Question

(2008·寧夏高考)天文學(xué)家將相距較近、僅在彼此的引力作用下運(yùn)行的兩顆恒星稱為雙星．雙星系統(tǒng)在銀河系中很普遍．利用雙星系統(tǒng)中兩顆恒星的運(yùn)動特征可推算出它們的總質(zhì)量．已知某雙星系統(tǒng)中兩顆恒星圍繞它們連線上的某一固定點(diǎn)分別做勻速圓周運(yùn)動，周期均為T，兩顆恒星之間的距離為r，試推算這個雙星系統(tǒng)的總質(zhì)量．(引力常量為G)

 

Answer 1

     

解析:設(shè)兩顆恒星的質(zhì)量分別為m₁、m₂，做圓周運(yùn)動的半徑分別為r₁、r₂，角速度分別是ω₁、ω₂.根據(jù)題意有

ω₁＝ω₂                                                                                                 ①

r₁＋r₂＝r                                                               ②

根據(jù)萬有引力定律和牛頓第二定律，有

G＝                                                                                    ③

G＝                                                                                    ④

聯(lián)立以上各式解得r₁＝                                             ⑤

根據(jù)角速度與周期的關(guān)系知ω₁＝ω₂＝                                     ⑥

聯(lián)立③⑤⑥式解得m₁＋m₂＝.