Solve the following :

Speed of particle at equator=distance/time $=\frac{2 \pi R}{T}$

$=\frac{2 \times 3.14 \times 6400 \times 10^{3}}{24 \times 60 \times 60}$

$=465.1 \mathrm{~m} / \mathrm{s}$

Acceleration of particle $=$

$\mathrm{A}_{\Gamma}=\frac{v^{2}}{R}$

$A_{r}=\frac{(465.1)^{2}}{\left(6400 \times 10^{3}\right)}$

$A_{r}=0.038 \mathrm{~m} / \mathrm{s}^{2}$