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4.0000e+002 7.0294e+006 -5.9973e+002 3.7911e-001 9.6951e-004
5.0000e+002 6.9622e+006 -7.4393e+002 4.7697e-001 9.8832e-004
6.0000e+002 6.8808e+006 -8.8389e+002 5.7693e-001 1.0118e-003
7.0000e+002 6.7856e+006 -1.0183e+003 6.7950e-001 1.0404e-003
8.0000e+002 6.6773e+006 -1.1456e+003 7.8520e-001 1.0744e-003
9.0000e+002 6.5568e+006 -1.2639e+003 8.9459e-001 1.1143e-003
1.0000e+003 6.4250e+006 -1.3708e+003 1.0083e+000 1.1605e-003
1.1000e+003 6.2831e+006 -1.4634e+003 1.1269e+000 1.2135e-003
1.2000e+003 6.1329e+006 -1.5384e+003 1.2512e+000 1.2737e-003
10
6
m. This occurs
The spacecraft hits the earth when
r
equals
R
e
=
6
.
378 14
×
between
t
1000 and1100 s.Amore accurate valueof
t
canbeobtainedbypolynomial
interpolation. If no great precisionis needed,linear interpolationwill do.Letting
1000
=
+
t
be the timeofimpact, wecan write
r
(1000
+
t
)
=
R
e
Expanding
r
in a two-term Taylor series, we get
+
=
r
(1000)
r
(1000)
t
R
e
+
−
10
3
10
6
10
3
6
.
4250
×
1
.
3708
×
t
=
6378
.
14
×
fromwhich
t
=
34
.
184s
Thus the timeofimpact is 1034.2 s.
The coordinate
of the impact site can be estimatedinasimilar manner. Using
again twoterms of the Taylor series, wehave
θ
˙
θ
(1000
+
t
)
=
θ
(1000)
+
θ
(1000)
t
+
1
10
−
3
(34
=
1
.
0083
.
1605
×
.
184)
00
◦
=
.
=
.
1
0480 rad
60
PROBLEM SET 7.1
1.
Given
y
+
x
2
4
y
=
y
(0)
=
1
compute
y
(0
1) using one step of the Taylor series method of order (a)two and
(b)four. Compare the result with the analytical solution
.
31
32
e
−
4
x
1
4
x
2
1
8
x
1
32
y
(
x
)
=
+
−
+
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