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NEET PHYSICSGRAVITATIONMedium

Question

The minimum energy required to launch a satellite of mass mm from the surface of the earth of mass MM and radius RR in a circular orbit at an altitude of 2R2R from the surface of the earth is:

A

2GmM3R\frac{2GmM}{3R}

B

GmM2R\frac{GmM}{2R}

C

GmM3R\frac{GmM}{3R}

D

5GmM6R\frac{5GmM}{6R}

Step-by-Step Solution

  1. Initial Energy (EiE_i): At the surface of the Earth, the satellite is at rest (ignoring Earth's rotation) at a distance RR from the center. Its energy is purely potential. Ei=Ui=GMmRE_i = U_i = -\frac{GMm}{R}
  2. Final Energy (EfE_f): The satellite is in a circular orbit at an altitude h=2Rh = 2R. The orbital radius is r=R+h=R+2R=3Rr = R + h = R + 2R = 3R. The total energy of an orbiting satellite is given by E=GMm2rE = -\frac{GMm}{2r}. Ef=GMm2(3R)=GMm6RE_f = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R}
  3. Energy Required (ΔE\Delta E): The energy required to launch the satellite is the difference between the final total energy and the initial energy. ΔE=EfEi\Delta E = E_f - E_i ΔE=(GMm6R)(GMmR)\Delta E = \left( -\frac{GMm}{6R} \right) - \left( -\frac{GMm}{R} \right) ΔE=GMm6R+6GMm6R\Delta E = -\frac{GMm}{6R} + \frac{6GMm}{6R} ΔE=5GmM6R\Delta E = \frac{5GmM}{6R}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from GRAVITATION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSGRAVITATIONminimumenergyrequiredlaunchsatellite

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