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.
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.
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).
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.
Fα m1 and m2
From Kepler’s Laws and the fact that the planets moved in ellipses he could mathematically determine that
Fα1 / r2
He put these two ideas together to state:
Fαm1 m2 / r2
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, thenwouldn’titsfreefall 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!”
Isay “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= 4π2/GM
Warning! Be Careful about r !!!
1.At the surface of a planet,
2.For a satellite in orbit around a planet,
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? Thesummit 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.
(4027N ; 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)
Earth6.38 x 10 65.98 x 10 24
Moon1.74 x 10 67.36 x 10 22
Sun6.96 x 10 81.99 x 10 30
Venus6.052 x 10 64.87 x 10 24
Mars3.4 x 10 66.42 x 10 23
Jupiter7.15 x 10 71.9 x 10 27
Average
Earth-Moon distance3.84 x 10 8
Average
Earth-Sundistance1.496 x 10 11
Average
Mars-Sun distance2.28 x 10 11