If I have two spheres of a known volume $V$ and radius $r$ and I make one bigger sphere out of them, what will the new sphere's radius be?
For example:
Sphere $1$: $r_1=2$, $V_1=8$.
Sphere $2$: $r_2=2$, $V_2=8$.
New sphere: $r_n=?$, $V_n=16$.
Edit:
I need to use it in programming. The units are the same for all measurments and don't matter to me. I found out that the ratio is 1:8 but I don't know how to apply this. How does it work the other way if I divide the sphere in half? What if the spheres are different volume?
My work:
$r_n$=$r_1$+1/8$r_2$
Then
$v_n$ ≠ $v_1$+$v_2$
Code:
if(p1.r>p2.r)
{
p1.r=p2.r/8 + p1.r;
}
else
{
p1.r=p1.r/8 + p2.r;
}
For anyone interested visual help:
http://dharman.eu/planets/index.html
Best Answer
A sphere with radius $r$ will have volume $V=(4/3)\pi r^3$. Evidently you're using different units for volume and radius, but it turns out that it doesn't matter for this problem, so we'll say that $V=kr^3$. Suppose you have two spheres with radii $r_1, r_2$. Their combined volumes will be $$ k(r_1)^3+k(r_2)^3 = k((r_1)^3+(r_2)^3) $$ If they are combined to make a big sphere with radius $R$ we'll have $$ k((r_1)^3+(r_2)^3) = kR^3 $$ and so we'll have $$ R^3 = (r_1)^3+(r_2)^3 $$ and so, taking third roots, $$ R=((r_1)^3+(r_2)^3)^{1/3} $$ In your example, with $r_1=r_2=2$, you'd have the radius of the new sphere $$ R=(2\cdot 2^3)^{1/3}=2^{1/3}\cdot 2\approx 2.5198 $$