Question

If you dropped a ball while standing on the surface of the sun at what rate would it accelerate toward the ground

Answer 1

The expression for the gravitational acceleration is:

$$g = G\dfrac{M}{R^2}$$

where $G = 6.674\times10^{-11} N\cdot m^2/kg^2$ is the gravitational constant, $M = 1.989\times 10^{30}kg$ is the mass of the Sun and $R = 0.7\times 10^{9} m$ is the radius of the Sun. Thus, obtain:

$$g = G\dfrac{M}{R^2} = 6.674\times10^{-11} \dfrac{1.989\times 10^{30}}{ 0.7^2\times 10^{18}} \approx 270.9m/s^2$$

Answer. 270.9 m/s^2.