Radius of a planet, given the density

[s.dyseman]

Hello! I'd like to thank everyone in advance for any help I receive!

Homework Statement

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?

Homework Equations

density=m/v

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

 

[BvU]

From 3.28*10^25kg to 3.28*10^22g is a bit strange. 1 kg = 1000 g, so I would expect 10^28 .
Everything else is just fine.

 

[nil1996]

Hi s.dyseman

welcome to PF

You have just gone wrong at conversions.

 

[s.dyseman]

Thanks everyone! I was able to correctly answer the problem after you pointed out my mistake.Thanks again