Henry Braun 

Footnote 18 Project 

MAT 294 3:40 

Kepler's Laws 

Johannes Kepler created these three laws of planetary motion in order to explain the astronomical data gathered by Tycho Brahe.  

 

 

 

 

 

The textbook establishes the following notation: 

 

 

 

 

 

 

 

 

 

 

First law: 

 

The first law was proven in section 13.4 of Calculus: Early Transcendentals: 

r = ed/(1+e*cos(theta)) 

This is the polar equation for an ellipse with one focus at the origin, in this case the sun. e is the eccentricity of the planet's orbit and ed is its semi-latus rectum. 

 

Second Law: 

 

h = r^2(uu') 

u = r / r = `<,>`(cos*theta, sin*theta) 

u' = `<,>`(-sin*theta*(diff(theta(t), t)), cos*theta*(diff(theta(t), t)))(since theta is a function of t, the chain rule must be used.) 

(u x u') =  

 

h=r^2*(diff(theta, t))*k 

h = r^2*(diff(theta, t)) 

A(t) is the area swept out by r(t) in the time interval [t0,t].  dA is the area swept out by r(t) over an infinitesimal angle d*theta. 

 

We can see that the area dA swept out of by r(t) over an infinitesimal angle d*theta is dA = 1/2*r^2*d*theta = 1/2*r^2*(diff(theta(t), t))*dt 

Therefore: 

 

The fundamental theorem of Calculus then shows that: 

 

Since h is a constant, this proves Kepler's second Law. 

 

Third Law: 

 

T is the planet's period, that is, the time it takes for the planet to complete one elliptical orbit.  We also know A to be an ellipse with major and minor axes 2a and 2b. From this it follows that: 

`πab` = A and A = int(1/2*h, t = 0 .. T) and int(1/2*h, t = 0 .. T) = 1/2*hT 

T = 2*`πab`/h 

h = 2*`πab`/T 

From the definitions of e and d, we know that: 

ed = h^2/GM and h^2/GM = h^2/GM 

 

 

Since ed is also the semi-latus rectum l of the orbit, we also know that: 

 

al = b^2  (definition of the semi-latus rectum) 

ed = l and l = b^2/a 

equating the two definitiions for ed, we get: 

h^2/GM = ed and ed = b^2/a 

Substituting h = 2*`πab`/T, this*becomes; -1 

4*`πab`^2/(T^2*GM) = 4*Pi^2*a^2*b^2/(T^2*GM) and 4*Pi^2*a^2*b^2/(T^2*GM) = b^2/a 

T^2 = 4*Pi^2*a^3/GM 

 

The square of the period is proportional to the cube of the length of the planet's major axis.  Kepler's 3rd law is proved. 

 

Major axis of the Earth's Orbit 

The length of Earth's orbit's major axis can easily be found using the proof of Kepler's 3rd law and Maple: 

 

M := 0.1990000000e31 

0.1990000000e31 

G := 0.6670000000e-10 

0.6670000000e-10 

T must be converted into seconds for the units to work out correctly: 

T := 31557600.00 

31557600.00 

thirdlaw := T^2 = 4*Pi^2*a^3/(G*M) 

0.9958821178e15 = 0.3013568593e-19*Pi^2*a^3 

majoraxis := 2*solve(thirdlaw, a^3)^(1/3)*Units:-Unit(m) 

(Typesetting:-mprintslash)([0.2992074238e12*Unit(m)], [0.2992074238e12*Units:-Unit(m)]) 

The Earth's Major Axis is `*`(2.99, 100000000000)m long. 

 

Radius of a Geosynchrous orbit 

A satellite in a geosynchronous orbit around the earth will stay in exactly the same position above the planet; that is, it will have a period of exactly 24 hours.  This also implies that it will have a circular orbit, that is, a = b = r. 

 

M := 0.5980000000e25 

0.5980000000e25 

 

G := 0.6670000000e-10 

0.6670000000e-10 

T must be converted into seconds for the units to work out correctly: 

T := 86400 

86400 

Rearth := 6370000.00*Units:-Unit(m) 

(Typesetting:-mprintslash)([6370000.00*Unit(m)], [6370000.00*Units:-Unit(m)]) 

thirdlaw := T^2 = 4*Pi^2*a^3/(G*M) 

7464960000 = 0.1002843060e-13*Pi^2*a^3 

Rsatellite := solve(thirdlaw, a^3)^(1/3)*Units:-Unit(m) 

(Typesetting:-mprintslash)([42250474.31*Unit(m)], [42250474.31*Units:-Unit(m)]) 

altitude := Rsatellite-Rearth 

(Typesetting:-mprintslash)([35880474.31*Unit(m)], [35880474.31*Units:-Unit(m)]) 

A geostationary satellite must orbit 3.59*10 m above the surface of the Earth.