Question

A particle of mass \(m\) is projected with a velocity \(v=KV_{e}\) (\(K\lt 1\)) from the surface of the earth. The maximum height above the surface reached by the particle is (\(V_{e}\) = escape velocity, \(R\) is radius of earth)

• A. \(R\left(\frac{K}{1+K}\right)^{2}\)
• B. \(\frac{R^{2}K}{1+K}\)
• C. \(\frac{RK^{2}}{1-K^{2}}\)
• D. \(R\left(\frac{K}{1-K}\right)^{2}\)

Hint:

Apply conservation of energy between the surface and the highest point using the escape velocity expression.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Escape velocity is \(V_e=\sqrt{\dfrac{2GM}{R}}\). Mechanical energy is conserved during the flight.

Step 2: Energy conservation
At the surface: \(\dfrac{1}{2}mK^2V_e^2-\dfrac{GMm}{R}\). At the top: \(-\dfrac{GMm}{R+h}\).
Using \(V_e^2=\dfrac{2GM}{R}\): \(\dfrac{GMm}{R}K^2-\dfrac{GMm}{R}=-\dfrac{GMm}{R+h}\).

Step 3: Simplify
\(\dfrac{1-K^2}{R}=\dfrac{1}{R+h}\), so \(R+h=\dfrac{R}{1-K^2}\).

Step 4: Result
\(h=\dfrac{R}{1-K^2}-R=\dfrac{RK^2}{1-K^2}\), which is option (C).

Final Answer:
The maximum height is R K squared divided by 1 minus K squared. \[ \boxed{\text{(C) }\dfrac{RK^{2}}{1-K^{2}}} \]