Find the odd one out:
(1)
Change in value of 'g' at surface, change in value of 'g' at height, change in value
of 'g' at depth, change in value of 'g' on thickness.
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Answers
Answer:
Let, g
h
be the acceleration due to gravity at height h.
g
x
be the acceleration due to gravity at depth x.
The acceleration due to gravity on the surface of Earth is
g=
R
2
GM
where,R is the radius o Earth,M is mass of Earth and G is gravitational constant.
∴g∝
R
2
1
⇒g
h
∝
(R+h)
2
1
Therefore
g
g
h
=
(R+h)
2
R
2
g
g
h
=
(1+
R
h
)
2
1
g
g
h
=(1+
R
h
)
−2
g
g
h
=(1−
R
2h
)
g
h
=g−
R
2gh
g−g
h
=
R
2gh
Also, the acceleration due to gravity at depth x,
g
x
=
3
4
Gρ(R−x)
⇒g
x
∝(R−x)
Therefore
g
g
x
=
R
R−x
g
g
x
=1−
R
x
g
x
=g−
R
gx
g−g
x
=
R
gx
If the change in the value of g at height h above earth surface is the same as that at depth x i.e.
g−g
h
=g−g
x
R
2gh
=
R
gx
∴x=2h
Explanation: