360 CHAPTER 17. SOME PHYSICAL APPLICATIONS

and this yields

x2/

(ε2 p2

1− ε2

)+

(y+

pε2

1− ε2

)2

/

(ε2 p2

(1− ε2)2

)= 1. (17.12)

Now note this is the equation of an ellipse and that the diameter of this ellipse is

2ε p(1− ε2)

≡ 2a. (17.13)

This follows becauseε2 p2

(1− ε2)2 ≥ ε2 p2

1− ε2 .

Now let T denote the time it takes for the planet to make one revolution about the sun. Itis left as an exercise for you to show that the area of an ellipse whose long axis is 2a andwhose short axis is 2b is πab. This is an exercise in trig. substitutions and is a little tediousbut routine. Using this formula, and 17.3 the following equation must hold.

area of ellipse︷ ︸︸ ︷π

ε p√1− ε2

ε p(1− ε2)

= Tc2

Therefore,

T =2c

πε2 p2

(1− ε2)3/2

and so

T 2 =4π2ε4 p4

c2 (1− ε2)3

Now using 17.10, recalling that k = GM, and 17.13,

T 2 =4π2ε4 p4

kε p(1− ε2)3 =4π2 (ε p)3

k (1− ε2)3 =4π2a3

k=

4π2a3

GM.

Written more memorably, this has shown

T 2 =4π2

GM

(diameter of ellipse

2

)3

. (17.14)

This relationship is known as Kepler’s third law.

17.4 The Angular Velocity VectorLet (i(t) ,j (t) ,k (t)) be a right handed system of unit basis vectors. Thus k (t) = i(t)×j (t) and each vector has unit length. This represents a moving coordinate system. Weassume that i(t) ,j (t) ,k (t) are each continuous having continuous derivatives, as many asneeded for the following manipulations for t in some open interval. The various rules ofdifferentiation of vector valued functions will be used to show the existence of an angularvelocity vector.

360 CHAPTER 17. SOME PHYSICAL APPLICATIONSand this yields3 ep? per 2 ep?=1. 17.12/ (525) + (r+ a) / (1-22) ( )Now note this is the equation of an ellipse and that the diameter of this ellipse is2Ep= 2a. 17.13(1 —€?) “ ( )This follows becauseep? . 2p?Now let T denote the time it takes for the planet to make one revolution about the sun. Itis left as an exercise for you to show that the area of an ellipse whose long axis is 2a andwhose short axis is 2b is zab. This is an exercise in trig. substitutions and is a little tediousbut routine. Using this formula, and 17.3 the following equation must hold.area of ellipse1 Ep Ep _7eJl—e2 (1-€?) 2Therefore,2 2 2edc(1—22)3?and so> 4ne4 p42 (1-22)Now using 17.10, recalling that k = GM, and 17.13,7 4n?e*pt An? (ep)? Ana _ 4n7a3~ kep(1—e2)>—k(1—-e2)8 sk GM "Written more memorably, this has shown7 4n? (Sn of ates ; (17.14)~ GM 2This relationship is known as Kepler’s third law.17.4 The Angular Velocity VectorLet (4(t),7 (t),&(t)) be a right handed system of unit basis vectors. Thus k(t) = i(t) xj(t) and each vector has unit length. This represents a moving coordinate system. Weassume that z(t) ,7 (t) ,& (t) are each continuous having continuous derivatives, as many asneeded for the following manipulations for tf in some open interval. The various rules ofdifferentiation of vector valued functions will be used to show the existence of an angularvelocity vector.