How to find the radius of a planet given its internal composition?

Consider a planet very similar to Earth, with the same radius and the same mass. In a very simplified model of its interior, 70 % of its mass is magnesium silicate and the remaining 30 % is iron. This latter, being denser than the first, it will sink towards the center, forming the planetary core.

Consider another planet with the same mass as the previous one, but now, iron represents 50 % of its mass and the other 50 % corresponds to magnesium silicate. It is logical to think that this planet will be smaller than the previous one, since iron is denser than magnesium silicate and, therefore, a certain amount of iron occupies a smaller volume than the same amount of magensium silicate. But the question is: how small would it be?

What i did was:

If it is known that the nucleus is a sphere of 100 % iron, its density would be the same as that of iron, and its mass, that is half the mass of the planet, is also known. Knowing the density and mass, it can be found the volume:

$d=\frac{m}{v}$

Solving for $v$:

$v=\frac{m}{d}$

I followed the same procedure to find the volume of the mantle. Then, by adding both volumes, the volume of the planet is found, and with him, the radius:

$v=\frac{4}{3}Ï€r^{3}$

Solving for $r$:

$r=\frac{\sqrt[3]{6vπ^{2}}}{2π}$

By following this procedure, the radius of the planet turned out to be 45 % larger than the radius of the original planet, which is illogical. I would like to know: is this the procedure to follow in this case or am I doing something wrong?

posted over 5 years ago

URIZEN‭

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