Acceleration of sun

If the centripetal acceleration is $a$ , the tangential speed is $v$ , and the radial distance is $r$ , we have $a = \dfrac {v^2}{r}.$ We also know that, if $T$ is the period, $v = \dfrac {2\pi r}{T}$ , so $\begin{aligned} a &= \dfrac {\dfrac {4\pi^2 r^2}{T^2}}{r} \\&= \dfrac {4\pi^2 r}{T^2}. \end{aligned}$ Substituting the given numbers, we have $a = \boxed {1.62 \times 10^{-10} \, \text{m/s}^2}$ .

$a=\frac{v^2}{r}=\frac{(\frac{2\pi r}{T})^2}{r}=\frac{4\pi^2r}{T^2}=\frac{4\pi^2(2.55×10^{20})}{(250×10^{20}×365×24×60×60)^2}=1.62×10^{-10}m/s^2$

We use the formula $a = \frac{v^2}{r}$ . We first find $v$ in $ms^{-1}$ . Let $r = 2.55E20$ . Then, $v = \frac{2\pi r}{250000000 \cdot 365 \cdot 24 \cdot 60 \cdot 60} = 203223.2691$ . Thus, subtituting this into the formula gives $a = \frac{v^2}{r} = 1.62E-10$ .

Use the equation $a=\frac{v^{2}}{r}=\frac{4\pi^{2}}{T^{2}}.r$ in which r is the distance from the sun to the center Milky Way and T is the period of sun's move around the center of our galaxy, with detail number given in the problem.

ACELERATION IS rW2

The acceleration of the sun toward the center of the galaxy is equivalent to its circular velocity squared over it's distance from the center of the galaxy. Circular velocity = (2 pi 2.55E20)/250 million years = 203223 m/s Acceleration hence follows as 1.62E-10