Question

The ratio of values of g at height R/2 to that at depth R/2 is . ( R is radius of earth)

Answer 1

Answer:

The ratio of g is 8 : 9

Explanation:

The value of g at a height h = R/2 above the earth is

g₁ =  g / [1 + h/R]²  = g/ [1 + 1/2]² = g/(9/4) = 4g/9

The value of g at a depth d = R/2 is

g₂ = g x ( 1 - d/Re) = g x ( 1 - 1/2) = g/2

∴ g₁/g₂ = (4g/9) /(g/2) =8/9

Answer 2

Answer:

The ratio of values of g at a particular height R/2 to that at a depth of R/2 is found to be 8:9

Explanation:

Here we have been given to find the values of g at a height of R/2 and at a depth of R/2

Here R is the radius of the earth.

The value of acceleration due to the gravity (g') at a height of about h is given by the following formula:

$g'= \frac{gR^{2} }{(R+h)^{2} }$

g = acceleration due to gravity on Earth.

Therefore at h = R / 2, we have the acceleration due to gravity ($g_{h}$) is as below:

∴ $g_{h} = \frac{gR^{2} }{(R+\frac{R}{2} )^{2} }$

⇒ $g_{h} = \frac{gR^{2} }{(\frac{3R}{2} )^{2} }$

⇒ $g_{h} = \frac{4gR^{2} }{9R^{2} }$

⇒ $g_{h} = \frac{4g}{9}$

The value of acceleration due to the gravity (g) at a depth of about d is given by the following formula:

$g_{1} = g (1 - \frac{d}{R} )$

At d = R/ 2 the value of the acceleration due to gravity ($g_{d}$) is as below:

∴ $g_{d} = g (1 - \frac{R/2}{R} )$

⇒ $g_{d} = g (1 - \frac{R}{2R} )$

⇒ $g_{d} = g (1 - \frac{1}{2} )$

⇒ $g_{d} = \frac{g}{2}$

Therefore their ratio is as below:

∴ $\frac{g_{h} }{g_{d} } = \frac{4g}{9} / \frac{g}{2}$

⇒ $\frac{g_{h} }{g_{d} } = \frac{4*2}{9}$

⇒ $\frac{g_{h} }{g_{d} } = \frac{8}{9}$

Therefore the required ratio is 8:9