Mass of Earth : 6*10^24 Radius : 6400km Calculate 'g' at the top of Mt Everest. Height of Everest : 8848m
Answers
Decrease in magnitude of g
≈
0.0273
m
s
2
Explanation:
Let
R
→
Radius of the Earth to sea level
=
6369
k
m
=
6369000
m
M
→
the mass of the Earth
h
→
the height of the tallest spot of
Mt Everest from sea level
=
8857
m
g
→
Acceleration due to gravity of the Earth
to sea level
=
9.8
m
s
2
g
'
→
Acceleration due to gravity to tallest
spot on Earth
G
→
Gravitational constant
m
→
mass of a body
When the body of mass m is at sea level, we can write
m
g
=
G
m
M
R
2
...
...
.
.
(
1
)
When the body of mass m is at the tallest spot on Everst, we can write
m
g
'
=
G
m
M
(
R
+
h
)
2
...
...
(
2
)
Dividing (2) by (1) we get
g
'
g
=
(
R
R
+
h
)
2
=
(
1
1
+
h
R
)
2
=
(
1
+
h
R
)
−
2
≈
1
−
2
h
R
(Neglecting higher power terms of
h
R
as
h
R
<<
1
)
Now
g
'
=
g
(
1
−
2
h
R
)
So change (decrease) in magnitude of g
Δ
g
=
g
−
g
'
=
2
h
g
R
=
2
×
8857
×
9.8
6369000
≈
0.0273
m
s
2
Answer link
Eddie
Oct 11, 2016
≈
−
.027
m
s
−
2
Explanation:
Newton's Law for Gravitation
F
=
G
M
m
r
2
And
g
is computed at the earth's surface
r
e
as follows:
m
g
e
=
G
M
m
r
2
e
So
g
e
=
G
M
r
2
e
if we were to compute different
g
's we would get
g
e
v
e
r
e
s
t
−
g
s
e
a
=
G
M
(
1
r
2
e
v
e
r
e
s
t
−
1
r
2
s
e
a
)
G
M
=
3.986005
×
10
14
m
3
s
−
2
≈
3.986005
×
10
14
⋅
(
1
(
6369000
+
8857
)
2
)
−
1
6369000
2
)
≈
−
.027
m
s
−
2
Using differentials to double check:
g
e
=
G
M
r
2
e
⇒
ln
(
g
e
)
=
ln
(
G
M
r
2
e
)
=
ln
(
G
M
)
−
2
ln
(
r
e
)
d
g
e
g
e
=
−
2
d
r
e
r
e
d
g
e
=
−
2
d
r
e
r
e
g
e
=
−
2
⋅
8857
6369000
⋅
9.81
=
−
0.027
m
s
−
2
Explanation:
M= 6*10^24kg
r = 6400km =6.4*10^6m
h =8848m
we know that
g = GM/(r+h)^2
=6.673×10^-11×6×10^24/(6.4×10^6+8848)^2
= 9.75m/s^2(approx).
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