Question

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

Answer 1

Case-1: Near to the surface

$kx=mg$ (where, $g=\frac{GM}{R^2}$)
So, $kx=m\frac{GM}{R^2}$ ....(1)

Case-2: Near to the hight of $800km.$
$g_{efft.}=\frac{GM}{(R+h{)}^2}$
so, $k{x}^{\prime }=m\frac{GM}{(R+h{)}^2}$ ....(2)

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

Here, given $R=6400km,h=800km$
$\frac{x^{\prime }}{x}=\frac{(6400{)}^2}{(6400+800{)}^2}{x}^{\prime }=0.79x$
and $x=1cm$
$\therefore x^{\prime }=0.79cm$