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M
= (1.93
×
10
24
kg + 4.01
×
10
24
kg ) = 5.94
×
10
24
kg.
The first term is the mass of the core and the second is the mass of the mantle. Thus,
()
(
)
11
3
2
24
2
2
6
6.67 10
m /s kg 5.94 10 kg
==
9
.
8
4
m
/
s
.
6.345 10 m
g
a
−
×⋅
×
×
(c) A point 25 km below the surface is at the mantlecrust interface and is on the surface
of a sphere with a radius of
R
= 6.345
×
10
6
m. Since the mass is now assumed to be
uniformly distributed the mass within this sphere can be found by multiplying the mass
per unit volume by the volume of the sphere:
33
(/) ,
ee
M
RRM
=
where
M
e
is the total
mass of Earth and
R
e
is the radius of Earth. Thus,
3
6
24
24
6
=
5.98 10 kg = 5.91 10 kg.
6.37 10 m
M
§·
×
××
¨¸
×
©¹
The acceleration due to gravity is
(
)
11
3
2
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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, Acceleration, Force, Gravity, Mass

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