The paths of two small satellites, X and Y, of equal mass of 6.00 kg each, are s
ID: 1500167 • Letter: T
Question
The paths of two small satellites, X and Y, of equal mass of 6.00 kg each, are shown below. They orbit around a massive star, as illustrated, with M = 5.20×1029 kg. The orbits are in the plane of the paper and are drawn to scale.
In the statements below KE is kinetic energy, PE is potential energy, and |L| is magnitude of the angular momentum.
(greater than less than equal to) The speed of X at p is .... that at c
(greater than less than equal to) The PE of Y at s is .... the PE of X at r.
(greater than less than equal to) At p, the KE of Y is .... that of X.
(greater than less than equal to) The |L| of Y at s is .... that at n.
(greater than less than equal to) At p, the PE of Y is .... that of X.
Explanation / Answer
1. The speed of X at p is Greater Than that at c. Reason - As satellite moves away from star, it loses Speed.
2. The PE of Y at s is Less Than the PE of X at r. Reason - As the satellite moves away from star, it loses kinetic energy and gains potential energy and as it moves back towards star it gains kinetic energy (because it gains speed) and loses potential energy. Here distance of s is smaller than r, therefore speed of satellite at s is greater than at r , thus, P.E at s is Less Than the P.E at r.
3. At p, the KE of Y is Less than that of X. Reason - V = sqrt(GM [2/r-1/a]) where a is semi major axis and r is distance of satelitte from star. Therefore at p speed of satelitte in X orbital is greater than speed of satelitte in Y orbital ,thus KE of Y is Less than that of X.
4. The |L| of Y at p is Equal to that at s. Reason - Conservation of Angular Momentum in orbit . (Keplers Law)
5. At p, the PE of Y is Equal to that of X. Reason -P.E = (- GMm/r), In this case for both the satellite , we have same values of M, m and r .Therefore P.E is same.
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