IAO 2003, Теория, Група B — Задача 6

• 582000 km

  "Height of the sky" comparable to lunar distance, order 6×10⁵ km.

It is evident that if the anvil impacted the earth after 9 days, the height of the sky is rather great and it is impossible to use the formula $h = gt^2/2$, since acceleration is not constant. The anvil must therefore have followed an elliptical orbit with the center of the Earth at one focus and the apogee of an orbit near to the place, where Hephaestus dropped it. This ellipse must also have been very much prolate: it is hardly likely that Hephaestus gave the anvil a speed of more than a few meters per second by his careless treatment. Thus, we can consider the motion of the anvil for $t = 9$ days as motion on an almost degenerate ellipse ($e \approx 1$) from the point of apogee up to the surface of the Earth. Apparently the complete period of revolution on such an ellipse is more than $2t = 18$ days, comparable to the period of revolution of Moon around Earth ($T_{Л} = 27.3$ days); that is, the major semiaxis of our degenerate ellipse is of the same order as distance from the Earth to the Moon ($a_{Л} = 384$ thousand km). Even without calculations, then, we can already answer the second question of the problem: the "height of the sky" is comparable to distance to the Moon. But it tells us more: that the size of the Earth may be neglected in comparison with "the height of the sky" $h$ in the solution of our problem. So we may consider that "the height of the sky" $h = 2a$ is the major axis of our ellipse. Using Kepler's 3rd Law, and comparing our orbit with an orbit of Moon we obtain the result

or of order 600 thousand km.