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This question already has an answer here:

N[(1 - 2*6.674*6*10^13/(6371000*299792458^2))^(0.5), 20]

1.

What I expect:

0.99999999930065988290

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marked as duplicate by Lukas Lang, Bob Hanlon, Edmund, anderstood, m_goldberg Jan 5 at 1:21

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  • 2
    $\begingroup$ N[(1 - 2*(6.674 // Rationalize)*6*10^13/(6371000*299792458^2))^(0.5 // Rationalize), 20] will provide your expected result $\endgroup$ – Bob Hanlon Jan 4 at 1:05