What is the impact of radiant pressure by the sun on cosmic dust?

  • Thread starter Thread starter unscientific
  • Start date Start date
  • Tags Tags
    Pressure Sun
AI Thread Summary
The discussion revolves around calculating the impact of solar radiant pressure on cosmic dust particles. The initial calculations provided an answer significantly off by orders of magnitude, prompting a reevaluation of the density used in the equations. It was noted that the density of air (1.22 kg/m3) was incorrectly applied instead of a more appropriate density for solid cosmic dust, which is around 2000 kg/m3. After correcting for density, the recalculated answer aligned with expected values. The importance of accurate density in such calculations is emphasized for achieving valid results.
unscientific
Messages
1,728
Reaction score
13

Homework Statement




I've solved part (a), but part (b) puzzles me as I am of 103 orders of magnitude away from the answer.



The Attempt at a Solution



Intensity on dust particle = 1600 W m-2
Earth-Sun Distance = (8*60)(3*108) = 1.44 * 1011m
Mass of Sun = 1.99 * 1030 kg
g-field strength = 6.4 * 10-3 N/kg
Force acting on dust particle = Mg = (4/3)∏R3g

Radiant force = P/c = (IA)/c = (I*∏R2)/c


(I*∏R2)/c = (4/3)∏R3g

R = (3I)/(4cρ*6.4*10-3)

The answer is 2*10-7 m
 

Attachments

  • sun1.jpg
    sun1.jpg
    27.7 KB · Views: 373
Physics news on Phys.org
unscientific said:

Homework Statement

I've solved part (a), but part (b) puzzles me as I am of 103 orders of magnitude away from the answer.

The Attempt at a Solution



Intensity on dust particle = 1600 W m-2
Earth-Sun Distance = (8*60)(3*108) = 1.44 * 1011m
Mass of Sun = 1.99 * 1030 kg
g-field strength = 6.4 * 10-3 N/kg
Force acting on dust particle = Mg = (4/3)∏R3g

Radiant force = P/c = (IA)/c = (I*∏R2)/c(I*∏R2)/c = (4/3)∏R3g

R = (3I)/(4cρ*6.4*10-3)

The answer is 2*10-7 m

Did you include the density of the dust particle in the calculation? You missed out \rho in a couple of steps, but put it into the last formula, so I'm not sure. But if you take the usual density estimates for cosmic dust into account, the result is definitely within the ballpark.
 
Curious3141 said:
Did you include the density of the dust particle in the calculation? You missed out \rho in a couple of steps, but put it into the last formula, so I'm not sure. But if you take the usual density estimates for cosmic dust into account, the result is definitely within the ballpark.

Yes, I took the density of dust as the density of air, 1.22 kg/m3..
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top