In January 2006, astronomers reported the discovery of a planet comparable in size to the earth orbiting another star and having a mass of about 5.5 times the earth's mass.

It is believed to consist of a mixture of rock and ice, similar to Neptune. Take mass Earth=5.97x10^24 kg and radius Earth=6.38x10^6 m. If this planet has the same density as Neptune (1.76 g/cm^3), what is its radius expressed in kilometers?

R = 5.28 103 km

Explanation:

The definition of density is

ρ = m / V

V = m / ρ

Where m is the mass and V the volume of the body

The volume of a sphere is

V = 4/3 π r³

Let's replace

4/3 π r³ = m / ρ

R = ∛ ¾ m / ρ π

The mass of the planet is

M = 5.5 Me

R = ∛ ¾ 5.5 Me / ρ π

Let's reduce the density to SI units

ρ = 1.76 g / cm³ (1 kg / 10³ g) (10² cm / 1 m) ³

ρ = 1.76 10³ kg / m³

Let's calculate

R = ∛ ¾ 5.5 5.97 10²⁴ / (1.76 10³ pi)

R = ∛ 0.14723 10²¹

R = 0.528 10⁷ m

R = 0.528 104 km

R = 5.28 103 km