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Answer in Brief

**Answer the following question in detail.**

Derive an expression for variation in gravitational acceleration of the Earth at with latitude.

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#### Solution

- Latitude is an angle made by the radius vector of any point from the center of the Earth with the equatorial plane.
- The Earth rotates about its polar axis from west to east with uniform angular velocity ω as shown in the figure.

Hence, every point on the surface of the Earth (except the poles) moves in a circle parallel to the equator. - The motion of a mass m at point P on the Earth is shown by the dotted circle with the center at O′.
- Let the latitude of P be θ and the radius of the circle be r.

∴ PO' = r

∠EOP = θ, E being a point on the equator

∴ ∠OPO' = θ

In Δ OPO', cos θ = `"PO'"/"PO" = "r"/"R"`

∴ r = R cos θ - The centripetal acceleration for the mass m, directed along PO' is,

a = rω^{2}

∴ a = rω^{2 }cos θ

The component of this centripetal acceleration along PO, i.e., towards the centre of the Earth is,

`"a"_"r" = "a" cos theta`

∴ `"a"_"r" = "R"omega^2 cos theta xx cos theta`

`"a"_"r" = "R"omega^2cos^2theta` - Part of the gravitational force of attraction on P acting towards PO is utilized in providing this component of centripetal acceleration. Thus, the effective force of gravitational attraction on m at P can be written as,

mg' = mg - mRω^{2}cos^{2}θ

Thus, the effective acceleration due to gravity at P is given as,

g' = g - Rω^{2}cos^{2}θ

Concept: Variation in the Acceleration Due to Gravity with Altitude, Depth, Latitude and Shape

Is there an error in this question or solution?

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