The force, and therefore the weight, and therefore the profit by weight, is inversely proportional to the square of the distance. Gary probably understated that distance. The larger the distance, the less the profit.
No, the larger the difference between the squares of the two distances, the greater the profit. Perhaps an illustration of the situation considered will be helpful here:
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The mass of the gold (m) and the moon (M) should hopefully not change throughout the experiment. The difference between R1 and R2 will be affected by 2 things: the shape of the orbit (which I'm ignoring here, since it doesn't have to do with your misled objection) and the radius of the earth. In the case of R2, including the radius of the earth
diminishes the distance between the gold and the moon, but in case of R1 it
increases it. As such, the difference between the distances becomes greater, and thus the difference between the two results of
increases (note that G, M and m are constant, so R
2 is the only relevant factor) also increases, and thus the difference between weights measured increases. This, in our hypothetical scenario (we calculate price from weight, we buy at the lowest weight available and sell at the highest weight available) means that including the earth's radius in the calculation does indeed increase our calculated profit. Omitting the radius of the earth would place the piece of gold right at its core, thus reducing the difference between the two distances, and, conversely, the cost.