**1. The problem statement, all variables and given/known data**

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 mEarth=5.97×10^24kg and rEarth=6.38×10^6m.

If this planet has the same density as Neptune (1.76 g/cm3), what is its radius expressed in kilometers?

**2. Relevant equations**

density=m/v

volume(of a sphere)=4/3(pi)r^3

**3. The attempt at a solution**

I calculated the mass of the new planet (n) as (5.5)5.97*10^24kg=3.28*10^25kg

I set the given density of 1.76 g/cm3 equal to the found mass of 3.28*10^22g divided by the formula for the volume of a sphere, shown below:

1.76 g/cm3 = 3.28*10^22g/[4/3(pi)r^3]

I simplify until I reach r^3=4.46*10^21 cm3, then take the cubic root.

The answer I reach is 16456067.27 cm or 164.56 km (165 to three significant figures). I’m not sure why this answer is incorrect.

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