Finding the Density of a given object

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Homework Help Overview

The discussion revolves around calculating the density of Earth using its mass and volume. The original poster presents a method involving the formula for density and the volume of a sphere, but expresses uncertainty about their calculations and the resulting answer.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate density by dividing mass by volume, using the radius of Earth to find volume. They question the correctness of their approach and results. Other participants raise concerns about unit conversions and the clarity of the calculations presented.

Discussion Status

Participants are actively engaging with the original poster's calculations, suggesting that clarity in unit usage is necessary. There is a focus on verifying the conversion of units and ensuring that all steps in the calculation are clearly stated. No consensus has been reached yet.

Contextual Notes

The original poster initially omitted the units for mass, which led to confusion. There are indications that the problem may involve complexities related to unit conversions, particularly from cubic kilometers to cubic meters.

iJpawn
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The density of an object equals its mass divided by its volume. The mass of Earth is 6x10e+24 and its radius is 4000 miles (1.61km = 1mile). What is the density of the Earth in kg/m^3?V=4/3pi(r)^3. D=M/VConverting 4000 to Km, I get 6400. After plugging it into the Volume equation, I end up with approximately 1x10e+12. I divide 6x10e+24 by 1x10e+12, and my result is 6x10e^12. Is this the correct way to do the given problem? My answer sheet from my professor does not state this as any of the answers, and I can't think of any other way to approach this equation.
 
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The answer should be in kg/m^3.
 
iJpawn said:
The density of an object equals its mass divided by its volume. The mass of Earth is 6x10e+24 and its radius is 4000 miles (1.61km = 1mile). What is the density of the Earth in kg/m^3?V=4/3pi(r)^3. D=M/VConverting 4000 to Km, I get 6400. After plugging it into the Volume equation, I end up with approximately 1x10e+12. I divide 6x10e+24 by 1x10e+12, and my result is 6x10e^12. Is this the correct way to do the given problem? My answer sheet from my professor does not state this as any of the answers, and I can't think of any other way to approach this equation.
Your calculation is hard to follw if you do not state the units at every point. The answer is much too high.
I suspect you erred in converting cu km to cu m.
What units are the given mass in?
 
The mass of Earth is given in 6x10e+24kg, sorry for not posting it earlier!
 
iJpawn said:
The mass of Earth is given in 6x10e+24kg, sorry for not posting it earlier!
Ok, but repost your calculation showing units at every step. Omit no steps.
 

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