Two Earth's satallites move in a common plane along circular orbits. The orbitasl radius of one satellite r = 7000 km while that of the other satellite is Delta r = 70 km less. What time interval separates the periodic approaches of the satellites to each other over the minimum distance ?

Two Earth's satallites move in a common plane along circular orbits. T

1.	Two Earth's satallites move in a common plane along circular orbits. The orbitasl radius of one satellite r = 7000 km while that of the other satellite is Delta r = 70 km less. What time interval separates the periodic approaches of the satellites to each other over the minimum distance ?
Answer» Solution :Now for first satellite which is revolving about the earth (mass M and radius r) the orbital speed is `v = SQRT((GM)/(r ))` Let `T_(1)` and `T_(2)` be the TIME period for first and second satellites respectively. Then we know that `T_(1) = (2pi r)/(v) = (2pi r)/(sqrt(GM))r^(3//2)` and `T_(2) = (2pi)/(sqrt(GM))(r-Delta r)^(3//2)` As second satellite is revolving in a radius `(r-Delta r)`. Know the period INTERVAL `(T_(1)-T_(2))` is given by `T_(1)-T_(2)=(2pi)/(sqrt(GM))[r^(3//2)-r^(3//2)(1-(Delta r)/(r ))^(3//2)]` `= (2pi)/(sqrt(GM)) r^(3//2)[1-(1-(3)/(2)(Delta r)/(r ))]` `= (2pi)/(sqrt(GM)) r^(3//2)((3)/(2)(Delta r)/(r ))`.