Question

A particle of mass ‘m’ is projected with a velocity v=kV_e(k < 1) from the surface of the earth. (V_e=escape velocity) The maximum height above the surface reached by the particle is

• A. \(\frac{Rk^2}{1-k^2}\)
• B. \(R(\frac{k}{1-k})^2\)
• C. \(R(\frac{k}{1+k})^2\)
• D. \(\frac{R^2k}{1+k}\)

Answer 1

The Correct Option is A

To solve this problem, we need to determine the maximum height reached by a particle projected upwards from the Earth's surface with an initial velocity defined in terms of escape velocity. Let's break down the solution step-by-step:

The given data includes: 

• Mass of the particle \( m \)
• Escape velocity \( V_e \)
• Initial velocity \( v = kV_e \) where \( k < 1 \)

The escape velocity from the Earth's surface is given by:

\(V_e = \sqrt{\frac{2GM}{R}}\)

where:

• \( G \) is the gravitational constant
• \( M \) is the mass of the Earth
• \( R \) is the radius of the Earth

The initial kinetic energy of the particle is:

\(\frac{1}{2}mv^2 = \frac{1}{2}m(kV_e)^2 = \frac{1}{2}mk^2V_e^2\)

The initial gravitational potential energy at the Earth's surface is:

\(-\frac{GMm}{R}\)

At the maximum height \( h \) above the Earth's surface, the particle will momentarily stop, and all initial kinetic energy is converted into gravitational potential energy.

Equating the initial total energy to the total energy at the maximum height, we have:

\(\frac{1}{2}mk^2V_e^2 - \frac{GMm}{R} = -\frac{GMm}{R + h}\)

From the expression of escape velocity, substitute \( V_e^2 = \frac{2GM}{R} \) to get:

\(\frac{1}{2}mk^2 \cdot \frac{2GM}{R} - \frac{GMm}{R} = - \frac{GMm}{R + h}\)

Simplifying this equation:

\(mk^2 \frac{GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R + h}\)

Further simplification gives:

\(GMm \left(\frac{k^2}{R} - \frac{1}{R}\right) = -\frac{GMm}{R + h}\)

This can be simplified to:

\(GMm \left(\frac{k^2 - 1}{R}\right) = -\frac{GMm}{R + h}\)

Cancelling GMm from both sides and solving for \( h \), we get:

\(R + h = \frac{R}{1-k^2}\)

Hence, the maximum height \( h \) above the Earth's surface is:

\(h = \frac{R}{1-k^2} - R = \frac{Rk^2}{1-k^2}\)

Therefore, the correct answer is: \(\frac{Rk^2}{1-k^2}\)