Two planets of equal mass orbit a much massive star (figure). Planet `m_(1)` moves in circular orbit of radius `1 xx 10^(8) km` with period `2 yr`. Planet `m_(2)` moves in an elliptical orbit with closed distance `r_(1) = 1 xx 10^(8) km` and farthest distance `r_(2) = 1.8 xx 10^(8) km`, as shows.

(a) Using the fact that the mean radius of an elliptical orbit is the length of the semi-major axis, find the period of `m_(2)'s` orbit.
(b) Which planet has the greater speed at point `P`? Which has the greater total energy?
(c ) Compare the speed of planet `m_(2)` at `P` with that at `A`.

Answer

A, B, C, D

Explanation

Mean radius of planet,
`m_(2) = (r_(1) + r_(2))/(2) = 1.4 xx 10^(8) km`
Now, `T prop r^(3//2)`
`:.` Time period of `m_(2)` :
`T_(2) = T_(1) ((1.4 xx 10^(8))/(10^(8)))^(3//2)`
or `T_(2) = 2(1.4)^(3//2)`
`= 3.31 yr`
(b) For `m_(2)`, point `P` is perigee position. So, speed at this point is greater than orbit speed for circular orbit.
`:. U_(m2) = U_(m1)`
`:. E_(m2) gt E_(m1)`
`K_(m2) gt K_(m1)`
(c ) `nu r =` constant