Problem 48 - University Physics with Modern Physics

Question

A tiny particle of density \(2000 . \mathrm{kg} / \mathrm{m}^{3}\) is at the same distance from the Sun as the Earth is \(\left(1.50 \cdot 10^{11} \mathrm{~m}\right)\). Assume that the particle is spherical and perfectly reflecting. What would its radius have to be for the outward radiation pressure on it to be \(1.00 \%\) of the inward gravitational attraction of the Sun? (Take the Sun's mass to be \(\left.2.00 \cdot 10^{30} \mathrm{~kg} .\right)\)

Step-by-Step Solution

Answer

Answer: The radius of the particle should be approximately \(2.24 \times 10^{-6}\) meters.

Step 1: Calculate the mass of the particle

Using the given density of the particle, we can find the mass using the formula \(m = \frac{4}{3} \pi r^{3} \rho\), where \(r\) is the radius of the particle, and \(\rho\) is the density. Let \(m\) be the mass of the particle and let \(r\) be its radius. $$m = \frac{4}{3} \pi r^{3} (2000)$$

Step 2: Calculate the gravitational force between the Sun and the particle

We can use the equation for the gravitational force, which is given by \(F_G = \frac{GmM}{R^{2}}\), where \(G\) is the gravitational constant, \(M\) is the mass of the Sun, and \(R\) is the distance between the Sun and the particle. Here, \(G = 6.674 \times 10^{-11} \; \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}\), \(M = 2.00 \times 10^{30}\mathrm{~kg}\), and \(R = 1.50 \times 10^{11}\mathrm{~m}\). Substitute the values and replace \(m\) with the equation from Step 1 to get the gravitational force \(F_G\). $$F_G = \frac{6.674 \times 10^{-11} \cdot (\frac{4}{3} \pi r^{3} (2000)) \cdot (2.00 \times 10^{30})}{(1.50 \times 10^{11})^{2}}$$

Step 3: Calculate the outward radiation pressure on the particle

Let \(P_r\) be the outward radiation pressure. We need to find \(P_r\) such that it is equal to \(1.00\%\) of the gravitational force, i.e., \(P_r = 0.01 \times F_G\). Now to find \(P_r\), we first need to find the solar constant \(I\), which represents the incoming solar radiation per unit area at the Earth's distance from the Sun. \(I = 1361 \;\mathrm{W}/\mathrm{m}^2\). The outward radiation pressure is given by the formula \(P_r = \frac{I A}{c}\), where \(A\) is the cross-sectional area of the particle (\(A = \pi r^2\)) and \(c\) is the speed of light (\(c = 3.00 \times 10^8 \;\mathrm{m/s}\)). Substitute the values to get, $$P_r = \frac{1361 \cdot \pi r^2}{3.00 \times 10^8}$$ Now, set \(P_r\) equal to \(0.01 \times F_G\) and solve for radius \(r\).

Step 4: Solve for the radius of the particle

Equating \(P_r\) and \(0.01 \times F_G\), we get: $$\frac{1361 \cdot \pi r^2}{3.00 \times 10^8} = 0.01 \cdot \frac{6.674 \times 10^{-11} \cdot (\frac{4}{3} \pi r^{3} (2000)) \cdot (2.00 \times 10^{30})}{(1.50 \times 10^{11})^{2}}$$ Now, solve for \(r\): $$r \approx 2.24 \times 10^{-6} \;\mathrm{m}$$ So, the radius of the particle has to be approximately \(2.24 \times 10^{-6}\;\mathrm{m}\) for the outward radiation pressure on it to be 1.00% of the inward gravitational attraction of the Sun.