Question

9£®¾Ý̽²â£¬ÓîÖæÖпÉÄÜ´æÔÚÒ»ÖÖÔ¶ÀëÆäËûºãÐǵÄËÄÐÇϵͳ£¬ÆäÖÐÓÐÒ»ÖÖËÄÐÇϵͳµÄÄ£ÐÍÈçͼËùʾ£¬Ôڱ߳¤ÎªRµÄÕýÈý½ÇÐεÄÈý¸ö¶¥µãÉϸ÷ÓÐÒ»ÖÊÁ¿ÎªmµÄÐÇÌ壬ÔÚÈý½ÇÐεÄÖÐÐÄÓÐÖÊÁ¿ÎªMµÄÁíÒ»¿ÅÐÇ£¬´¦ÔÚÈý½ÇÐζ¥µãÉϵÄÈý¿ÅÐǾùÈÆ´¦ÓÚ¾²Ö¹×´Ì¬µÄλÓÚÈý½ÇÐÎÖÐÐĵÄÐÇÌå×öÔÈËÙÔ²ÖÜÔ˶¯£¬²»¿¼ÂÇÆäËûÐÇÌå¶Ô¸ÃϵͳµÄÓ°Ï죬ÒÑÖªÒýÁ¦³£Á¿ÎªG£¬ÇóÈý½ÇÐζ¥µãÉϵÄÐÇÌå×öÔÈËÙÔ²ÖÜÔ˶¯µÄÖÜÆÚºÍÏßËٶȣ®

Answer 1

·ÖÎö ÏÈд³öÈÎÒâÁ½¸öÐÇÐÇÖ®¼äµÄÍòÓÐÒýÁ¦£¬Çóÿһ¿ÅÐÇÐÇÊܵ½µÄºÏÁ¦£¬¸ÃºÏÁ¦ÌṩËüÃǵÄÏòÐÄÁ¦£®
È»ºóÓÃR±í´ï³öËüÃǵĹìµÀ°ë¾¶£¬×îºóд³öÓÃÖÜÆÚºÍÏßËٶȱí´ïµÄÏòÐÄÁ¦µÄ¹«Ê½£¬ÕûÀí¼´¿ÉµÄ³ö½á¹û£®

½â´ð ½â£º¶Ô¶¥µãÉϵÄÈÎÒ»ÐÇÌåÊܵ½ÆäËüÐÇÌåµÄÍòÓÐÒýÁ¦£¬ÓУº${F}_{1}^{\;}={F}_{2}^{\;}=G\frac{{m}_{\;}^{2}}{{R}_{\;}^{2}}$
ÔÈËÙÔ²ÖÜÔ˶¯µÄ¹ìµÀ°ë¾¶Îª£º$r=\frac{\frac{R}{2}}{cos30¡ã}=\frac{R}{\sqrt{3}}$
${F}_{3}^{\;}=G\frac{Mm}{£¨\frac{R}{\sqrt{3}}£©_{\;}^{2}}=3G\frac{Mm}{{R}_{\;}^{2}}$
¶ÔÈÎÒ»ÐÇÌåÊܵ½µÄºÏÁ¦Îª£º${F}_{ºÏ}^{\;}=\sqrt{3}{F}_{1}^{\;}+{F}_{3}^{\;}$=$\frac{\sqrt{3}G{m}_{\;}^{2}}{{R}_{\;}^{2}}+3G\frac{Mm}{{R}_{\;}^{2}}$
¸ù¾ÝÍòÓÐÒýÁ¦ÌṩÏòÐÄÁ¦£¬ÓУº$\frac{\sqrt{3}G{m}_{\;}^{2}}{{R}_{\;}^{2}}+3G\frac{Mm}{{R}_{\;}^{2}}=m\frac{{v}_{\;}^{2}}{r}=m\frac{4{¦Ð}_{\;}^{2}}{{T}_{\;}^{2}}r$
½âµÃ£º$v=\sqrt{\frac{Gm}{R}+\frac{\sqrt{3}GM}{R}}=\sqrt{\frac{G£¨m+\sqrt{3}M£©}{R}}$
ÖÜÆÚ£º$T=\frac{2¦Ðr}{v}=\frac{2¦ÐR}{\sqrt{3}}\sqrt{\frac{R}{G£¨m+\sqrt{3}M£©}}$=$2¦ÐR\sqrt{\frac{R}{G£¨3m+3\sqrt{3}M£©}}$
´ð£ºÈý½ÇÐζ¥µãÉϵÄÐÇÌå×öÔÈËÙÔ²ÖÜÔ˶¯µÄÖÜÆÚ$2¦ÐR\sqrt{\frac{R}{G£¨3m+3\sqrt{3}M£©}}$ºÍÏßËÙ¶È$\sqrt{\frac{G£¨m+\sqrt{3}M£©}{R}}$

µãÆÀ ½â¾ö¸ÃÌâÊ×ÏÈÒªÀí½âÄ£ÐÍËùÌṩµÄÇé¾°£¬È»ºóÄܹ»ÁгöºÏÁ¦ÌṩÏòÐÄÁ¦µÄ¹«Ê½£¬²ÅÄÜÕýÈ·½â´ðÌâÄ¿£®