Given, the time period of the first satellite,
T1 = 1h
The time period of the second satellite,
T2 = 8 h
The radius of the orbit of nearer satellite,
R1 = 2000 kmApplying the Kepler’s third law,
T2∝R3 ()2=()3 ⇒
=() ⇒
R2 = 8000 km
As we know the relation between the angular speed and timeperiod
ω= So, the angular speed of the nearer satellite to the orbit,
ω1=rad∕h and the angular speed of the farther satellite to the orbit,
ω2==rad∕h The speed of the nearer satellite to orbit,
v1=ω1R1=(2π)×2×103km∕h The speed of the farther satellite to the orbit
V2=ω2R2 =()×8000 =π×2×103km∕h Thus, the relative angular speed of the nearer satellite to the farthersatellite,
ω= =| 2π×2×103−π×2×103 |
| 8000−2000 |
==rad∕h Comparing with,
ω= The value of x = 3.