Question

（分）地球繞太陽的運(yùn)動(dòng)可看作是軌道半徑為R的勻速圓周運(yùn)動(dòng)，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(dá)地面的過程中，大約有30%的能量損耗。到達(dá)地面的太陽光由各種頻率的光子組成，每個(gè)光子不僅具有能量，還具有動(dòng)量，其能量與動(dòng)量的比值為c，c為真空中的光速。（在計(jì)算時(shí)可認(rèn)為每個(gè)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(shí)都會(huì)對(duì)物體產(chǎn)生壓強(qiáng)，光子對(duì)被照射物體單位面積上所施加的壓力叫做光壓，假設(shè)輻射到地面的太陽光被地面全部吸收，求太陽光對(duì)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(yùn)動(dòng)的軌道半徑R的平方成反比（P_S可認(rèn)為不變）。

Answer 1

（1） P_e=

（2）  

（3）設(shè)地球半徑為r，則地球受到的光輻射壓力為   

因?yàn)椋?i>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。

解析:

（1）太陽向各個(gè)方向均勻輻射，則  P_e＝＝··············································· ①

（2）在地面上取一個(gè)很小的截面積S，設(shè)在很短的時(shí)間間隔t內(nèi)，有N個(gè)光子垂直射入此面積，產(chǎn)生的光壓力為F，根據(jù)動(dòng)量定理  －Ft＝0－NP································································································ ②

根據(jù)光壓的定義  I＝ ····························································································· ③

根據(jù)光子能量E和動(dòng)量p的大小關(guān)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(lián)立解得  I＝ ································································································ ⑥

（3）設(shè)地球半徑為r，則地球受到的光輻射壓力為 F_e＝I·πr² ················································ ⑦

聯(lián)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