Question

Explain why the value of g is zero at the centre of the earth. I want the ans. from Navneet digest. plzzz... ​

Answer 1

Explanation:

First of all, let's recall the effect of depth on Acceleration due to gravity (the acceleration due to effect of Earth's gravity @here).

► Acceleration due to gravity decreases with depth.

It is given by,

$\large{ \boxed{ \rm{g' = g \bigg(1 - \frac{d}{R} \bigg) }}}$

Here,

• d = depth at which object is kept
• R = Radius of earth / any other celestial body.
• g' = Acceleration due to gravity at depth d
• g = Acceleration due to gravity at surface.

At the centre of the earth,

• g' = g0 (Let)
• d = R (depth = Radius)

Now, plug these in the above relation of acceleration due to gravity with depth of the object.

⇛ g0 = g( 1 - R/R)

⇛ g0 = g(1 - 1)

⇛ g0 = 0

It means, when the object is placed at the centre, and the depth becomes the radius. Thus, acceleration due to gravity = 0 [Value of g = 0]

And, we are done proving this !

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Answer 2

Explanation:

First of all, let's recall the effect of depth on Acceleration due to gravity (the acceleration due to effect of Earth's gravity here).

⭐Acceleration due to gravity decreases with depth.

$\large{ \boxed{ \red\rm{g' = g \bigg(1 - \frac{d}{R} \bigg) }}}$

Here,

d = depth at which object is kept

R = Radius of earth / any other celestial body.

g' = Acceleration due to gravity at depth d

g = Acceleration due to gravity at surface.

At the centre of the earth,

g' = g0 (Let)

d = R (depth = Radius)

Now, plug these in the above relation of acceleration due to gravity with depth of the object.

✡g0 = g( 1 - R/R)

✡g0 = g(1 - 1)

✡g0 = 0

It means, when the object is placed at the centre, and the depth becomes the radius. Thus, acceleration due to gravity = 0 [Value of g = 0]