.42 x 10 23
Jupiter
7.15 x 10 7
1.9 x 10 27
Average
Earth-Moon distance
3.84 x 10 8
Average
Earth-Sun distance
1.496 x 10 11
Average
Mars-Sun distance
2.28 x 10 11