IAO 2014, Теория, Група β — Задача 1. Dreams of the Polar Bear

Solution

It would seem that approaching 10 times closer to an object reduces its apparent magnitude by 5 magnitudes, and so approaching $1000=10^3$ times closer reduces the apparent magnitude by $3\cdot5=15$ magnitudes. The magnitude of the Moon would be equal to $-12.7-15=-27.7$?

Would it? It turns out that in this case the Moon shines 2.3 times brighter than the Sun! In fact, this cannot happen!

Actually, the rule of reducing the magnitude by 5 when approaching the object 10 times closer is valid only for point-like objects, or objects for which the small-angle approximation is acceptable, i.e., an approach 10 times closer to the object increases the apparent solid angle by a factor of 100. Indeed, the magnitude decrease of extended objects, like the lunar disc, is related to the increase of the object’s apparent solid angle while keeping brightness unchanged.

The solid angle of the visible surface of the Moon from the Earth is

When using the small-angle approximation, approaching $10^3$ times closer would increase the visible area by the factor of $10^6$ times, and it would become 67 rad². However, this solid angle cannot be larger than the solid angle of a hemisphere, $2\pi\ \mathrm{rad}^2=6.3\ \mathrm{rad}^2$. That is, in the calculations we have overestimated the area by a factor of at least $67/6.3=10.6$, and make an error in the magnitude of at least $2.5\log(10.6)=2.6$. Thus, the magnitude of the lunar disk cannot be less than $-27.7+2.6=-25.1$.

More precise calculations show that when approaching 1000 times closer to the surface of the Moon (approximately from 380 thousand km to 380 km), the apparent solid angle will be

i.e. it will increase by a factor of approximately $\Omega_2/\Omega_1=40000$ and (in this model of the same brightness of the lunar surface) the magnitude is

Official grading criteria

Solution: 7 pt in total, including:

• Understanding about impossibility to use approximation of point-like objects — 3 pt.
• Result approximately Ω₂/Ω₁=40000 — 2 pt.
• Formula and final calculations −12.7−2.5log(40000)≈−24.2 — 2 pt.

For any solution based on calculations in approximation of point-like objects or any answer resulting in larger than the Sun brightness, total mark (except mark for the picture) should not be larger than 1 pt for group α or 0.5 pt for group β.

Artistic picture of the Bear — 1 pt.

Jury note: evaluate the solution in essence rather than formal presence of sentences/formulae; award points when subsequent steps de facto use the required positions.