2.4 Variation of acceleration due to gravity $(g)$ due to rotation of earth

1.1 Characteristics of gravitational force
1.2 Universal gravitational constant
1.3 Principle of superposition of gravitation
1.4 Gravity
1.5 Acceleration due to gravity
1.6 Relation between $g$ and $G$

2.1 Variation of acceleration due to gravity $(g)$ due to shape of the earth
2.2 Variation of acceleration due to gravity $(g)$ due to altitude
2.3 Variation of acceleration due to gravity $(g)$ due to depth
2.4 Variation of acceleration due to gravity $(g)$ due to rotation of earth

3.1 Gravitational field due to a point mass
3.2 Gravitational field due to a uniform solid sphere
3.3 Gravitational field due to a uniform spherical shell
3.4 Gravitatioal field due to a uniform circular ring at a point on its axis

4.1 Gravitational potential due to a point mass
4.2 Gravitational potential due to a uniform solid sphere
4.3 Gravitational potential due to a uniform thin spherical shell
4.4 Gravitational potential due to a uniform ring at a point on its centre
4.5 Relation between gravitational field and gravitational potential

5.1 Gravitational potential energy for a system of particles
5.2 Gravitational potential energy of a body on earth's surface

6.1 Orbital speed of satellite
6.2 Time period of a satellite
6.3 Angular momentum of a satellite
6.4 Energy of a satellite
6.5 Types of satellite
6.6 Binding energy
6.7 Escape velocity
6.8 Weightlessness

7.1 Kepler's first law
7.2 Kepler's second law
7.3 Kepler's third law

Consider a body at a point with latitude on the surface of the earth.

Let,

$R$: Radius of the earth

$\omega $: Angular velocity of the earth about its own axis

$$\overrightarrow g ' = \overrightarrow g + {\omega ^2}\overrightarrow r $$ $$\left| {\overrightarrow g '} \right| = \sqrt {{g^2} + {r^2}{\omega ^4} + 2g{\omega ^2}r\cos \left( {\pi - \theta } \right)} $$ $$\left| {\overrightarrow g '} \right| = \sqrt {{g^2} + {r^2}{\omega ^4} - 2g{\omega ^2}r\cos \theta } $$

As, $\left( {\omega < < 1} \right)$. So, $\left( {{\omega ^2} \approx 0} \right)$.

So, $$g' = {\left( {{g^2} - 2gr{\omega ^2}\cos \theta } \right)^{\frac{1}{2}}}$$

Also, $$r = R\cos \theta $$ So, $$g' = {\left( {{g^2} - 2gR{\omega ^2}{{\cos }^2}\theta } \right)^{\frac{1}{2}}}$$$$g' = g{\left( {1 - \frac{{2R{\omega ^2}{{\cos }^2}\theta }}{g}} \right)^{\frac{1}{2}}}$$ Using binomial expansion, $$g' \approx g\left( {1 - \frac{{R{\omega ^2}{{\cos }^2}\theta }}{g}} \right)$$$$g' = g - R{\omega ^2}{\cos ^2}\theta $$

The value of acceleration due to gravity is minimum at the equator and maximum at the poles. Thus $g$ increases as we move from equator to the pole.