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AARGH!! Physics Help!!
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Kaiser
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AARGH!! Physics Help!!
The density of a certain planet varies with radial distance as:
ρ( r)= ρ0(1- αr/R0)
where R0= 14.13×106 m is the radius of the planet, ρ0= 3200.0 kg/m3 is its central density, and α = 0.10. What is the total mass of this planet ?

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(This post was last modified: 04/10/2009 06:17 PM by Kaiser.)
04/10/2009 06:15 PM
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OmegaEye
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RE: AARGH!! Physics Help!!
my guess is to multiply both sides by V®=(4/3)*pi*r^3
where r=R0
so that V®*ρ( r)=m® and the total mass equal m(R0)
blah blah don't feel like writing the rest but you get the idea

04/10/2009 07:08 PM
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xBu
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RE: AARGH!! Physics Help!!
My brain hurts.......
04/10/2009 11:02 PM
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ZiNgA BuRgA
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RE: AARGH!! Physics Help!!
I forgot how to do integral calculus >_>
05/10/2009 12:07 AM
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feinicks
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RE: AARGH!! Physics Help!!
I gave up on physics after high school.... Now I just like the thories and none of the calculations liked with it...

However, that equation is familiar to my memories.... *tears*

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05/10/2009 12:10 AM
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krystabegnalie
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RE: AARGH!! Physics Help!!
in a fun way, I've tried googling the first sentence, and there is a similar question, but different values

http://www.physicsforums.com/showthread.php?t=119878

but meh ive almost forgot how to integrate equations, lol

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05/10/2009 12:20 AM
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1-R
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RE: AARGH!! Physics Help!!
Kaiser Wrote:The density of a certain planet varies with radial distance as:
ρ( r)= ρ0(1- αr/R0)
where R0= 14.13×106 m is the radius of the planet, ρ0= 3200.0 kg/m3 is its central density, and α = 0.10. What is the total mass of this planet ?

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08/10/2009 12:05 AM
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ZiNgA BuRgA
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RE: AARGH!! Physics Help!!
Eh, this actually looks easier than I thought (if I remember my maths right, lol).

p(r) = (3200.0 kg/m3)(1 - 0.1r/(14.13×106 m))
  = 2.136495346446074 kg/m4 × (1 - 0.1r m)
  = 2.136495346446074 kg/m3 - 0.2136495346446074r kg/m3

Mass:
m(r) = p(r) × 4/3 × pi × r3
  = (2.136495346446074 × 4/3 × pi × r3) - (0.2136495346446074 × 4/3 × pi × r4)
  = 8.94933076953892r3 - 0.894933076953892r4

Integrate m(r):
m'(r) = 8.94933076953892r4 ÷4 - 0.894933076953892r5 ÷5
  = 2.23733269238473 r4 - 0.1789866153907784 r5
m'(0) = 0
m'(1497.78) = 2.23733269238473 × 5032596467778.001 - 0.1789866153907784 × 7537722337508535 = 11259592604939.64 - 1349151408946119 = -1337891816341179

Mass = m'(0) - m'(1497.78) = 1337891816341179 kg

Hope I got it right >_>
08/10/2009 01:51 AM
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Assassinator
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RE: AARGH!! Physics Help!!
ZiNgA BuRgA Wrote:Eh, this actually looks easier than I thought (if I remember my maths right, lol).

p(r) = (3200.0 kg/m3)(1 - 0.1r/(14.13×106 m))
  = 2.136495346446074 kg/m4 × (1 - 0.1r m)
  = 2.136495346446074 kg/m3 - 0.2136495346446074r kg/m3

Mass:
m(r) = p(r) × 4/3 × pi × r3
  = (2.136495346446074 × 4/3 × pi × r3) - (0.2136495346446074 × 4/3 × pi × r4)
  = 8.94933076953892r3 - 0.894933076953892r4

Integrate m(r):
m'(r) = 8.94933076953892r4 ÷4 - 0.894933076953892r5 ÷5
  = 2.23733269238473 r4 - 0.1789866153907784 r5
m'(0) = 0
m'(1497.78) = 2.23733269238473 × 5032596467778.001 - 0.1789866153907784 × 7537722337508535 = 11259592604939.64 - 1349151408946119 = -1337891816341179

Mass = m'(0) - m'(1497.78) = 1337891816341179 kg

Hope I got it right >_>

You didn't remember all those formulas right? Otherwise you're way superior to me... I pretty much cannot remember anything from back then...
08/10/2009 02:03 AM
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ZiNgA BuRgA
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RE: AARGH!! Physics Help!!
I remembered Mass = Volume × Density, and from the above, Volume = 4/3 × pi × radius (I remember seeing something like that before)
Didn't remember how to integrate, but I recall it something to do with adding or taking a power from the variable.  Playing with an example, I managed to figure it out though.

The rest is just derivation...

EDIT: or maybe not...  meant to integrate m(r) or p(r) ?  I've really forgotten how to do this T_T
(This post was last modified: 08/10/2009 05:28 AM by ZiNgA BuRgA.)
08/10/2009 04:49 AM
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