火星與水星繞日公轉半徑之比為4比1

火星與水星繞日公轉半徑之比為4比1, r1/r2=4, 求1.兩者週期之比為何r1=火星半徑, r2=水星半徑T1=火星週期, T2=水星週期Kepler第3定律:(T1/T2)^2=(r1/r2)^3=4^3 =>T1/T2=4^(3/2)=8/1.........ans
2.兩者公轉速率比為何v1/v2=?v1/v2=(2*pi*r1/T1)/(2*pi*r2/T2)=(r1/r2)*(T2/T1)=4*1/8 =1/2........ans
3.兩者公轉角速率比為何w1/w2=?w1/w2=(v1/r1)/(v2/r2)=(v1/v2)*(r2/r1)=(1/2)*(1/4)=1/8........ans
4.兩者掠掃面積之速率比為何A1/A2=?A1/A2=r1^2*w1/(r2^2*w2)=(r1/r2)^2*(w1/w2)=(4^2)*(1/8)=2/1

1. Let r1 and r2 be the orbital radii of Mars (火星) and Mercury (水星) respectively.
Hence, r1/r2 = 4

Let T1 and T2 be the periods of Mars and Mercury respectively.
Using Kepler's Third Law: (T1/T2)^2 = (r1/r2)^3
i.e. (T1/T2)^2 = 4^3
T1/T2 = 4^(3/2) = 8
T1:T2 = 1:8

2. Let v1 and v2 be the speeds of revolution of Mars and Mercury respectively
(v1/v2) = (2.pi.r1/T1)/(2.pi.r2/T2) = (r1/r2).(T2/T1) = 4 x 1/8 = 1/2
v1:v2 = 1:2

3. Let w1 and w2 be the angular speeds of Mars and Mercury respectively
(w1)/(w2) = (v1/r1)/(v2/r2) = (v1/v2).(r2/r1) = (1/2).(1/4) = 1/8
w1:w2 = 1:8

4. (r1)^2.(w1)/[(r2)^2.(w2)]
= (r1/r2)^2.(w1/w2)
= (4^2).(1/8)
= 2
hence, ratio of rates of area swept = 2:1