Question

地球繞太陽的運動可看作是軌道半徑為R的勻速圓周運動，太陽源源不斷地向四周輻射能量，太陽光的總輻射功率為P_S，太陽光在穿過太空及地球大氣層到達地面的過程中，大約有30%的能量損耗。到達地面的太陽光由各種頻率的光子組成，每個光子不僅具有能量，還具有動量，其能量與動量的比值為c，c為真空中的光速。（在計算時可認為每個光子的頻率均相同）
（1）求射到地面的太陽光在垂直于太陽光方向的單位面積上的輻射功率P_e；
（2）輻射到物體表面的光子被物體吸收或反射時都會對物體產生壓強，光子對被照射物體單位面積上所施加的壓力叫做光壓，假設輻射到地面的太陽光被地面全部吸收，求太陽光對地面的光壓I；
（3）試證明：地球表面受到的太陽光輻射壓力，和地球繞太陽做圓周運動的軌道半徑R的平方成反比（P_S可認為不變）。

Answer 1

（1）P_e=
（2） 
（3）設地球半徑為r，則地球受到的光輻射壓力為  
因為：I＝   所以：  
由于式中P_s、r、c均為常量，可見地球所受的光輻射壓力和地球到太陽的距離R的平方成反比。

（1）太陽向各個方向均勻輻射，則 P_e＝＝··············································· ①
（2）在地面上取一個很小的截面積S，設在很短的時間間隔t內，有N個光子垂直射入此面積，產生的光壓力為F，根據動量定理 －Ft＝0－NP································································································ ②
根據光壓的定義 I＝ ····························································································· ③
根據光子能量E和動量p的大小關系 P＝ ··································································· ④
在地球軌道處、垂直于太陽光方向的單位面積上的太陽輻射功率 P_e＝ ······························· ⑤
聯立解得 I＝ ································································································ ⑥
（3）設地球半徑為r，則地球受到的光輻射壓力為F_e＝I·πr² ················································ ⑦
聯立⑥⑦解得 F_e＝·πr²＝· ····································································· ⑧
式中P_s、r、c均為常量，可見地球所受的光輻射壓力和地球到太陽的距離R的平方成反比············· ⑨