Question

A particle hanging from a spring stretches it by 1 cm at earth's surface. How much the same particle stretches the spring at a place 1600 km above the surface of earth (r = 6400 km)

Answer 1

spring stretched due to act of it gravitational force . let if mass of spring is m then Fg = mg° ( weight )

e.g Fs = Fg

Kx = mg° ----------(1)

but we know,

g is varies with height from earth surface. e.g

g = g°/( 1 + h/r)²

where h is height from the surface of earth .

at h = 800 km , and r = 6400 km

g =g°/( 1+ 800/6400)²

=g°/( 1 + 1/8)²

=g°/(9/8)²

=64g°/81

now, spring stretched act by gravitational force = m64g°/81

F"s = F"g

Kx" = 64mg°/81 --------(2)

equation (1) and (2)

x/x" = 1/(64/81)

x" = 64x/81

hence 64/81 times stretched spring at h = 800 km

e.g stretching of spring = 64×1/81 cm=64/81 cm