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The acceleration of free fall is ____ $m / s^{2}$ .

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Detailed Solution

The acceleration of free fall is

9.8

m / s^{2}

The gravitational force on an object of mass

m

due to the Earth is given by

F = \frac{Gm m_{e}}{r^{2}}

…(1)
Where

F, G, m_{}, m_{e},

and

r

represents the force, universal gravitational constant, the mass of object, the mass of the earth and distance, respectively.
Therefore, the acceleration produced in that object will be

\Rightarrow a = \frac{F}{m}

…..(2)
From equations (1) and (2),

\Rightarrow a = \frac{Gm m_{e}}{r^{2}} \times \frac{1}{m}

\Rightarrow a = \frac{G m_{e}}{r^{2}}

….(3)
We know that

G = 6.7 \times 10^{- 11}

And

m_{e} = 6 \times 1 0^{24}

Distance

r = 6.4 \times 1 0^{6}

By putting the above values in equation (3),

\Rightarrow a = \frac{(6.7 \times 10^{- 11}) (6 \times 10^{24})}{{(6.4 \times 10^{6})}^{2}}

\Rightarrow a = 9.8 m / s^{2}

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