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Answers
Answer:
Let a body P of mass m be situated at a depth h below the earth's surface.
The gravitational force of attraction on a body inside a spherical shell is always zero. Therefore, the body P experiences gravitational attraction only due to the inner solid sphere. The mass of this sphere is
M
′
=volume×density=
3
4
π(R
e
−h)
3
ρ
where ρ is mean density of the earth. Therefore, according to Newton's law of gravitation, the force of attraction on the body P is
(R
e
−h)
2
GM
′
m
=G
(R
e
−h)
2
3
4
π(R
e
−h)
3
ρm
=
3
4
πG(R
e
−h)ρm
This force must be equal to the weight of the body mg
′
, where g
′
is the acceleration due to gravity at a depth h below the surface of the earth. Thus,
mg
′
=
3
4
πG(R
e
−h)ρm....(i)
Similarly, if a body be at the surface of the earth (h=0) where acceleration due to gravity is g, then
mg=
3
4
πGR
e
ρm....(ii)
Dividing eq.(i) by (ii) we have
⇒
g
g
′
=
R
e
R
e
−h
⇒g
′
=g(1−
R
e
h
)