地球绕太阳的运动可看作是轨道半径为R的匀速圆周运动，太阳源源不断地向四周辐射能量，太阳光的总辐射功率为PS，太阳光在穿过太空及地球大气层到达地面的过程中，大约有30%的能量损耗。到达地面的太阳光由各种频率的光子组成，每个光子不仅具有能量，还具有动量，其能量与动量的比值为c，c为真空中的光速。（在计算时可认为每个光子的频率均相同）
（1）求射到地面的太阳光在垂直于太阳光方向的单位面积上的辐射功率Pe；
（2）辐射到物体表面的光子被物体吸收或反射时都会对物体产生压强，光子对被照射物体单位面积上所施加的压力叫做光压，假设辐射到地面的太阳光被地面全部吸收，求太阳光对地面的光压I；
（3）试证明：地球表面受到的太阳光辐射压力，和地球绕太阳做圆周运动的轨道半径R的平方成反比（PS可认为不变）。
（1）Pe=
（2）
（3）设地球半径为r，则地球受到的光辐射压力为
因为：I＝ 所以：
由于式中Ps、r、c均为常量，可见地球所受的光辐射压力和地球到太阳的距离R的平方成反比。
（1）太阳向各个方向均匀辐射，则 Pe＝＝··············································· ①
（2）在地面上取一个很小的截面积S，设在很短的时间间隔t内，有N个光子垂直射入此面积，产生的光压力为F，根据动量定理 －Ft＝0－NP································································································ ②
根据光压的定义 I＝ ····························································································· ③
根据光子能量E和动量p的大小关系 P＝ ··································································· ④
在地球轨道处、垂直于太阳光方向的单位面积上的太阳辐射功率 Pe＝ ······························· ⑤
联立解得 I＝ ································································································ ⑥
（3）设地球半径为r，则地球受到的光辐射压力为Fe＝I·πr2 ················································ ⑦
联立⑥⑦解得 Fe＝·πr2＝· ····································································· ⑧
式中Ps、r、c均为常量，可见地球所受的光辐射压力和地球到太阳的距离R的平方成反比············· ⑨