Question

18．如圖所示，α星和β星組成的雙星系統(tǒng)的“晃動(dòng)”周期為T（實(shí)際是環(huán)繞運(yùn)動(dòng)，不過(guò)人們往往只能看到它們?cè)诨蝿?dòng)），α星的晃動(dòng)范圍為D_α，β星的晃動(dòng)范圍為D_β，試求α星和β星的質(zhì)量．

Answer 1

分析 這是一個(gè)雙星的問(wèn)題，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(dòng)，它們之間的萬(wàn)有引力提供各自的向心力，兩顆恒星有相同的角速度和周期，結(jié)合牛頓第二定律和萬(wàn)有引力定律解決問(wèn)題．

解答 解：由圖知α星的軌道半徑${r}_{α}^{\;}=\frac{{D}_{α}^{\;}}{2}$；β星的軌道半徑${r}_{β}^{\;}=\frac{{D}_{β}^{\;}}{2}$，α星和β星之間的距離為$L={r}_{α}^{\;}+{r}_{β}^{\;}=\frac{{D}_{α}^{\;}+{D}_{β}^{\;}}{2}$
根據(jù)萬(wàn)有引力提供向心力$G\frac{{m}_{α}^{\;}{m}_{β}^{\;}}{{L}_{\;}^{2}}={m}_{α}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{α}^{\;}={m}_{β}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{β}^{\;}$
解得：${m}_{α}^{\;}=\frac{4{π}_{\;}^{2}{r}_{β}^{\;}}{G{T}_{\;}^{2}}{L}_{\;}^{2}$=$\frac{{π}_{\;}^{2}{D}_{β}^{\;}（{D}_{α}^{\;}+{D}_{β}^{\;}）_{\;}^{2}}{2G{T}_{\;}^{2}}$
${m}_{β}^{\;}=\frac{{π}_{\;}^{2}{D}_{α}^{\;}（{D}_{α}^{\;}+{D}_{β}^{\;}）_{\;}^{2}}{2G{T}_{\;}^{2}}$
答：α星的質(zhì)量為$\frac{{π}_{\;}^{2}{D}_{β}^{\;}（{D}_{α}^{\;}+{D}_{β}^{\;}）_{\;}^{2}}{2G{T}_{\;}^{2}}$，β星的質(zhì)量為$\frac{{π}_{\;}^{2}{D}_{α}^{\;}（{D}_{α}^{\;}+{D}_{β}^{\;}）_{\;}^{2}}{2G{T}_{\;}^{2}}$

點(diǎn)評(píng) 本題是雙星問(wèn)題，與衛(wèi)星繞地球運(yùn)動(dòng)模型不同，兩顆星都繞同一圓心做勻速圓周運(yùn)動(dòng)，關(guān)鍵抓住條件：相同的角速度和周期．