Question

Assuming earth to be a sphere of uniform density, the acceleration due to gravity at a point
100 km below, the earth surface is( given r= 6380 X 103)
a) 9.66 m/s2 b) 7.64 m/s2
c) 5.06 m/s2
d) 3.10 m/s2​

Answer 1

Answer:

g below height

d = g' = g[1 – (d / R)]

g' = 9.8[1 / {(100) / (6400)}]

given d = 100km

R = 6400 km

g' = 9.66 m/s2