If earth-moon distance is 3.8×10^5km,calculate the time (in days)it takes the moon to circle the earth .Mass of the earth is 5.98×10^24kg.

Suppose v is the velocity of the moon. Now centrifugal force will be $\frac{m{v}^2}{r}$. Where m is the mass of moon. Now this force will be equal to the force of attraction between earth and moon and hence we have
$$\begin{gathered} \frac{GMm}{r^2}=\frac{m{v}^2}{r} \\ {v}^2=\frac{GM}{r} \\ v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{6.6\times {10}^{-11}\times 5.98\times {10}^{24}}{3.8\times {10}^8}}=1019.1 \text{m/s} \end{gathered}$$
Here M is the mass of earth and $r=3.8\times {10}^5 \text{km}=3.8\times {10}^8 \text{m}$ is the distance between earth and moon.
Now the total path moon need to travel to complete one revolution is $2\mathrm{\pi r}=2\mathrm{\pi }\times 3.8\times {10}^8=2.4\times {10}^8$.
So time taken is $t=\frac{2.4\times {10}^8}{1019.1}=2.3\times {10}^6\text{ s }= \frac{2.3\times {10}^6}{86400}=27.1 \text{day}.$