Question

The ice-rich planetesimals formed Jovian planet cores, which have masses around 10 times the mass of the Earth. Assuming such an iceball has a density of about twice that of frozen water on Earth, 2 g/cm3 , what would its radius be

Answer 1

To find the radius of a planetesimal with 10 times Earth's mass and a density of 2 g/cm³, convert the density to kg/m³, calculate the volume, and then solve for the radius using the volume of a sphere formula. The result is approximately 3,370 kilometers.

Step-by-step explanation:

To calculate the radius of an ice-rich planetesimal that has a density of about 2 g/cm³ and a mass 10 times that of the Earth, we will use the formula for volume based on mass and density (Volume = Mass / Density), and then solve for the radius using the formula for the volume of a sphere (Volume = 4/3 πr³). Knowing the Earth's mass is approximately 5.972 × 10²⁴ kg, the planetesimal would have a mass of 10 times that, or approximately 5.972 × 10²⁵ kg.

First, we need to convert the density into kg/m3. Since 1 g/cm³ equals 1000 kg/m3, a density of 2 g/cm3 is equivalent to 2000 kg/m³. Now, we can calculate the volume V = Mass / Density = (5.972 × 1025 kg) / (2000 kg/m3) = 2.986 × 10²² m3.

Next, we solve the volume formula for the radius: V = 4/3 πr³ => r³ = V / (4/3 π) => r = ∛(V / (4/3 π)). Plugging our volume into this equation gives us a radius of approximately 3.37 × 10³ kilometers (3,370 kilometers).

Answer 2

The radius of the ice ball is determined as 1.924 x 10⁴ km.

How to calculate the radius of the ice ball?

The volume of the ice ball is calculated by using the following formula.

Volume = mass / density

mass of earth = 5.97 x 10²⁷ g

mass of the ice ball = 10 x 5.97 x 10²⁷ g = 5.97 x 10²⁸ g

Volume of the ice ball = (5.97 x 10²⁸ g) / (2 g/cm³)

Volume of the ice ball = 2.985 x 10²⁸ cm³

The radius of the ice ball is calculated as follows;

V = ⁴/₃ πr³

r³ = (3V) / 4π

r = ∛ (3V/4π)

r = ∛[ (3 x 2.985 x 10²⁸) / 4π]

r = 1.924 x 10⁹ cm

r = 1.924 x 10⁴ km