Question:
a particle hanging from a spring stretches itby 2 cm at the surface of the earth.at a place 3600 km above the earth surface how much will the particle stretch the spring
Answer:
The particle hanging from the spring experiences gravitational force
Hence, F = mg
This force stretches it by some distance x Hence, F = mg = Kx -------(1)
In (1) x = 1 cm and g = 9.8 m/s2
Now at a place, 3600Km above the surface of the earth, the gravitational force will be different Hence, F1 = mg1 = Kx1 -------(2)
We need to find g1 first. We know
g1 = gR2 / (R + h)2 as R and h are comparable in Km units
where h is height from the surface of earth and R is the radius of the earth
at h = 3600 km , and R = 6400 km
g1 = g (6400)2 / (6400 + 3600)2 = g (0.64)2
mg = Kx ---(1)
mg1 = Kx1 ---(2)
Divide (1) by (2)
g / g1 = x / x1
g / (0.64)2 g = x / x1
As x = 2 cms, x1 = 2 / (0.64)2 = 4.88 cms
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