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(g) To find the period, we use Eq. 1334 but replace
r
with
a
. The result is
23
2
6
3
3
11
3
2
24
44
(
6
.
6
3
1
0
m
)
5.37
10 s
89.5 min.
(6.67
10
m /s kg)(5.98
10 kg)
a
T
GM
ππ
−
×
==
=
×
≈
×⋅
×
(h) The orbital period
T
for Picard’s elliptical orbit is shorter than Igor’s by
0
5540 s 5370 s 170 s
TTT
Δ= −=
−
=
.
Thus, Picard will arrive back at point
P
ahead of Igor by 170 s – 90 s = 80 s.
68. The orbital radius is
6
6370 km 400 km
6770 km
6.77 10 m.
E
rR h
=+
=
+
=
=
×
(a) Using Kepler’s law given in Eq. 1334, we find the period of the ships to be
2
6
3
3
0
11
3
2
24
(
6
.
7
7
1
0
m
)
5.54
10 s
92.3 min.
(6.67
10
m /s kg)(5.98
10 kg)
r
T
GM
−
×
=
×
≈
×
(b) The speed of the ships is
6
32
0
3
0
22
(
6
.
7
7
1
0
m
)
7.68 10 m/s
5.54 10 s
r
v
T
×
=×
×
.
(c) The new kinetic energy is
2
3
2
1
0
0
11
1
(0.99 )
(2000 kg)(0.99) (7.68 10 m/s)
5.78 10 J.
2
Km
v
m
v
=
×
=
×
(d) Immediately after the burst, the potential energy is the same as it was before the burst.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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