Respuesta :
Answer:
Approximately 9.796 m/s^2
Explanation:
The equation for the force that causes the acceleration can be written as.
Fg = G * M * m ÷ d^2
where
d is the distance from the center of the earth to the falling object. This distance is equal to the sum of the radius of the earth and the height of the object.
a = Fg/m
a = G * M ÷ d^2
G = 6.67 * 10^-11
M = 5.98 * 10^24
G * M = 6.67 * 10^-11 * 5.98 * 10^24 = 3.98866 * 10^14
9.8 = 3.98866 * 10^14 ÷ d^2
d^2 = 3.98866 * 10^14 ÷ 9.8
d = √(3.98866 * 10^14 ÷ 9.8)
This is approximately 6.38 * 10^6 meters.
For the top of the building, d = 1300 + √(3.98866 * 10^14 ÷ 9.8)
a = 3.98866 * 10^14 ÷ [1300 + √(3.98866 * 10^14 ÷ 9.8)]^2
Which is approximately 9.796 m/s^2.
When gravity is the sole force on the object then it undergoes free fall. The free fall at the top of the 1300 m tall tower will be [tex]9.796 \;\rm m/s^{2}[/tex].
What is free fall?
Free-fall acceleration is the downward force applied on the object when the only force present is the gravitational force.
The force can be given as,
[tex]\rm F_{g} = \dfrac{G \times M \times m } {d^{2}}[/tex]
Where, distance from the centre = d
And,
[tex]\begin{aligned}\rm a &=\rm \dfrac{F_{g}}{m}\\\\&= \rm \dfrac{G \times M} {d^{2}}\end{aligned}[/tex]
Here,
G = [tex]6.67 \times 10^{-11}[/tex]
M = [tex]5.98 \times 10^{24}[/tex]
Substituting values in the equation:
[tex]\begin{aligned} 9.8 &= \rm \dfrac{(6.67 \times 10^{-11}) \times ( 5.98 \times 10^{24})}{d^{2}}\\\\\rm d^{2} &=\sqrt{(\dfrac{3.98866 \times 10^{14}}{ 9.8})} \\\\&= 6.38 \times 10^{6}\;\rm meters\end{aligned}[/tex]
For the top of a 1300 m -tall tower, [tex]\rm d = 1300 + \sqrt{(\dfrac{3.98866 \times 10^{14} }{ 9.8}) }[/tex]
And,
[tex]\begin{aligned}\rm a &= \dfrac{3.98866 \times 10^{14} }{[1300 + \sqrt{(\dfrac{3.98866 \times 10^{14} }{ 9.8})} ]^{2} }\\\\&= 9.796 \rm \; m/s^{2}\end{aligned}[/tex]
Therefore, the acceleration is [tex]9.796 \;\rm m/s^{2}[/tex].
Learn more about free fall here:
https://brainly.com/question/6035428
