Derive an expression for the various of acceleration due to gravity with height 'h' from the surface of earth?
Answers
Answer:
Explanation:Let a body PP of mass mm be situated at a depth hh below the earth's surface.
The gravitational force of attraction on a body inside a spherical shell is always zero. Therefore, the body PP experiences gravitational attraction only due to the inner solid sphere. The mass of this sphere is
M'=volume\times density=\cfrac { 4 }{ 3 } \pi { \left( { R }_{ e }-h \right) }^{ 3 }\rhoM
′
=volume×density=
3
4
π(R
e
−h)
3
ρ
where \rhoρ is mean density of the earth. Therefore, according to Newton's law of gravitation, the force of attraction on the body PP is
\cfrac { GM'm }{ { \left( { R }_{ e }-h \right) }^{ 2 } } =G\cfrac { \cfrac { 4 }{ 3 } \pi { \left( { R }_{ e }-h \right) }^{ 3 }\rho m }{ { \left( { R }_{ e }-h \right) }^{ 2 } }
(R
e
−h)
2
GM
′
m
=G
(R
e
−h)
2
3
4
π(R
e
−h)
3
ρm
=\cfrac { 4 }{ 3 } \pi G\left( { R }_{ e }-h \right) \rho m=
3
4
πG(R
e
−h)ρm
This force must be equal to the weight of the body mg'mg
′
, where g'g
′
is the acceleration due to gravity at a depth hh below the surface of the earth. Thus,
mg'=\cfrac { 4 }{ 3 } \pi G\left( { R }_{ e }-h \right) \rho m....(i)mg
′
=
3
4
πG(R
e
−h)ρm....(i)
Similarly, if a body be at the surface of the earth (h=0(h=0) where acceleration due to gravity is gg, then
mg=\cfrac { 4 }{ 3 } \pi G{ R }_{ e }\rho m....(ii)mg=
3
4
πGR
e
ρm....(ii)
Dividing eq.(i)(i) by (ii)(ii) we have
\Rightarrow \cfrac { g' }{ g } =\cfrac { { R }_{ e }-h }{ { R }_{ e } }⇒
g
g
′
=
R
e
R
e
−h
\Rightarrow g'=g\left( 1-\cfrac { h }{ { R }_{ e } } \right)⇒g
′
=g(1−
R
e
h
)
Answer:
Answer:
Explanation:Let a body PP of mass mm be situated at a depth hh below the earth's surface.
The gravitational force of attraction on a body inside a spherical shell is always zero. Therefore, the body PP experiences gravitational attraction only due to the inner solid sphere. The mass of this sphere is
M'=volume\times density=\cfrac { 4 }{ 3 } \pi { \left( { R }_{ e }-h \right) }^{ 3 }\rhoM
′
=volume×density=
3
4
π(R
e
−h)
3
ρ
where \rhoρ is mean density of the earth. Therefore, according to Newton's law of gravitation, the force of attraction on the body PP is
\cfrac { GM'm }{ { \left( { R }_{ e }-h \right) }^{ 2 } } =G\cfrac { \cfrac { 4 }{ 3 } \pi { \left( { R }_{ e }-h \right) }^{ 3 }\rho m }{ { \left( { R }_{ e }-h \right) }^{ 2 } }
(R
e
−h)
2
GM
′
m
=G
(R
e
−h)
2
3
4
π(R
e
−h)
3
ρm
=\cfrac { 4 }{ 3 } \pi G\left( { R }_{ e }-h \right) \rho m=
3
4
πG(R
e
−h)ρm
This force must be equal to the weight of the body mg'mg
′
, where g'g
′
is the acceleration due to gravity at a depth hh below the surface of the earth. Thus,
mg'=\cfrac { 4 }{ 3 } \pi G\left( { R }_{ e }-h \right) \rho m....(i)mg
′
=
3
4
πG(R
e
−h)ρm....(i)
Similarly, if a body be at the surface of the earth (h=0(h=0) where acceleration due to gravity is gg, then
mg=\cfrac { 4 }{ 3 } \pi G{ R }_{ e }\rho m....(ii)mg=
3
4
πGR
e
ρm....(ii)
Dividing eq.(i)(i) by (ii)(ii) we have
\Rightarrow \cfrac { g' }{ g } =\cfrac { { R }_{ e }-h }{ { R }_{ e } }⇒
g
g
′
=
R
e
R
e
−h
\Rightarrow g'=g\left( 1-\cfrac { h }{ { R }_{ e } } \right)⇒g
′
=g(1−
R
e
h
)