Question

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?

Answer 1

Answer:

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