Radius of a planet with the same density as Neptune

[PhyiscsisNeat]

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 (5.97 x 10^24 kg). It is believed to consist of a mixture of rock and ice, similar to Neptune. If this planet has the same density as Neptune (1.76g/cm^3), what is the radius expressed a) in kilometers, b) in miles"

What I know is that the mass is ~5.97 x 10^24(5.5) = 3.2835 x 10^26 and I tried to convet to grams and got 3.2835 x 10^29 grams. Density is 1.76 g/cm^3[/B]

Homework Equations

The question didn't provide equations so I resorted to google. I know that density = mass/volume and I found the equation for the volume of a sphere to be V = 4/3pir[/B]

The Attempt at a Solution

I started by multiplying the mass of the Earth by 5.5 and got 3.2835 x 10^26 kg, I then converted to grams for division to be compatible with the density of Neptune. So in grams I believe it is 3.2835 x 10^29.

Rearranging d=m/v I get v=m/d. I divide 3.2835 x 10^29 g by 1.76 g/cm^3 and get 1.865625 x 10^29 g/cm^3.

I then take this value for volume and plug it into V=4/3pir.

1.865625 x 10^29 = 4/3pir (this is the same as V = 4pir/3 ??)

I multiply both sides by 3 to remove the denominator of 3 and get 5.596875 x 10^29 = 4pir

I divide both sides by 4pi to isolate r and I get 4.453851611 x 10^28 = r

I apply a cube root to the radius to get 1.645 x 10^29 cm

This is where I get confused. The radius I am getting is no where near the radius of Neptune, despite having the same density? 1.645 x 10^29 cm is an absurd number, so I consulted with wolfram alpha and got a more reasonable number of 3.544 x 10^9.

My calculator says that the cube root of 4.453 x 10^28 is 1.645 x 10^29. Wolfram alpha says that it is 3.544 x 10^9.

Just because the planet has the same density as Neptune, does that mean the radius should be in the same ballpark? My current answer for the radius in km is 35,440,000.

I think I am comfortable enough with algebra to more terms around in these equations to get what I want, but the numbers involved are screwing me up? It's very possible that I made mistakes converting as well. Not sure what the deal is with the differences in cube roots from my calculator vs wolfram alpha.

Any help is GREATLY appreciated.[/B]

 

[PhyiscsisNeat]

Current answer in kilometers is 3,544,000 km.

 

[haruspex]

V = 4/3pir

$\frac 43\pi r^3$

got 3.2835 x 10^26 kg

Try that again.

 

[QuantumQuest]

I started by multiplying the mass of the Earth by 5.5 and got 3.2835 x 10^26 kg

Check this again.

I then take this value for volume and plug it into V=4/3pir

The formula reads $V = \frac{4}{3}\pi r^3$

EDIT: I say the same things with haruspex, we just did the post almost at the same time and had no time to see it.

 

[PhyiscsisNeat]

Okay, I multiplied 5.97 x 10²⁴ by 5.5 and I'm still getting 3.2835 x 10²⁶ :(

Edit: Thank you for the correct formula.

 

[QuantumQuest]

Okay, I multiplied 5.97 x 1024 by 5.5 and I'm still getting 3.2835 x 1026

It is $32.835 \times 10^{24}$. Now, do you see the mistake?

 

[PhyiscsisNeat]

It is $32.835 \times 10^{24}$. Now, do you see the mistake?

Yes, I think so. The order of magnitude stays the same (the same as the Earth) and I'm just multiplying 5.97 by 5.5 to get 32.835. Perhaps I should read the instructions for the calculator. If you're multiplying something that is expressed in scientific notation by another number, is it proper to multiply the two numbers first and then add the 10ⁿ after? Because multiplying 5.97 x 10²⁴ in my calculator by 5.5 is giving me 10²⁶.

 

[QuantumQuest]

Yes, I think so. The order of magnitude stays the same (the same as the Earth) and I'm just multiplying 5.97 by 5.5 to get 32.835. Perhaps I should read the instructions for the calculator. If you're multiplying something that is expressed in scientific notation by another number, is it proper to multiply the two numbers first and then add the 10n after? Because multiplying 5.97 x 1024 in my calculator by 5.5 is giving my 1026.

Yes, you can multiply $5.97$ by $5.5$ first and then just write "times" the power of $10$ you have. Then, if you want to write it as $3.2835$ you must increase the exponent of the power of $10$ accordingly.

 

[haruspex]

multiplying 5.97 x 10²⁴ in my calculator by 5.5 is giving me 10²⁶.

What would multiplying 5.97 x 10²⁴ by 10 give? Should multiplying by 5.5 instead give more or less?

 

[PhyiscsisNeat]

Okay, I am back and happy to report that I was having some calculator technical difficulties. Here is my current attempt. Thank you for the responses so far.

First I multiply 5.97 x 10²⁴ kg by 5.5 to get 3.2835 x 10²⁵ kg. I multiply 3.2835 x 10²⁵ by 1000 to get 3.2835 x 10²⁸ grams.

Now that I have the mass in grams, I rearrange d=m/v to get v=m/d and plug the mass and density into get a volume of 1.8657 x 10²⁸.

I set v = 1.8657 x 10²⁸ into the equation V=4/3pir³ and multiply both sides by 3 to remove the denominator and I get...

5.5971 x 10²⁸/4pi = r³. I now have 4.454 x 10²⁷ = r³.

I apply a cube root to get r = 1.645 x 10⁹ cm and I divide that by 1000 to convert to km and I am left with a radius of 1,645,000 km.

Did I screw anything up?

 

[QuantumQuest]

Okay, I am back and happy to report that I was having some calculator technical difficulties. Here is my current attempt. Thank you for the responses so far.

First I multiply $5.97 \times 10^{24}$ kg by 5.5 to get $3.2835 \times 10^{25}$ kg. I multiply $3.2835 \times 10^{25}$ by 1000 to get $3.2835 \times 10^{28}$ grams.

Now that I have the mass in grams, I rearrange d=m/v to get v=m/d and plug the mass and density into get a volume of $1.8657 \times 10^{28}$.

I set v = $1.8657 \times 10^{28}$ into the equation $V = \frac{4}{3}\pi r^3$ and multiply both sides by 3 to remove the denominator and I get...

$\frac{5.5971 \times 10^28}{4pi} = r^3$. I now have $4.454 \times 10^{27} = r^3$.

I apply a cube root to get $r = 1.645 \times 10^9$ cm and I divide that by 1000 to convert to km and I am left with a radius of 1,645,000 km.

Did I screw anything up?

Obviously the radius you found can't be correct. Be careful at your conversion from $cm$ to $km$ for the radius $r$ in the end.
Also, you're slightly off to the volume: it is $1.865625 \times 10^{28} cm^3$.
As a final comment, you can convert the density in S.I. units right off the bat and work with S.I units, making the whole thing a little easier.

Note: I used some latex inside the quote in order things to be more clear.

 

[PhyiscsisNeat]

Hey, okay, good catch on the conversion, I derped :(

Looking at my metric system sheet, centimeters are 10^-2 and kilometers are10³. I divided my answer in centimeters by 1000 to try to put it into km. Should I have divided by 10,000 instead? If I do that, that gives me a radius of 16,453.18 km.

I am newb :(

 

[QuantumQuest]

Should I have divided by 10,000 instead?

No, you must divide by $10^5$. In order to understand it think how many $cm$ is 1 $km$. The answer you give is correct in order but to get there you have to divide by $10^5$. Check again.

 

[PhyiscsisNeat]

Ahhh this is frustrating. Thank you for your time.