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Sagot :
Answer:
For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.
Explanation:
The force of gravity comes from Newton's second law with the force the universal attraction
F = ma
F = [tex]G \frac{m_1 M}{(R_e +h)^2}[/tex]
we substitute
[tex]G \frac{m_1 M}{ (R_e+ h)^2}[/tex] = m₁ a
where Re is the radius of the Earth 6.37 106 m
a = [tex]G\frac{M}{R_e^2} \ ( 1 + \frac{h}{R_e})^{-2}[/tex]
In general, the height is much less than the radius of the earth, therefore the term ha / Re is very small and we can use a series expansion leaving only the first fears.
(1 + x)⁻² = 1 -2x + [tex]\frac{2 \ 1}{2!}[/tex] x²
we substitute
a = g₀ ([tex]1 - 2 \frac{h}{R_e}[/tex] )
with
g₀ = [tex]G \frac{M}{R_e^2}[/tex]
let's launch the expression.
* For small height compared to the radius of the earth we can neglect the last term
g = g₀
* For height comparable to the radius of the Earth
g = g₀ [tex](1 - \frac{2h}{Re} )[/tex]
We see that the acceleration of gravity is decreasing.
For which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.
The student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.
The given problem is based on the concept of gravity and gravitational force. The force of gravity comes from Newton's second law with the force the universal attraction as,
F = ma
[tex]F=G\dfrac{mM}{(R+h)^{2}}\\\\\\ma = G\dfrac{mM}{(R+h)^{2}}[/tex]
Here, a is the linear acceleration, m is the mass of object, M is the mass of Earth, R is the radius of Earth and h is the height from where the objects will be dropped. Then,
[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}[/tex]
In general, the height is much less than the radius of the earth, therefore the term h/ R is very small, hence can be neglected.
[tex]a = \dfrac{GM}{R^{2}}\\\\a=g = \dfrac{GM}{R^{2}}[/tex]
g is the gravitational acceleration.
For small height compared to the radius of the earth we can neglect the last term as,
a = g
And for the height comparable to radius of Earth,
[tex]a = \dfrac{GM}{R^{2}} \times(1+h/R)^{-2}\\\\a=g \times(1+h/R)^{-2}[/tex]
Clearly, the acceleration of gravity is decreasing, for which the reasoning of the boy is correct for small heights, but as height increases his analysis is not correct.
Thus, we can conclude that the student's reasoning gone wrong when the analysis is undertaken for the increasing heights, to drop the object.
Learn more about the gravitational force here:
https://brainly.com/question/15647838
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