 As a rule of thumb, the system is likely to be long-lived if "m"1 > 100"m"2 > 10,000"m"3 (in which "m"1, "m"2, and "m"3 are the masses of the star, planet, and trojan).More formally, in a three-body system with circular orbits, the stability condition is 27("m"1"m"2 + "m"2"m"3 + "m"3"m"1) < ("m"1 + "m"2 + "m"3)2.So the trojan being a mote of dust, "m"3→0, imposes a lower bound on of
25+√621/2
≈ 24.9599.