Physics - Gravitation Question with Solution | TestHub

Acceleration due to gravity,

$g=\frac{GM}{r^2}$

$\therefore g\propto \frac{1}{r^2}$

When the body is hanged on a spring

$F=-kx=mg$.....(i)

Acts on it, where$x$is extension in spring. Let$800 km$above the Earth's surface, the stretch in the length of spring is$x\prime$and value of$g$is$g\prime$.

$g=\frac{GM}{R^2}$and$g^{\prime }=\frac{GM}{{\left(R+h\right)}^2}$

So,$g^{\prime }=\frac{g}{{\left(1+\frac{h}{R}\right)}^2}$....(ii)

Where,$R=$radius of Earth

From Equation (i), we get

$kx=mg$$\Rightarrow x\propto g$......(iii)

Therefore,$\frac{x\prime }{x}=\frac{g\prime }{g}$

From Equations (ii) and (iii), we get

$\frac{x\prime }{x}=\frac{\frac{g}{{\left(1+\frac{h}{R}\right)}^2}}{g}=\frac{g}{g{\left(1+\frac{800\times {10}^3}{6400\times {10}^3}\right)}^2}$

$=\frac{g}{g{\left(1+\frac{1}{8}\right)}^2}$

$=\frac{g}{g{\left(\frac{9}{8}\right)}^2}=\frac{g\times 64}{g\times 81}=\frac{64}{81}$

$x^{\prime }=\frac{64}{81}x$$\left[x=3 cm\right]$

$\Rightarrow x^{\prime }=\frac{64}{81}\times 3=\frac{64}{27}$

$\Rightarrow x^{\prime }=\frac{64}{27}\mathrm{ cm}$