<p>Is the gravitational field through the central axis = 0? Here's the question:</p>
<p>The International Space Station is currently under construction. Eventually, simulated earth gravity may become a reality on the space station. What would the gravitational field though the central axis be like under these conditions?</p>
<p>A. zero
B. g/4
C. g/2
D. 3g/4
E. 1g</p>
<p>Answer is A. But shouldn't the answer be infinity because if R->0 in this equation: F= GMm/r^2, then it tends toward infinity?</p>
<p>The key phrase is “simulated gravity,” produced by putting a spin on the space station. This force, though interpreted by a person as a gravitational attraction toward the outer hull, is not the effect of gravitation attraction of two bodies, but an effect of centrifugal force acting on the person. You need to use a different formula to calculate this which can be found on wikipedia at [Artificial gravity](<a href=“Artificial gravity - Wikipedia”>Artificial gravity - Wikipedia). The closer you get to the central axis, the weaker this force, until at the central axis, there is no centrifugal force, only rotational acceleration.</p>
<p>Question is indeed very poorly worded. Centrifugal “force” is an effect of centripetal force on an object. </p>
<p>Another way of interpreting the question is the answer is 0 because there is no “gravity field” on a spinning space station in free fall any more than there is on any object in free fall (orbit) around another object. </p>
<p>I assumed this question is not from an actual exam.</p>
<p>Let me word this ridiculous question in a way that makes more sense:</p>
<p>The International Space Station is currently under construction. Eventually, simulated earth gravity may become a reality on the space station. What would be the gravitational acceleration towards the center for someone walking on it?</p>
<p>The trick here is that there is no gravitational acceleration, it is radial acceleration.</p>