11. A particle hanging from a spring stretches by 1 cm at earth's surf

1.	11. A particle hanging from a spring stretches by 1 cm at earth's surface. How much will the same particlestretch the spring at a place 800 km above the earth's surface? Radius of the earth is 6400 km. [0.8cm]
Answer» 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"gKx" = 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

11. A particle hanging from a spring stretches by 1 cm at earth's surface. How much will the same particlestretch the spring at a place 800 km above the earth's surface? Radius of the earth is 6400 km. [0.8cm]

Answer»

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"gKx" = 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

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