Answer:
[tex]g=274\ m/s^2[/tex]
Explanation:
Mass of the Sun, [tex]M=1.99\times 10^{30}\ kg[/tex]
The radius of the Sun, [tex]r=6.96\times 10^8\ m[/tex]
We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :
[tex]g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2[/tex]
So, the value of acceleration due to gravity on the Sun is [tex]274\ m/s^2[/tex].
The acceleration due to gravity, in meters per second squared, on the surface of the Sun is [tex]296.88 \;m/s^2[/tex].
Given the following data:
Gravitational constant = [tex]6.67 \times 10^{-11}[/tex]
To calculate the acceleration due to gravity, in meters per second squared, on the surface of the Sun:
From the law of gravitational force, we have the formula:
[tex]g = \frac{Gm}{r^2}[/tex]
Where:
Substituting the given parameters into the formula, we have;
[tex]g = \frac{6.67 \times 10^{-11} \times 1.99 \times 10^{30}}{(6.69 \times 10^8)^2} \\\\g= \frac{1.33 \times 10^{20} }{4.48 \times 10^{17}} \\\\g=296.88 \;m/s^2[/tex]
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