A star of mass M and radius R is moving with velocity v through a cloud of particles of density ?. If all the particles that collide with the star are trapped by it, show that the mass of the star will increase at a rate $$ \frac{\mathrm{d} M}{\mathrm{~d} t}=\pi \rho v\left(R^{2}+\frac{2 G M R}{v^{2}}\right) $$ Given that \(M=10^{31} \mathrm{~kg}\) and \(R=10^{8} \mathrm{~km}\), find how the effective crosssectional area compares with the geometric cross-section \(\pi R^{2}\) for velocities of \(1000 \mathrm{~km} \mathrm{~s}^{-1}, 100 \mathrm{~km} \mathrm{~s}^{-1}\) and \(10 \mathrm{~km} \mathrm{~s}^{-1}\).

First, consider a small time interval dt. During this time interval, the star travels a distance v*dt through the cloud of particles. The volume of the space through which the star passes is given by \(V = A \cdot v \cdot dt\), where A is the effective cross-sectional area of the star. Then, the mass of particles trapped by the star during this interval is dm=\(\rho \cdot V = \rho \cdot A \cdot v \cdot dt\). Now we can find the mass increase rate by dividing dm by dt:$$\frac{dM}{dt} = \rho \cdot A \cdot v$$ The effective cross-sectional area (A) depends on both the geometrical cross-section \(\pi R^2\) and the gravitational pull of the star, which can attract nearby particles. To account for the gravitational component, we need to know how far a particle can be from the star and still get attracted to it. This can be found by equating the gravitational force between the particle and the star to the centripetal force acting on the particle. $$F_{gravitational} = F_{centripetal}$$ $$\frac{GMm}{r^2} = m\frac{v^2}{r}$$ Solving for r, we get: $$r = \frac{2GM}{v^2}$$ Therefore, we can find the effective cross-sectional area (A) by considering the circular region with a radius equal to the sum of the star's radius (R) and the distance (r) from the star where the particle can still be attracted: $$A = \pi \left(R + \frac{2GM}{v^2}\right)^2$$ Now, substitute A in the mass increase rate formula: $$\frac{dM}{dt} = \pi \rho v\left(R^2 + \frac{2GM}{v^2}\right)$$

To compare the effective cross-sectional area with the geometric cross-section for the given velocities, we first calculate the actual effective cross-sectional area for each velocity: $$A_{1000} = \pi \left(R + \frac{2GM}{(1000 \cdot 10^3)^2}\right)^2$$ $$A_{100} = \pi \left(R + \frac{2GM}{(100 \cdot 10^3)^2}\right)^2$$ $$A_{10} = \pi \left(R + \frac{2GM}{(10 \cdot 10^3)^2}\right)^2$$ Then, we calculate the geometric cross-section: $$A_{geometric} = \pi R^2$$ Lastly, we compare the ratios: $$\frac{A_{1000}}{A_{geometric}}= \frac{\pi \left(R + \frac{2GM}{(1000 \cdot 10^3)^2}\right)^2}{\pi R^2}$$ $$\frac{A_{100}}{A_{geometric}}= \frac{\pi \left(R + \frac{2GM}{(100 \cdot 10^3)^2}\right)^2}{\pi R^2}$$ $$\frac{A_{10}}{A_{geometric}}= \frac{\pi \left(R + \frac{2GM}{(10 \cdot 10^3)^2}\right)^2}{\pi R^2}$$ These ratios represent how the effective cross-sectional area compares with the geometric cross-section for velocities of 1000 km/s, 100 km/s, and 10 km/s.