What is the mathematical relationship between the escape velocity (Ve) and the orbital velocity (Vo) of a satellite revolving very close to the surface of the Earth? MCQ with Answer and Explanation
What is the mathematical relationship between the escape velocity (Ve) and the orbital velocity (Vo) of a satellite revolving very close to the surface of the Earth?
A. Vo = sqrt(2) * Ve
B. Ve = 2 * Vo
C. Ve = sqrt(2) * Vo
D. Ve = Vo
Answer: Option C
Solution (By JKExamLibrary)
The orbital velocity of a satellite very close to Earth is Vo = sqrt(gR). The escape velocity from Earth's surface is Ve = sqrt(2gR). Therefore, substituting Vo into the escape velocity equation yields Ve = sqrt(2) * Vo. A satellite requires a 41.4% increase in speed to escape orbit completely.
Explanation:
In free fall under gravity alone, mechanical energy (kinetic + potential) is conserved, assuming no air resistance. Potential energy decreases while kinetic energy increases, sum constant. Law of conservation of mechanical energy. Non-conservative forces absent.
Explanation:
For equal distances: average speed = total distance/total time = 2d/(d/v₁ + d/v₂) = 2v₁v₂/(v₁+v₂). This is harmonic mean, not arithmetic. Memory tip: Equal distances ⇒ harmonic mean of speeds; equal times ⇒ arithmetic mean. Frequently tested kinematics concept distinguishing average speed calculation methods.
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