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NEET PHYSICSMedium

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=Ef−Ei\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}
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