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1.4 Energiya gravitacionnogo vzaimodeistviya
My videli, chto energiya gravitacionnogo vzaimodeistviya dlya dvuh mass
i
ravna
. Na sluchai
tochechnyh mass vyrazhenie dlya
obobshaetsya
sleduyushim obrazom:










Dlya tochechnyh mass neobhodimo bylo otbrasyvat' energiyu samodeistviya, ogovarivaya
pravilo summirovaniya. V sploshnoi srede samodeistvie ne uchityvaetsya avtomaticheski.
Po poryadku velichiny
, i samodeistvie elementa
est'
, t.e. velichina bolee vysokogo poryadka, chem
energiya vzaimodeistviya s ostal'nymi massami, kotoraya
.
Ispol'zuem teper' vyrazhenie dlya sfericheski-simmetrichnogo raspredeleniya
i vychislim gravitacionnuyu energiyu. Imeem:
Eto vyrazhenie mozhno znachitel'no uprostit'. Vvedem vspomogatel'nuyu funkciyu




Takim obrazom, integral ot pervogo chlena v vyrazhenii (1.3) raven integralu ot vtorogo, i okonchatel'no poluchim






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