# Rearrange the following equation to make G the subject where r>0 and M>0: 8pi^2/G^3M = T^2/r^3?

##### 1 Answer

Nov 28, 2017

#### Explanation:

#"one way is to use the method of "color(blue)"cross-multiplication"#

#• " given "a/b=c/drArrad=bc#

#(8pi^2)/(G^3M)=(T^2)/(r^3)#

#rArrG^3MT^2=8pi^2r^3#

#"divide both sides by "MT^2#

#(G^3cancel(MT^2))/cancel(MT^2)=(8pi^2r^3)/(MT^2)#

#rArrG^3=(8pi^2r^3)/(MT^2)#

#color(blue)"take the cube root of both sides"#

#root(3)(G^3)=root(3)((8pi^2r^3)/(MT^2))#

#rArrG=root(3)((8pi^2r^3)/(MT^2))to(T!=0)#