a. What is the free-fall acceleration at the surface of the sun?

b. What is the free-fall acceleration toward the sun at the distance of the earth?

Free-fall acceleration is the acceleration experienced by a freely falling body towards a body with gravitational attraction.

The formula of free-fall acceleration is given by : $g=GMR2.$

Known parameters:

Mass of sun: localid="1648480621147">1.98×1030kg.

Radius of sun localid="1648480643127" $\mathrm{Rs}=6.96\times {10}^8\mathrm{m}.$

Distance from earth to sun: localid="1648480650999" $Rse=1.49\times {10}^{11}\mathrm{m}.$

Universal gravitation constant: localid="1648480663071" $\mathrm{G}=6.67\times {10}^{-11}\mathrm{N}·\mathrm{m}2/\mathrm{kg}2.$

Therefore, the free-fall acceleration at the surface of Sun is :

$$\begin{gathered} gs=GMsRs2 \\ =(6.67\times {10}^{-11}\mathrm{N}·\mathrm{m}2/\mathrm{kg}2)(1.98\times {10}^{30}\mathrm{kg})(6.96\times {10}^8\mathrm{m})2. \\ =273\mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$

The free-fall acceleration at the surface of the sun is$273\mathrm{m}/{\mathrm{s}}^2.$

The free-fall acceleration at the surface of earth due to Sun is :

$$\begin{gathered} \text{gse=GMsRse}2 \\ =(6.67\times {10}^{-11}\mathrm{N}·\mathrm{m}2/\mathrm{kg}2)(1.98\times {10}^{30}\mathrm{kg})(1.49\times {10}^{30}\mathrm{m})2 \\ =6.23\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2. \end{gathered}$$

The free-fall acceleration at the surface of earth due to Sun is$6.23\times {10}^{-3}\mathrm{m}/{\mathrm{s}}^2.$