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Accretion rate of Earth

by interdinghy
Tags: accretion, earth, rate
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Dec5-12, 02:51 PM
P: 13
1. The problem statement, all variables and given/known data

i) Estimate (in kg/yr) the rate of Earth accreting inter-planetary dust.
Assume a particle number density of n=1e-8/m3, a particle mass density of 5 gm/cc, a particle radius 50 microns and a relative velocity equal to Earth's orbital velocity (so the dust particles are, on average, at rest with respect to the Sun -- which would be expected if dust were deposited by long period comets with a uniform distribution of inclinations.

ii) How many years would it take Earth to increase its mass by 50%?

2. Relevant equations

3. The attempt at a solution

I talked with my professor and he told me to consider a flat circle moving through the cloud, and he said it would gain a certain amount of pass per second, then to consider the volume of the cylinder it would make.

Doing this, I found the area of a circle with the radius of the Earth to be about 1*10^8 km^2 accounting for significant figures. Then to make it a cylinder, I multiplied by an unknown height h in kilometers.

Next I multiplied this by the particle mass density:

[itex](110^{8})(h)km^{3} 5 g/cm^{3}[/itex] and got [itex]510^{23}[/itex] grams

I looked up the speed at which Earth is orbiting the sun, and found it to be [itex]9.39810^{8}[/itex] km/year, so I set [itex]h = 9.39810^{8}km[/itex] and plugged it into what I had.

My final answer was that Earth accrued [itex]510^{29}[/itex] kilograms of mass per year and this is just ridiculous.

I don't think it's right to view the Earth in this problem as a flat circle, and I definitely don't see how it's right to see the volume it creates as a cylinder since the Earth isn't moving in a perfect straight line all the time. Also, he gave two values that I never used, so I'm sure I'm doing something wrong there too.

Can someone maybe clear up what he was trying to tell me or explain a better way to approach this problem?

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Dec5-12, 04:40 PM
P: 11,928
5g/cm^3 is the density of the dust particles, not the density of material in your volume.
The flat circle for earth is fine, but you have to consider the low density of dust particles and their small mass.
Dec5-12, 04:48 PM
P: 378
Quote Quote by interdinghy View Post
[itex](110^{8})(h)km^{3} 5 g/cm^{3}[/itex] and got [itex]510^{23}[/itex] grams
That last should be 51023g/m h.

Your methodology is sound, but your density is wrong. The density you're given is typical for the density of a rock - or a piece of rock dust. There are also two pieces of information you haven't used - a radius and the number of dust grains per cubic meter. Can you put them all together?

Dec5-12, 05:04 PM
P: 13
Accretion rate of Earth

So would it be right to say that since the particle number density is [itex]100cm^{-3}[/itex]:

[itex]\frac{m}{v} = \frac{5g}{1cc}[/itex]

Then there must be 5 grams of mass per 100 particles? That gives .05g per particle which is different but I don't think that completely fixes the answer, and I still can't see where I'd need the radius of the particles.
Dec5-12, 05:08 PM
P: 378
The number density tells you that there are 100 particles per cubic centimeter, yes. But what is the mass of an individual particle? Hint: they're almost certainly pretty close to spherical.
Dec5-12, 05:21 PM
P: 13
then is this right?

[itex]\frac{4}{3}\pi r^{3} = \frac{4}{3}\pi 50\mu m^{3} [/itex]

[itex] = 510^{-7}cc[/itex]

[itex]\frac{5g}{1cc} = \frac{xg}{5.010^{-7}}[/itex]

[itex]x = 5(5.010^{-7})[/itex]

[itex]x = 2.6 10^{-6}g =[/itex] the mass per particle

Because this seems a lot more like what the mass of a particle should be.
Dec5-12, 05:27 PM
P: 378
Looks good. Carry on...
Dec5-12, 05:53 PM
P: 13
Then I'd adjust the mass to [itex]310^{-6}[/itex] and do what I was doing before,

particle number density mass per particle

[itex](\frac{110^{17}}{km^3}) (310^{-6}g)[/itex]

[itex]= \frac{0.3kg}{m^3} [/itex] the amount of mass the Earth gains per [itex]m^{3}[/itex]

Then I multiply this by the volume of the Earth-cylinder

[itex]110^{14}(h)m^{3} \frac{0.3kg}{m^3} [/itex], h becomes unitless

[itex] = 310^{13}(h) kg[/itex]

to make it for a year, I use the orbital speed from my first post, to get

[itex] h = (9.39810^{8})[/itex]

[itex]310^{13} (9.39810^{8}[/itex]) = [itex]310^{22}kg[/itex] rounded for sig digits. This still seems incredibly high?
Dec5-12, 06:11 PM
P: 80
The particle number density is 10/km^3 not 1*10^17. (it was 1e-8/m^3 not 1e8)
Dec5-12, 07:04 PM
P: 13
Quote Quote by HossamCFD View Post
The particle number density is 10/km^3 not 1*10^17. (it was 1e-8/m^3 not 1e8)
This change gave me [itex]3,000,000 \frac{kg}{yr}[/itex] adjusting for sig. figures, which is a much more believable answer.

Doing the math for part ii tells me that it'll take ~[itex]10^{18}[/itex] years for the Earth to increase its mass by 50%, a much less worrisome time frame.

Thank all of you for your time.

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