Problem P. 4491. (December 2012)

P. 4491. A spaceship is orbiting around the Earth. The distance between the spaceship and the surface of the Earth is the same as the radius of the Earth. Operating the rockets on the spaceship for a short time, the direction of the velocity of the spaceship is changed such that the angle between the actual radius drawn to the spaceship and the velocity is =150^o. The speed and the height of the spaceship above the Earth surface is unchanged.

a) By what amount will the spaceship go further away from the surface of the Earth?

b) Without altering again the path of the spaceship, at what speed would the spaceship hit the surface of the Earth?

(Neglect the effect of air and the rotation of the Earth.)

(5 pont)

Deadline expired on January 10, 2013.

Sorry, the solution is available only in Hungarian. Google translation

Megoldásvázlat. \(\displaystyle a)\) 2,73 földsugárnyira, kb. 17 ezet km-nyire.

\(\displaystyle b)\) \(\displaystyle \sqrt{3Rg/2}\approx 9{,}7\) km/s.

Statistics:

49 students sent a solution.
5 points:	Antalicz Balázs, Asztalos Bogdán, Balogh Menyhért, Barta Szilveszter Marcell, Bingler Arnold, Blum Balázs, Czipó Bence, Dávid Bence, Fehér Zsombor, Fekete Panna, Forrai Botond, Garami Anna, Janzer Barnabás, Jenei Márk, Juhász Péter, Kaprinai Balázs, Kenderes Anett, Kollarics Sándor, Kovács 444 Áron, Majoros Péter, Olosz Balázs, Öreg Botond, Papp Roland, Reitz Angéla, Sal Kristóf, Sárvári Péter, Szabó 928 Attila, Sztilkovics Milán, Trócsányi Péter, Ürge László, Vajda Balázs, Váli Tamás.
4 points:	Buttinger Milán, Dinev Georgi, Filep Gábor.
3 points:	7 students.
2 points:	3 students.
1 point:	3 students.
0 point:	1 student.

Problems in Physics of KöMaL, December 2012