Thomas Calculus 13th [Solutions]

960 Chapter 13 Vector-Valued Functions and Motion in Space
5. 2t 2 0, r 2cos 4t r 8sin 4t r 32cos 4t
v 8sin 4t ur
2cos 4 t (2) u 8 sin 4t ur
4 cos 4t
u
2
a 32cos 4t 2cos 4 t (2) ur
2cos 4t 0 2 8sin 4 t (2) u
32cos 4t 8cos 4t ur
0 32sin 4t u 40 cos 4t ur
32 sin 4t
u
6.
2
0 0
r v 2 GM ( e 1) GM ( e 1)
e 1 v0 v0
;
GM r r
Circle: e
0 v GM
0 r
0 0
Ellipse: 0 e 1 GM v 2GM
r 0 r
0
0 0
Parabola: e 1 v 2GM
0 r0
Hyperbola: e 1 v 2GM
0 r
0
7. r GM v 2 GM GM
2
v
r
v r
which is constant since G, M, and r (the radius of orbit) are constant
8.
1 1 ( ) 1 ( ) ( ) ( ) 1 ( ) ( )
( ) ( ) A r t t r t t r t r t r t t r t
A r t t r t r( t) r( t) r( t) 1 r( t) r( t)
2 t 2 t 2 t 2 t t
1 r( t t) r( t) 1 ( ) ( )
( ) dA
r t t r t
lim ( ) 1 d r
( ) 1 d r
r t r t r t r( t)
1 r r
2 t dt
t 0
2 t 2 dt 2 dt 2
9.
2 2 4 2 4
2 2
2 2 2 4 2 4
r0 v0
T a 1 e T a 1 e a 1 1 (from Equation 5)
r v
GM
2 2 2 2
0 0 r0 v0 r0 v0
2 4 2 2 2 4
2 4 2
2 4 2 4 0 0
2
0 0 0 0 2 4 a 2GM r v
a r v r v 0 0 0 0 2 4 2 0 0
2 a GMr v r v GM r v
4 a
2 2 2 2 2 2 2 2 2 2
2
0 0 0 0 0
0
4 4 2
r v G M GM r v G M r G M
r GM GM
2 4
4 a 1 2
2a
GM
(from Equation 10)
2 3 2 2
2
T 4 a T 4
GM 3
a GM
10. For each of the planets listed we form the ratio
2 3
T / a
.
2
(4 ) / ( GM )
The values we obtain are
Mercury 1.00225
Venus 1.00288
Mars 1.00252
Saturn 1.00019
These values are all close to 1, so they support Keplers third law.
11. Solve Keplers third law for a and double this result:
12. Solve Keplers third law for a and double this result:
2
1/3
(365.256 days)
10
2 29.925 10 m
2
(4 ) / ( GM )
2
1/3
(84 years)
10
2 573.95 10 m
2
(4 ) / ( GM )
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