Question

derive an expression for acceleration due to gravity at depth ​

Answer 1

Answer:

Let M be mass of the earth, R be the radius of the earth

g

d

be gravitational acceleration at depth

′

d

′

from the earth surface

g be gravitational acceleration on the earth surfaces.

p be the density of the earth.

′

p

′

be the point inside the earth at depth

′

d

′

from earth surfaces.

∴CS−CP=d, ∴CP=R−d-----------(1) (since CS=R)

g=

R

2

GM

∴g=

R

2

G

3

4

πR

3

p

∴g=

3

4GπRp

--------------(2)

gd= acceleration due to gravity at depth

′

d

′

g

d

=

Cp

2

G×MassofthespherewithradiusCP

∴g

d

=

CP

2

G

3

4

πCP

3

ρ

∴g

d

=

3

4GπCPρ

-----------(3)

Dividing eq. (3) by eq(2)

g

g

d

=

R

CP

=

R

R−d

∴g

d

=g(1−

R

d

)

Answer 2

Answer :-

According to the gravitational law

$\sf F = \frac{GMm}{R^2}$ - i

According to Newton's second law of motion -

F = ma - ii

From equation i and ii :-

$\sf ma = \frac{GMm}{R^2}$

$\sf a = \frac{GM}{R^2}$

This a is known as acceleration due to gravity.

$\boxed{\sf a = g = \frac{GM}{R^2}}$