Given:
The mass of the sun is
[tex]M\text{ = 1.99}\times10^{30}\text{ kg}[/tex]
The mean distance from the earth is
[tex]d=\text{ 1.5}\times10^{11}\text{ m}[/tex]
The mean radius of the sun is
[tex]r=\text{ 6.96}\times10^8\text{ m}[/tex]
To find the freefall acceleration toward the sun at the distance of the earth's orbit.
Explanation:
The total distance from the earth's orbit is
[tex]\begin{gathered} R\text{ = r+d} \\ =6.96\times10^8+1.5\text{ }\times10^{11} \\ =1.50696\text{ }\times10^{11}\text{ m} \end{gathered}[/tex]
The free-fall acceleration due to gravity is given by the formula
[tex]g\text{ =}\frac{GM}{R^2}[/tex]
Here, the universal gravitational constant is
[tex]G\text{ = 6.67 }\times10^{-11}\text{ m}^3\text{ kg}^{-1}s^{-2}[/tex]
On substituting the values, the free-fall acceleration will be
[tex]\begin{gathered} g=\frac{6.67\times10^{-11}\times1.99\times10^{30}}{(1.50696\times10^{11})^2} \\ =5.84\times10^{-3}\text{ m/s}^2 \end{gathered}[/tex]
Final Answer: The freefall acceleration toward the sun at the distance of the earth's orbit is 5.84 x 10^(-3) m/s^2
