Question

19．宇宙中兩個相距較近的星球可以看成雙星，它們只在相互間的萬有引力作用下，繞兩球心連線上的某一固定點(diǎn)做周期相同的勻速圓周運(yùn)動．根據(jù)宇宙大爆炸理論，雙星間的距離在不斷緩慢增加，設(shè)雙星仍做勻速圓周運(yùn)動，則下列說法正確的是（　�。�

• A． 雙星做圓周運(yùn)動的動能均減小
• B． 雙星做圓周運(yùn)動的半徑均增大
• C． 雙星相互間的萬有引力變大
• D． 雙星做圓周運(yùn)動的周期均增大

Answer 1

分析 雙星做勻速圓周運(yùn)動具有相同的角速度，靠相互間的萬有引力提供向心力，應(yīng)用萬有引力定律與牛頓第二定律求出雙星的軌道半徑關(guān)系，從而確定出雙星的半徑如何變化，以及得出雙星的角速度和周期的變化．

解答 解：根據(jù)萬有引力提供向心力，雙星做勻速圓周運(yùn)動具有相同的角速度和周期，$G\frac{{m}_{1}^{\;}{m}_{2}^{\;}}{{L}_{\;}^{2}}={m}_{1}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{1}^{\;}={m}_{2}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{2}^{\;}$，得${m}_{1}^{\;}{r}_{1}^{\;}={m}_{2}^{\;}{r}_{2}^{\;}$，且${r}_{1}^{\;}+{r}_{2}^{\;}=L$，得${r}_{1}^{\;}=\frac{{m}_{2}^{\;}}{{m}_{1}^{\;}+{m}_{2}^{\;}}L$，${r}_{2}^{\;}=\frac{{m}_{1}^{\;}}{{m}_{1}^{\;}+{m}_{2}^{\;}}L$
A、根據(jù)萬有引力提供向心力，$G\frac{{m}_{1}^{\;}{m}_{2}^{\;}}{{L}_{\;}^{2}}={m}_{1}^{\;}\frac{{v}_{1}^{2}}{{r}_{1}^{\;}}$，即${E}_{k1}^{\;}=\frac{1}{2}{m}_{1}^{\;}{v}_{1}^{2}=G\frac{{m}_{1}^{\;}{m}_{2}^{\;}{r}_{1}^{\;}}{2{L}_{\;}^{2}}$=$G\frac{{m}_{1}^{\;}{m}_{2}^{2}}{2L（{m}_{1}^{\;}+{m}_{2}^{\;}）}$，
同理$G\frac{{m}_{1}^{\;}{m}_{2}^{\;}}{{L}_{\;}^{2}}={m}_{2}^{\;}\frac{{v}_{2}^{2}}{{r}_{2}^{\;}}$，即${E}_{k2}^{\;}=\frac{1}{2}{m}_{2}^{\;}{v}_{2}^{2}=G\frac{{m}_{1}^{\;}{m}_{2}^{\;}{r}_{2}^{\;}}{2{L}_{\;}^{2}}$=$G\frac{{m}_{1}^{2}{m}_{2}^{\;}}{2L（{m}_{1}^{\;}+{m}_{2}^{\;}）}$，因?yàn)殡p星距離增大知雙星做圓周運(yùn)動的動能均減小，故A正確
B、根據(jù)萬有引力提供向心力，$G\frac{{m}_{1}^{\;}{m}_{2}^{\;}}{{L}_{\;}^{2}}={m}_{1}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{1}^{\;}={m}_{2}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{2}^{\;}$，得${m}_{1}^{\;}{r}_{1}^{\;}={m}_{2}^{\;}{r}_{2}^{\;}$，且${r}_{1}^{\;}+{r}_{2}^{\;}=L$，得${r}_{1}^{\;}=\frac{{m}_{2}^{\;}}{{m}_{1}^{\;}+{m}_{2}^{\;}}L$，${r}_{2}^{\;}=\frac{{m}_{1}^{\;}}{{m}_{1}^{\;}+{m}_{2}^{\;}}L$，由題意知雙星間距離增加，雙星做圓周運(yùn)動的半徑均增大，故B正確；
C、雙星間的距離在緩慢增大，根據(jù)萬有引力定律$F=G\frac{{m}_{1}^{\;}{m}_{2}^{\;}}{{r}_{\;}^{2}}$，知萬有引力減小，故C錯誤；
D、根據(jù)$G\frac{{m}_{1}^{\;}{m}_{2}^{\;}}{{L}_{\;}^{2}}={m}_{1}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{1}^{\;}={m}_{2}^{\;}\frac{4{π}_{\;}^{2}}{{T}_{\;}^{2}}{r}_{2}^{\;}$，得${m}_{1}^{\;}=\frac{4{π}_{\;}^{2}{r}_{2}^{\;}{L}_{\;}^{2}}{G{T}_{\;}^{2}}$，${m}_{2}^{\;}=\frac{4{π}_{\;}^{2}{r}_{1}^{\;}{L}_{\;}^{2}}{G{T}_{\;}^{2}}$，${m}_{1}^{\;}+{m}_{2}^{\;}=\frac{4{π}_{\;}^{2}{L}_{\;}^{3}}{G{T}_{\;}^{2}}$，得$T=\sqrt{\frac{4{π}_{\;}^{2}{L}_{\;}^{3}}{G（{m}_{1}^{\;}+{m}_{2}^{\;}）}}$，因?yàn)殡p星間距離增大，雙星做圓周運(yùn)動的周期均增大，故D正確；
故選：ABD

點(diǎn)評 解決本題的關(guān)鍵知道雙星靠相互間的萬有引力提供向心力，應(yīng)用萬有引力定律與牛頓第二定律即可正確解題，知道雙星的軌道半徑比等于質(zhì)量之反比