Question

If earth is supposed to be a sphere of radius R, if g₃₀ is value of acceleration due to gravity at lattitude of 30° and g at the equator, the value of g-g₃₀ is(a) $\frac{1}{4}\omega^{2}R$(b) $\frac{3}{4}\omega^{2}R$(c) $\omega^{2}R$(d) $\frac{1}{2}\omega^{2}R$

Answer 1

Answer:

B) 3/4 Rω

²

Explanation:

Radius of the earth = R (Given)

Value of acceleration = g30 (Given)

Angle of latitude = 30°(Given)

Acceleration due to gravity at latitude  λ  is expressed as -

g' = g−Rω²  cos²λ

Thus at the degree of 30° it will be -

g30  =  g−Rω²  cos²30°

= g-3/4 Rω²

Thus the value of g-g₃₀ is = 3/4 Rω²

Answer 2

Answer:

1/4 ω^2 R

Explanation:

Please refer to the picture for the detailed solution.