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NEET PHYSICSGRAVITATIONMedium

Question

A rocket is fired vertically upward with a speed of v=ve2v = \frac{v_e}{\sqrt{2}} from the Earth's surface, where vev_e is escape velocity on the surface of Earth. The distance from the surface of Earth upto which the rocket can go before returning to the Earth is: (given, the radius of Earth R=6400R = 6400 km)

A

1600 km

B

3200 km

C

6400 km

D

12800 km

Step-by-Step Solution

  1. Principle of Conservation of Energy: The total mechanical energy of the rocket is conserved. Let mm be the mass of the rocket and MM be the mass of the Earth. The initial point is the surface (r=Rr=R) and the final point is the maximum height (r=R+hr=R+h) where velocity becomes zero. Ki+Ui=Kf+UfK_i + U_i = K_f + U_f 12mv2GMmR=0GMmR+h\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}
  2. Substitute Velocity: Given v=ve2v = \frac{v_e}{\sqrt{2}}. We know escape velocity ve=2GMRv_e = \sqrt{\frac{2GM}{R}}. Therefore, v2=ve22=12(2GMR)=GMRv^2 = \frac{v_e^2}{2} = \frac{1}{2} \left( \frac{2GM}{R} \right) = \frac{GM}{R}.
  3. Solve for h: Substitute v2v^2 into the energy equation: 12m(GMR)GMmR=GMmR+h\frac{1}{2}m \left( \frac{GM}{R} \right) - \frac{GMm}{R} = -\frac{GMm}{R+h} GMm2RGMmR=GMmR+h\frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{R+h} GMmR(121)=GMmR+h\frac{GMm}{R} \left( \frac{1}{2} - 1 \right) = -\frac{GMm}{R+h} GMm2R=GMmR+h-\frac{GMm}{2R} = -\frac{GMm}{R+h} 12R=1R+h\frac{1}{2R} = \frac{1}{R+h} 2R=R+h    h=R2R = R + h \implies h = R
  4. Calculation: Given R=6400R = 6400 km, the height reached is h=6400h = 6400 km.

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.

PHYSICSGRAVITATIONrocketverticallyupwardfracvesqrtearths

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