How to find Density/ Radius of planetesimals

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    Density Radius
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Homework Help Overview

The discussion revolves around calculating the density and radius of ice-rich planetesimals, specifically those that are hypothesized to be significantly more massive than Earth. The original poster seeks assistance in determining the radius of such a planetesimal given its density.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the mass of Earth multiplied by ten and applying the formula for density and the volume of a sphere to find the radius. There are questions about isolating the radius in the equations.

Discussion Status

Some participants have provided guidance on algebraic manipulation and have confirmed the approach being taken. There is an ongoing exploration of how to rearrange the equations to isolate the radius.

Contextual Notes

Participants are working under the assumption of a specific density for the planetesimals and are converting units from kilograms to grams as part of their calculations.

Cherrio
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The ice rich planetesimals that formed the cores of the Jovian planets were about 10 times more massive than Earth. Assuming such an ice ball has a density of 2g cm g-3, what would its radius be? How does this compare to Earth's radius.


I need Help =\
 
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What have you tried?
 
I was thinking of multiplying the Earths mass by 10 then using the fomula m=dv

and use the formula for volume of a sphere to get r? IDK
 
Sounds like a good plan.
 
okay so that would be m=dv


10 x 5.97 x 1024 Kg which is 5.97x 1025Kg

then convert from Kg to g 1kg= 1000g

5.97x 1025Kg x 1000 g= 5.97x 1028g
volume of a sphere= (4/3) pi r3

5.97x 1028g= (2g cm-3)((4/3)pir3)

im not to sure how to isolate R
 
You just need a bit of algebraic manipulation. It might help if you use the letters instead of numbers right until the end.
 
ok so it would be


m=d 4/3 pi r3

then m/ d(4/3)pi =r3 ?

then r = cubed root of that anwser?
 
Yes, that looks good.
 

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