Question

fill in the blank question
the higher a spacecraft’s orbit, the
the magnitude of free-fall acceleration, because of the
square relationship between ( r_e ) and ( r ).
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Explanation:

Free-fall acceleration follows the inverse square law from Newton's law of gravitation: $g \propto \frac{1}{r^2}$, where $r$ is the distance from the Earth's center. A higher orbit means a larger $r$, so $g$ decreases.

Answer:

smaller; inverse