How do you solve for d in #G=(Gme)/d^2#?

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Answer 1 · 2016-05-13T13:43:37

#d=+-sqrt(me)#

Turn everything upside down giving:

#1/G=d^2/(Gme)" ....an alternative at this stage is" 1/(cancel(G))=d^2/(cancel(G)me)#

Multiply both sides by #Gme#

#(1xxGme)/G=d^2xx(Gme)/(Gme)#

But #(Gme)/(Gme)=1#

#d^2= (cancel(G)me)/(cancel(G))#

Take the square root of both sides

#d=+-sqrt(me)#