The Earth takes one year to orbit the sun along a path of radius $1.5 \times 10^{11}\text{ m}$. At what tangential velocity is the Earth moving?
From the previous question I have worked out the period to be $31,536,000 \text{ s}$.
Here is my attempt:
$$\begin{align}v &= \frac{2\pi r}{t} \\ &= \frac{2\pi (1.5 \times 10^{11}\text{ m})}{31,536,000 \text{ s}} \\ &= 29885.77\text{ ms$^{-1}$}\end{align}$$
Is this correct ?
Yes, this answer is correct. In general, tangential velocity $v_t$ is given by $v_t=\omega{r}$, where $\omega$ is angular velocity, equal to $\frac{2\pi}{t}$ times the number of revolutions recorded in time $t$, which in this case is 1.