Two satellites `A` and `B` of equal mass move in the equatorial plane of the earth, close to earth's surface. Satellite `A` moves in the same direction as the of the rotation of the earth while satellite `B` moves in the opposite direction. Calclate the ratio of the kinetic energy of `B` of that of `A` in the reference frame fixed to the earth `(g=9.8ms^(-2)` and radius of the earth `=6.37xx10^(6)km)`

Explanation

Let `omega_(A)` and `omega_(B)` be the absolute anglar speeds of `A` and `B`. Since they are in the same orbit, their time periods must be the same i.e. `omega_(A)=omega_(B)`. Considering the dynamics of circular motion in the cases.
`momega_(A)^(2)R=(GMm)/(R^(2))impliesomega_(A)=sqrt(g/R)` (`:'GM=gR^(2))`
similarly `omega_(B)=sqrt(g/R)`
and `omega_(A)=omega_(B)=sqrt(9.8/(6.37xx10^(6)))=124xx10^(-5)rads^(-1)`
Now `omega_(AE)=omega_(A)-omega_(E)=124xx10^(-5)-7.3xx10^(-5)`
`=116.7xx10^(-5)rads^(-1)`
`omega_(BE)` (velocity of `B` relative to `E`)
`=omega_(B)-(-omega_(E))=omega+B+omega_(E)`
`=131.1xx10^(-5)rads^(-1)`
Therefore `(K_(B))/(K_(A))=(1/2momega_(BE)^(2)r^(2))/(1/2momega_(AE)^(2)r^(2))=(131.3^(2))/(116.7^(2))=1.27`