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Astronomy course....
- Thread starter iridium42
- Start date
hey guys. need to refresh on some stuff for an astronomy course. an example was given as 2GMr/D^3 = Gm/r^2 to get D = (2M/m)^1/3 r ....so how did we get to D = (2M/m)^1/3 r ? any help/suggestions would be greatly appreciated. thanks!
no real magic here, just algebra
\(\displaystyle \frac{2 G M r}{D^3}=\frac{G m}{r^2}\)
divide by G on both sides
\(\displaystyle \frac{2 M r}{D^3}=\frac{m}{r^2}\)
rearrange things a bit
\(\displaystyle 2 M r^3=D^3 m\)
divide by m and take the cube root
\(\displaystyle \sqrt[3]{\frac{2 M r^3}{m}}=D\)
and pull the r3 out of the cube root
\(\displaystyle r \sqrt[3]{\frac{2 M}{m}}=D\)
.hey guys. need to refresh on some stuff for an astronomy course.
an example was given as 2GMr/D^3 =
Gm/r^2 to get D = (2M/m)^1/3 r .... \(\displaystyle \ \ \ \) You must have grouping symbols around the exponent. And I recommend you put the "r" in front: D = r(2M/m)^(1/3) *
so how did we get to D = (2M/m)^1/3 r ? \(\displaystyle \ \ \ \) See * above.
any help/suggestions would be greatly appreciated. thanks!