The Milky Way galaxy rotates about its center with a period of about 200 million yr. The Sun is \(2 \times 10^{20} \mathrm{m}\) from the center of the galaxy. (a) What is the Sun's radial acceleration? (b) What is the net gravitational force on the Sun due to the other stars in the Milky Way?

To find the rotational velocity, we can use the formula for the circumference of a circle and divide that by the period of rotation. The formula is: $$v = \frac{2 \pi r}{T}$$ Where: - v is the rotational velocity - r is the distance from the center of the galaxy \((2 \times 10^{20} \mathrm{m})\) - T is the period of rotation \((200 \mathrm{million\;yr})\) First, we need to convert the period of rotation from years to seconds: $$T(s) = 200 \mathrm{million\;yr} \times \frac{365.25\;\mathrm{days}}{\mathrm{yr}} \times \frac{24\;\mathrm{h}}{\mathrm{day}} \times \frac{60\;\mathrm{min}}{\mathrm{h}} \times \frac{60\;\mathrm{s}}{\mathrm{min}}$$ Now, we can find the rotational velocity: $$v = \frac{2 \pi (2 \times 10^{20} \mathrm{m})}{T(s)}$$

Now that we have the rotational velocity, we can calculate the centripetal acceleration using the following formula: $$a_c = \frac{v^2}{r}$$ Replace v and r with the values from Step 1 and calculate the radial acceleration: $$a_c = \frac{v^2}{2 \times 10^{20} \mathrm{m}}$$

To find the gravitational force, we can use Newton's law of universal gravitation: $$F = \frac{G * m_1 * m_2}{r^2}$$ Where: - F is the force between the objects - G is the gravitational constant \((6.674 \times 10^{-11} \mathrm{N m^2/kg^2})\) - m_1 and m_2 are the masses of the two objects - r is the distance between the objects We can assume that the net gravitational force on the Sun is causing its centripetal acceleration. Therefore, we can equate the gravitational force to the centripetal force acting on the Sun: $$F = m_\mathrm{sun} * a_c$$ Where: - \(m_\mathrm{sun}\) is the mass of the Sun \((1.989 \times 10^{30}\;\mathrm{kg})\) Now, we can find the net gravitational force on the Sun using the radial acceleration (centripetal acceleration) value from Step 2: $$F = (1.989 \times 10^{30}\;\mathrm{kg}) * a_c$$