Kepler's Laws:

Johann Kepler was a contemporary of both Galileo and an experimentalist named Tycho Brahe (covering the period from 1546-1642; Isaac Newton was born in 1642).  Brahe's contribution to astronomy was that he charted the motion of the planets for over 20 years, providing a voluminous set of data on their orbital motion around the Sun.  Kepler studied this data and discovered that the planets followed three "rules" in their trip around the Sun:

(Isaac Newton would later use the 3rd rule to confirm his gravitational equation)

1. The planets follow an elliptical, not circular, path around the sun. The sun is located at one of the foci. Having said that, note that the actual orbit has a fairly low "eccentricity", that is, it is very close to being circular. The planets vary, but take our Earth for example:

At it's farthest point (apogee) it is 94,500,000 miles away from the sun. At it's closest, perigee, it is 91,500,000 miles away. Not a big difference - a little more than 3%.

2. The arc swept out by the radius to the earth covers equal areas in equal times.

In the above exaggerated ellipse, the planet must move faster when it is closer to the sun in order to cover more area. By the way, our summer occurs when we are farthest away!

Since the earth is tilted on its axis, it tilts away from the sun in winter (even though it is closer) and towards the sun in summer (when it is further away). The summer is determined from the fact that the Sun's rays strike the Earth more directly (for those of us in the Northern Hemisphere) which overwhelms the fact that we are further away.

3. The last of Kepler's laws is the most interesting. The ratio of a planet's distance from the Sun (cubed) to the time it takes to orbit the Sun (the period, squared) is the same for every Planet!!!

Kepler discovered this looking over Tycho Brahe's voluminous data on planetary positions. Let's see what that translates to mathematically speaking:

R³/T² = Constant

In fact, we can compare orbits and rotational periods of the planets since this "Constant" is the same for all of them. That is a way of determining the distance of a planet if it's period is known.

Example 1:

Earth takes 365 days to orbit the Sun (not a revelation). It's average distance from the Sun is 93,000,000 miles, or 1.00 AU (Astronomical Unit...!). If it takes Venus 225 days to orbit the Sun, how far from the Sun is Venus?

(R³/T²)_earth = (R³/T²)_venus

(1.00³/365²)_earth = (R³/225²)_venus

Now cross-multiply and you get,

R³ = 50625/133225

R³ = 0.38

R = 0.79 AU, or 67.4 million miles

Example 2:

OK. This is a little tougher than just setting up a ratio. And since many of you have questions on this, let's try a little example. (I'll let you work out the math).

Our galaxy (Milky Way) has an estimated mass of 4 x 10⁴¹ kg. The earth is located at a distance of 2.84 x 10²⁰ m. Can we determine the time it takes for the Sun to orbit the center of the Galaxy?

Well, yeh, but Kepler won't help us much here. Newton, however, will:

Since the gravitational attraction of the Galaxy MUST supply the centripetal force holding the Sun in place, we can equate these two forces:

Centripetal force: F = m_sunv²/r

Gravitational force: F = GM_galaxym_sun/r²

Let's equate them:

m_sunv²/r = GM_galaxym_sun/r²

v²/r = GM_galaxy/r²

But we need to input the Period, T. We can do this by noticing that,

v = 2pr/T

Now, substituting this in for v, we get

4p²r³/GT² = M_galaxy

or,

T² = 4p²r³/GM_galaxy

Try it out and see what you get…!!! Will we live to see one orbit?

CENTRIPETAL FORCE AND SATELLITES…

Newton's Law of Universal Gravitation,

F = GMm/r²

and the formula for centripetal force,

F = mv²/r

can combine to give you the velocity of the Moon in orbit around the Earth. Since both of these forces MUST be the same we can equate them (as in the above example). Let's assume the Moon's orbit is roughly circular,

Here's our data:

r_earth-moon = 3.82 x 10⁸ m

m_moon = 7.36 x 10²² kg

M_earth = 5.98 x 10²⁴ kg

G = 6.67 x 10-11 Nm²/kg²

T_moon (period of rotation) = 27.3 days = _______ seconds (we need this to find the velocity for the centripetal force formula)

Now equate the two forces to find out the velocity of the Moon in its orbit:

GM_earthm_moon/r² = mv²/r

GM_earth/r² = v²/r

v² = (GM_earth/r)

v = 1022 m/s, or about 2300 mph.

You can check that this is reasonable by knowing that the moon covers the circumference of a circle in its Period of 27.3 days:

v = (2pr/T)

v_moon = 1018 m/s

Well, that's within significant digits…

YET ANOTHER EXAMPLE:

Let's take a satellite that's permanently stationed above a point on the equator: this is called a geosynchronous satellite. To be stationed there always, it must rotate at the same rate as the Earth, or once every 24 hours (= 8.64 x 10⁴ sec). Now we need Kepler's help. OK, he's dead, but his third law can still help; we need to know how far away the satellite must be:

r³/T² = r³/T²

We can use the Moon as a reference satellite,

(3.82 x 10⁸m)³/(27.3 days)² = r³/(1 day)²

r = 4.2 x 10⁷ m

or about 26, 000 miles up!!