12.85:
a)
,
E
2
3
2
GM
r
T
π
=
therefore
(
29
(
29
.
1
1
)
(
E
2
1
E
2
3
E
2
3
E
3
2
3
2
2
3
2
2
3
2
GM
r
r
π
r
r
GM
r
π
r
r
GM
r
π
GM
π
T
r
r
T
T
∆
∆
∆
+
=
+
≈
+
=
∆
+
=
∆
+
Since
,
.
,
2
1
E
3
E

∆
π
=
=
∆
=
r
GM
v
T
v
v
r
r
GM
therefore
(
29
(
29
.
1
1
)
(
2
3
2
2
2
1
E
2
1
2
1
2
1
r
v
r
GM
r
GM
r
r
GM
v
v
r
GM
r
r
r
r
E
E
E
∆

=

≈
+
=
∆
+
=
∆

∆


∆


Since
.
,
2
E
2
3
T
r
GM
r
π
v
T
∆
π
=
∆
=
b) Note: Because of the small change in
r
, several significant figures are needed to
see the results. Starting with
GM
r
T
2
3
2
π
=
(Eq.(12.14)),
v
r
T
2
π
=
,
and
r
GM
v
=
(Eq.(12.12)) find the velocity and period of the initial orbit:
s,
m
10
672
.
7
m
10
776
.
6
)
kg
10
97
.
5
)(
kg
m
N
10
673
.
6
(
3
6
24
2
2
11
×
=
×
×
⋅
×
=

v
and
5
.
92
s
5549
2
=
=
=
v
r
T
min. We then can use the two derived equations to
approximate the
s.
m
05662
.
and
s,
1228
.
0
.
and
,
and
)
s
5549
(
)
m
100
(
s
m
10
672
.
7
m)
100
(
3
r
3
3
=
=
=
∆
=
=
∆
=
∆
=
∆
∆
∆
∆
×
π
∆
∆
π
T
r
π
T
r
π
v
π
v
T
v
T
v
T
Before the cable breaks, the shuttle will have traveled a distance
d, d
22
s
7
.
1324
)
s
m
05662
(.
)
m
75
(
So,
m.
75
)
m
100
(
)
m
125
(
2
2
=
=
=

=
min. It will take
22 minutes for the cable to break.
c) The ISS is moving faster than the space shuttle, so the total angle it covers in an
orbit must be
2
radians more than the angle that the space shuttle covers before they are
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 International Space Station, GM E

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