PrepRung
Sign inSign up free

JEE Advanced 2024 Paper 2 · Physics · Gravitation · medium

A satellite moves in a circular orbit of radius r0 around a planet with period T0. At one point of its path an instantaneous tangential thrust raises…

Numerical answer

A satellite moves in a circular orbit of radius r0r_0 around a planet with period T0T_0. At one point of its path an instantaneous tangential thrust raises its speed to 3/2\sqrt{3/2} times its circular-orbit speed. The satellite then moves in an elliptical orbit with the thrust point as perigee, and its new period is T=2n T0T=2\sqrt{n}\,T_0, where nn is a positive integer. The value of nn is
Show answer and solution

Answer: 2

Let vc2=GM/r0v_c^2=GM/r_0. After the thrust, v2=32vc2v^2=\tfrac32 v_c^2 at perigee r0r_0. Conservation of energy and angular momentum between perigee and apogee rar_a: 12v2−GMr0=12va2−GMra\tfrac12v^2-\dfrac{GM}{r_0}=\tfrac12v_a^2-\dfrac{GM}{r_a} and va=v r0/rav_a=v\,r_0/r_a. Writing ρ=ra/r0\rho=r_a/r_0: 34−1=34ρ2−1ρ\tfrac34-1=\tfrac{3}{4\rho^2}-\dfrac1\rho, i.e. −14=34ρ2−1ρ-\tfrac14=\tfrac{3}{4\rho^2}-\dfrac1\rho, so ρ2−4ρ+3=0\rho^2-4\rho+3=0, giving ρ=3\rho=3 (the root ρ=1\rho=1 is the perigee itself). The semi-major axis is a=r0+3r02=2r0a=\dfrac{r_0+3r_0}{2}=2r_0. By Kepler's third law T/T0=(a/r0)3/2=23/2=22T/T_0=(a/r_0)^{3/2}=2^{3/2}=2\sqrt2. Comparing with T=2n T0T=2\sqrt n\,T_0 gives n=2n=2.
Report a mistake

Original question written in the JEE Advanced 2024 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake