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JEE Main 2020 (7 Jan, Shift 1) · Satellite Energy & Weightlessness

Satellite Energy & Weightlessness question from JEE Main 2020 (7 Jan, Shift 1)

Q24JEE Main 2020 · 7 JAN · SHIFT 1Difficult

Satellite Energy & Weightlessness

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass m10\frac{m}{10} so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth):

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Answer: (c)

  1. Energy from surface to r=2Rr = 2R: 12u2−GMR=12v2−GM2R⇒v2=u2−GMR\frac{1}{2}u^2 - \frac{GM}{R} = \frac{1}{2}v^2 - \frac{GM}{2R} \Rightarrow v^2 = u^2 - \frac{GM}{R} (radial).
  2. After ejection the satellite (9m10\frac{9m}{10}) has only tangential speed v0=GM2Rv_0 = \sqrt{\frac{GM}{2R}} and no radial speed.
  3. Radial momentum: mv=m10vr⇒vr=10vm v = \frac{m}{10}v_r \Rightarrow v_r = 10v. Tangential: 0=9m10v0+m10vt⇒vt=−9v00 = \frac{9m}{10}v_0 + \frac{m}{10}v_t \Rightarrow v_t = -9v_0.
  4. K=12⋅m10(100v2+81v02)=m20(100u2−100GMR+812GMR)K = \frac{1}{2}\cdot\frac{m}{10}\left(100v^2 + 81v_0^2\right) = \frac{m}{20}\left(100u^2 - 100\frac{GM}{R} + \frac{81}{2}\frac{GM}{R}\right).
  5. K=5m(u2−119200GMR)K = 5m\left(u^2 - \frac{119}{200}\frac{GM}{R}\right).
  6. So the answer is (c).
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