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JEE Advanced 2018 Paper 2 · Physics · Gravitation · hard

A satellite moves in a circular orbit of radius R about a planet of mass M, with orbital speed v0=√(GM/R). At one instant its speed is changed…

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A satellite moves in a circular orbit of radius RR about a planet of mass MM, with orbital speed v0=GM/Rv_0=\sqrt{GM/R}. At one instant its speed is changed impulsively, without changing its direction, to the value listed in List-I. Match each case with the resulting orbit in List-II.

List-I
(P) New speed 1/2 v0\sqrt{1/2}\,v_0
(Q) New speed 2 v0\sqrt{2}\,v_0
(R) New speed 4/3 v0\sqrt{4/3}\,v_0
(S) New speed 3/2 v0\sqrt{3/2}\,v_0

List-II
(1) Elliptical orbit with farthest distance 2R2R from the planet
(2) Elliptical orbit with farthest distance 3R3R from the planet
(3) Elliptical orbit with closest distance R/3R/3 to the planet
(4) Parabolic (just escaping) path
(5) Hyperbolic path
  1. A. P→3, Q→4, R→1, S→2
  2. B. P→3, Q→4, R→2, S→1
  3. C. P→1, Q→4, R→3, S→2
  4. D. P→3, Q→5, R→1, S→2
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Answer: (A) P→3, Q→4, R→1, S→2

After the impulse the speed at r=Rr=R is kv0kv_0, which is tangential, so RR is an apsis. The total energy per unit mass is 12k2v02−v02=v022(k2−2)\tfrac12k^2v_0^2-v_0^2=\tfrac{v_0^2}{2}(k^2-2). Energy is zero when k2=2k^2=2, which is a parabola, so (Q) matches 4. For k2<2k^2<2 the orbit is bound; conservation of angular momentum and energy at the other apsis r′r' gives r′=k2R2−k2r'=\dfrac{k^2R}{2-k^2}. For (P), k2=1/2k^2=1/2: r′=1/23/2R=R/3r'=\tfrac{1/2}{3/2}R=R/3, and since r′<Rr'<R the point RR is the farthest distance and the closest is R/3R/3, matching 3. For (R), k2=4/3k^2=4/3: r′=4/32/3R=2Rr'=\tfrac{4/3}{2/3}R=2R, a farthest distance of 2R2R, matching 1. For (S), k2=3/2k^2=3/2: r′=3/21/2R=3Rr'=\tfrac{3/2}{1/2}R=3R, matching 2. Option 5 would need k2>2k^2>2 and is unused. So P→\to3, Q→\to4, R→\to1, S→\to2.
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