# The astronaut, leaving the spacecraft flying in a circular orbit around the Earth, threw 3 stones: one forward

**The astronaut, leaving the spacecraft flying in a circular orbit around the Earth, threw 3 stones: one forward, in the direction of motion, the other backward, and the third sideways, perpendicular to the orbital plane. The ship made one revolution, finding itself at the same point. In what position will stones 1, 2, 3 be relative to the ship?**

Before the astronaut threw stones, they flew with the spacecraft in a circular orbit at the first space velocity. After the throw, stone 1 gained additional speed and entered an elliptical orbit, where the throwing point became perigee. The semi-major axis of the orbit has become a little larger, which means that, according to Kepler’s law, the orbital period also increased slightly, and after the completion of the spacecraft’s revolution, stone 1 has not yet completed its revolution and, therefore, will be behind the ship. Stone 2, thrown backwards, will, on the contrary, go into a lower orbit, at which the throwing point will be the apogee, and after the turn of the ship will be in front. Thus, stones 1 and 2 are actually swapped.

Stone 3 will receive a lateral velocity increment, which will practically not affect the semi-major axis of the orbit and the orbital period. This is the only stone that an astronaut can easily catch on the next revolution of the spacecraft, while he will arrive from the side opposite to the direction of the throw.