A particle hanging from a spring stretches it by $1 cm$ at earth’s surface. How much will be the same particle stretch (in cm) the spring at a place $800 km$ above the earth’s surface? (Take the radius of the earth as $6400 km$ )

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Solution

Case-1: Near to the surface

$k x = m g$ (where, $g = \frac{G M}{R^{2}}$ )
So, $k x = m \frac{G M}{R^{2}}$ ....(1)

Case-2: Near to the hight of $800 k m .$
$g_{e f f t .} = \frac{G M}{(R + h)^{2}}$
so, $k x^{'} = m \frac{G M}{(R + h)^{2}}$ ....(2)

Now, dividing equation 2 by equation 1, we get
$\frac{k x^{'}}{k x} = \frac{m \frac{G M}{(R + h)^{2}}}{m \frac{G M}{R^{2}}} \frac{x^{'}}{x} = \frac{R^{2}}{(R + h)^{2}}$

Here, given $R = 6400 k m, h = 800 k m$
$\frac{x^{'}}{x} = \frac{(6400)^{2}}{(6400 + 800)^{2}} x^{'} = 0.79 x$
and $x = 1 c m$
$∴ x^{'} = 0.79 c m$

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