Question

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

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

Answer 1

Answer: (B)

0.64 cm

Answer 2

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

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

At a height h, mg` = k × x`

$\frac { g ^ { \prime } } { g } = \frac { x ^ { \prime } } { x } = \frac { R _ { e } ^ { 2 } } { \left( R _ { e } + h \right) ^ { 2 } } = \frac { ( 6400 ) ^ { 2 } } { ( 600 + 1600 ) ^ { 2 } }$

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

$\mathrm { x } ^ { \prime } = \frac { 16 } { 25 } \times \mathrm { x } = \frac { 16 } { 25 } \times 1 \mathrm {~cm} = 0.64 \mathrm {~cm}$