# Radiant pressure by sun

## Homework Statement

I've solved part (a), but part (b) puzzles me as im 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

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Curious3141
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## Homework Statement

I've solved part (a), but part (b) puzzles me as im 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.

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..