Question

Determine the height h above the surface where the value of g
(ii) falls by 1 % of the value at the surface.​

Answer 1

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