IAO 2007, Теория, Група β, руски — Задача 2. Космический парус.

Since the square of the sail is not varying the number of photons falling on the sail is reversly proportional to the square distance to the Sun and force of the light pressure is central, i.e. $F\sim1/R^2$. It means that the probe will have an orbit according to the Kepler’s laws but the “effectivity of gravitation” will be smaller. In other words the probe moves as in gravitational field of a Sun with reduced (from $M_0$ to $M_1$) mass. That is the orbit of the probe with the sail is an ellipse with perihelion at the Earths’ orbit (distance to the Sun is $R_0$) and aphelion (since it takes half of rotation around the Sun to reach it) in the asteroid belt (distance to the Sun is $R_1$). By using the II Kepler law, the law of conservation of energy for the points of perihelion and aphelion, and also the equation for the circular orbit (without sail), one can find (these procedures are standard and not displayed here) that the new “effective mass” of the Sun is:

and the force of gravitation became less by the value

Then we have to find a force of light pressure that leads to such an effect. The energy of 1 photon is $E=mc^2$, its momentum is $p=mc$, i.e. $p=E/c$ for every photon. The force of pressure is $\Delta P/\Delta t$ – the change of the moment of the system sail-probe due to all photons that fall to the sail in one time unit. According to the law of conservation of momentum $\Delta P/\Delta t$ is equal to the sum of varying of moment of all photons that came to the sail: $\Delta P/\Delta t=\sum(\Delta p/\Delta t)$. Since the sail is of the mirror-type the photons are reflected and change the moment to the opposite value, that is the change of the momentum of each of them $\Delta p\ \Delta p=2p_0=2E_0/c$. Thus,

Taking into account that the energy of the photons that fall in a unit of time to the sail remote to a distance R from the Sun is:

we can write,

This value is equal to $(1-R_0/R_1)/2$ part of gravitation, so:

It is easy to find (by using equations of the circular orbit of Earth rotation around the Sun) that $GM_0=4\pi^2R_0^3/T_0^2$, where $T_0$ is a sidereal period of the Earth. Thus,

Not very much.

Then let us estimate the thickness $d$ of this sail. The mass of the sail is $M_{\text{П}}=\rho dS$, $d=M_{\text{П}}/\rho S$. It is reasonable to propose that the sail has mirror characteristics due to the metallic substance. It may give us the interval of possible density of the substance $\rho$: from $2\cdot10^3$ kg/m³ till $2\cdot10^4$ kg/m³. Let us take the density of aluminum $\rho=2.7\cdot10^3$ kg/m³ for an estimate. It is evident that the mass of the sail is a part of the mentioned 1 ton of the probe-system. Let us take a half of this value for an estimation, $M_{\text{П}}=5\cdot10^2$ kg. So:

This value can be reached by the modern techniques.

Note for jury: The correct answers are in interval from 0.1 till 1.5 mkm.