Question

A body hanging from a spring stretches it by $1cm$ at the earth’s surface. How much will the same body stretch the spring at a place $16400km$ above the earth’s surface? (Radius of the earth = $6400km$)

• A. $1.28cm$
• B. $0.64cm$
• C. $3.6cm$
• D. $0.12cm$

Answer 1

Answer: (B)

In equilibrium, weight of the suspended body = stretching force.

At the earth’s surface, $mg=k\times x$

At a height $h,m{g}^{\prime }=k\times x^{\prime }$

$\frac{g^{\prime }}{g}=\frac{x^{\prime }}{x}=\frac{R_e^2}{{\left(R_e+h\right)}^2}$

$=\frac{{\left(6400\right)}^2}{{\left(6400+1600\right)}^2}$

$={\left(\frac{6400}{8000}\right)}^2=\frac{16}{25}$

$x^{\prime }=\frac{16}{25}\times x=\frac{16}{25}\times 1cm$

$=0.64cm$