M07M1

M07M1 is a short name for the first problem in the Classical Mechanics section of the May 2007 Princeton University Prelims. The problem statement can be found in the problems list. Here is the solution.

(a)

We use one of Kepler's laws, that says:

$$T^2= k a^3$$

where a is the semimajor axis of the ellipse and T is the period of the motion. To derive what the proportionality constant k is, we use the case of a circle:

$$ma\left(\frac{2\pi}{T}\right)^2=m\omega^2 a=\frac{GMm}{a^2}$$ $$a^3\frac{4\pi^2}{gR_0^2}=T^2$$

where $$R_0$$ is the radius of the earth.

$$T_{transfer}=\frac{T}{2}=\frac{\pi}{R_0}\sqrt{\frac{(R_0+R_1)^3}{8g}}$$

(b)

This problem has no relation with the previous one, as far as I can tell. Use the assumption that the shower of asteroids is much larger than the earth, but the asteroids themselves are much smaller than the earth. Then we can view this as a point scattering problem. Let m be the mass of each asteroid, d be the distance of closest approach, let b be the impact parameter, and let $$v_f$$ be the velocity at the point of closest approach. Then conservation of angular momentum and energy give:

$$mvb=mv_fd$$ $$\frac{mv^2}{2}=-\frac{GMm}{d}+\frac{mv_f^2}{2}$$

Solve the first equation for $$v_f$$ and plug into the second equation. The largest possible impact parameter such that the asteroid will hit the earth will be for the case when the distance of closest approach is exactly the radius of the earth. So plug $$d=R_0$$ in the second equation to obtain:

$$b_{max}=\sqrt{\frac{(R_0v^2+gR_0^2)^2-(gR_0^2)^2}{v^4}}$$ $$\sigma=\pi b_{max}^2=\pi R_0\frac{R_0v^2+2gR_0^2}{v^2} \quad Cross-section$$ $$N=n\sigma=n\pi R_0^2 \left(1+\frac{2gR_0}{v^2}\right)$$