InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Find the maximum and minimum distances of the planet `A` from the sun `S`, if at a ceration moment of times it was at a distance `r_(0)` and travelling with the velocity `upsilon_(0)`. With the angle between the radius vector and velocity vector being equal to `phi`. |
|
Answer» At minimum and maximum distance velocity vector `(V)` makes an angle of `90^(@)` with radius vector. Hence, from conservation of angular momentum, `mu upsilon_(0) r_(0) sin phi = mr upsilon` ..(i) Hence, `m` is the mass of the plant. From energy conservation law, it follows that `(m upsilon_(0)^(2))/(2) - (GMm)/(r_(0)) = (m upsilon^(2))/(2) - (GMm)/(r)` ..(ii) Hence, `M` is the mass of the sun. Solving Eqs. (i) and (ii) for `r`, we fet two values of `r`, one is `r_(max)` and another is `r_(min)`. So, `r_(max) = (r_(0))/(2 - K) (1 + sqrt(1 - K(2 - K) sin^(2) phi))` and `r_(min) = (r_(0))/(2 - K) (1 - sqrt(1 - K(2 - K) sin^(2) phi))` Here, `K = (r_(0)^(2) upsilon_(0)^(2))/(GM)` |
|