A body is taken to a height of (√2-1)Re above the earth's surface, where Re is the radius of earth. I f it is dropped from this height, its acceleration will be
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Explanation:
Rearranging (Figure), we have
M
E
=
g
R
2
E
G
=
9.80
m/s
2
(
6.37
×
10
6
m
)
2
6.67
×
10
−
11
N
⋅
m
2
/kg
2
=
5.95
×
10
24
kg.
The volume of a sphere is proportional to the radius cubed, so a simple ratio gives us
M
M
M
E
=
R
3
M
R
3
E
→
M
M
=
(
(
1.7
×
10
6
m
)
3
(
6.37
×
10
6
m
)
3
)
(
5.95
×
10
24
kg
)
=
1.1
×
10
23
kg.
We now use (Figure).
g
M
=
G
M
M
r
2
M
=
(
6.67
×
10
−
11
N
⋅
m
2
/kg
2
)
(
1.1
×
10
23
kg
)
(
1.7
×
10
6
m
)
2
=
2.5
m/s
2
Significance
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