# Numerical Problem Related to Acceleration Due to Gravity.

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If the value of $g$ is same at depth $d$ inside the earth and height $h$ above it, then:

$1$. $d=h$

$2$. $d=2h$

$3$. $d=\dfrac {h}{2}$

$4$. $d=3h$.

My Attempt:

The value of $g$(acceleration due to gravity) at depth $d$ is :
$$g'=(1-\dfrac {d}{R})\times g$$
where, $R$ is the radius of the earth.

Again, The value of $g$(acceleration due to gravity at height $h$ is:
$$g=\dfrac {GM}{(R+h)^2}$$
where, $M$ is the mass of the earth.

How do I proceed now?

1 vote

You are right so far, use binomial theorem ,
$g=GM/(R+h)^2$
$g=GM/R^2 (1+h/R)^2$

let $g_0$ be acceleration on surface.
hence , $g=g_0 (1+h/R)^{-2}$

now using binomial theorem,
$g=g_0 (1-2h/R)$

Hence if $g$ is to be same at the height and depth then equate the two equatiobs ,
$g=g_0 (1-2h/R) =g_0 (1-d/R)$
$2h=d$

answered May 7, 2017 by (2,320 points)
@physicsapproval, How did you use the Binomial Theorem here:
$$g=g_o (1-\dfrac {2h}{R})$$???
well, if you are not familiar with the binomial expansions then you may have a look on web.
Although you may come across this frequenly  of form
$(1+x)^n = 1+xn +\frac{n (n-1)}{2} x^2 +.........$

However since $x <<1$ So we neglect higher order terms, Hence,

$( 1+x)^n = 1+nx$ ( its a result worth remembering)

In question above we have treated $h <<R$ and so $h/R = x$ (suppose) then $x <<1$
@Albert Einstein is the question clear to you or it requires even more explanation
to make it more clear ?