Positioning the space station to a position of [tex]1.2\cdot R_{E}[/tex] from Earth represents a higher free-fall acceleration.
According to Newton's law of gravitation, the gravitational force ([tex]F[/tex]), in newtons, is inversely proportional to the square of the distance between the center of the Earth and the space station ([tex]R[/tex]), in meters. In addition, the gravitational force is directly proportional to the free-fall acceleration ([tex]g[/tex]), in meters per square second, experimented by the space station.
[tex]g \propto \frac{1}{R_{E}^{2}} [/tex] (1)
Thus, positioning the space station to a position of [tex]1.2\cdot R_{E}[/tex] from Earth represents a higher free-fall acceleration. [tex]\blacksquare[/tex]
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