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1
Lecture 13
1.1
Variable gravitational acceleration
Newton&s law of gravitation states that the gravitational force between two
masses
M
and
m
is given by
F
=
GMm
r
2
;
where
G
is an empirical constant:
G
= 6
:
6726
&
10
&
11
N
&
m
kg
2
±
:
Let
M
E
be the mass of Earth. Let a body of mass
m
is falling toward Earth
(the air resistance is neglected). Then
F
=
GM
E
m
(
R
E
+
r
)
2
, where
r
is the distance from the body to the Earth surface and
G
= 6
:
6726
&
10
&
11
N
²
m
kg
2
³
;
M
E
= 5
:
97
&
10
24
kg
;
R
E
= 6
:
378
&
10
6
m:
1
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a
r
t
h
s
u
f
c
e
Center of Earth
r
R
E
If the body is close to the surface of Earth, then
r
&
R
E
and
(
R
E
+
r
)
2
±
R
2
E
.
So,
F
=
mg
=
&
GM
E
R
E
2
±
m
, whence
g
=
GM
E
R
E
2
=
(6
:
6726
²
10
&
11
) (5
:
97
²
10
24
)
(6
:
378
²
10
6
)
2
= 9
:
792 7
±
9
:
8
m
sec
2
:
2
If the body is far away from the surface of the earth, then
r
cannot be
neglected and
g
(
r
) =
GM
E
(
R
E
+
r
)
2
=
(6
:
6726
&
10
&
11
) (5
:
97
&
10
24
)
(6
:
378
&
10
6
+
r
)
2
=
3
:
983 5
±
10
14
(
r
+ 6
:
378
±
10
6
)
2
is not constant:
r
g
(
r
)
100
m
9
:
792 2
m
sec
2
1000
m
9
:
789 5
m
sec
2
1
;
000
;
000
m
= 1000
km
7
:
317 9
m
sec
2
100
;
000
;
000
m
= 100
;
000
km
3
:
520 2
±
10
&
2
m
sec
2
1.2
Escape velocity
What initial velocity is needed to escape the Earth? Here we give the answer
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This note was uploaded on 01/26/2012 for the course MATH 285 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff
 Differential Equations, Equations

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