Question

A particle hanging from a spring stretches it by 1 cm at earth's surface. Radius of the earth is 6400 km. At a place 800 km above the earth's surface, the same particle will stretch the spring by

• A. 0.79 cm
• B. 1.2 cm
• C. 4 cm
• D. 17 cm

Answer 1

The correct option is A 0.79 cm

The extension is $x=\frac{mg}{k}$

At earth's surface, $x=\frac{1}{100}=\frac{m\times \frac{GM}{R^2}}{k}$..(1)

At 800 km,above the earth's surface,

$x^{\prime }=\frac{n\times \frac{GM}{(R+800{)}^2}}{k}$ ..(2)

(2) $÷$ (1)

$\frac{x^{\prime }}{\frac{1}{100}}=\frac{\frac{GM}{{(R+800)}^2}}{\frac{GM}{R^2}}=\frac{R^2}{{(R+800)}^2}$

$\Rightarrow x^{\prime }\times 100=\frac{x^{\prime }}{{(1+\frac{800}{R})}^2}=\frac{x^{\prime }}{{(1+\frac{800}{6400})}^2}=\frac{1}{1.265}=0.79\times {10}^{-2}m=0.79cm$