• No products in the cart.

Profile Photo

Losing Seconds

Taking the earth’s radius as \(6400 Km\) and assuming that the value of \(g\) inside the earth is proportional to the distance from the earth’s centre, at what depth below the earth’s surface would a pendulum which beats seconds at the earth’s surface lose \(5\) min in a day?


Let us assume \(g\) is the acceleration due to gravity at depth \(d\) below the surface of the earth and \(g\) is the acceleration due to gravity on the surface.

Let the corresponding time periods be \(T_0\) and \(T\).
$$ g=g_0(1-\frac{d}{R}) $$
where \(g\) and \(g_0\) are the acceleration due to gravity at depth respectively, and \(R\) is the radius of the earth
$$ T=T_0\sqrt{\frac{g_0}{g}}=T_0(1-d/R)^{-1/2}$$ $$=T_0(1+\frac{d}{2R})
Time registered for the whole body will be proportional to the time period. Thus,
$$ \frac{T}{T_0}=\frac{t}{t_0}=1+d/2R$$
$$ = \frac{86400}{86400-300}=1+\frac{d}{2R}$$
\(R=6400Km\), we find \(d=44.6Km\)

October 12, 2017

No comments, be the first one to comment !

    Leave a Reply

    Your email address will not be published. Required fields are marked *



    GOOGLECreate an Account