Question

A perticle hanging from a spring stretches it by 1 cm at earth's surface.How much time will the same particle stretch the spring at a place 800 km above the earth's surface?[The radius of earth = 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

Answer 2

Hi Arunima,
___________________________________________________________

x= 1cm=0.01m

h= 800 km

R=6400km

x'=?

g'/g=x'/x

g'/g=$\frac{ 6400^{2} }{ (6400+800)^{2} } = \frac{64}{81}$
$\frac{x'}{x} = \frac{64}{81}$
x'=0.64/81 m(stretching of particle)
g'/g=x'/x
g'=x' X g / x
T=2 Pi  √x'/g'
T=2 pi √x/g
T=2X 3.14 (√0.001)( we take it negligible)
so T= 6.28 seconds