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 calculate the maximum height reached by a particle projected with an initial velocity \( v = kV_e \) from the Earth's surface. Here, \( V_e \) is the escape velocity, and \( k < 1 \) is a given constant. We are given four options, and we need to verify which one is correct.

First, let's recall the formula for escape velocity \( V_e \):

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

where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the Earth's radius.

Now, according to the problem, the initial kinetic energy of the particle is:

\(KE_{initial} = \frac{1}{2} mv^2 = \frac{1}{2} m (kV_e)^2 = \frac{1}{2} m k^2 V_e^2\)

The potential energy at the Earth's surface is:

\(PE_{initial} = -\frac{GMm}{R}\)

At the maximum height \((h)\), the final kinetic energy is zero because the particle momentarily comes to rest, and the potential energy is:

\(PE_{final} = -\frac{GMm}{R+h}\)

By conservation of energy, the initial total energy should be equal to the final total energy:

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

Substituting \( V_e^2 = \frac{2GM}{R} \) into the equation:

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

Simplifying further:

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

Dividing through by \( m \), we have:

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

Rearranging terms gives:

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

We can cancel \( GM \) from both sides:

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

Cross-multiplying yields:

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

Expanding and simplifying:

\(Rk^2 - R - k^2Rh + R^2h = -R^2\)

\(R^2 - R^2k^2 = -k^2Rh + R^2h\)

Reorganize to solve for \( h \):

\(h(R^2 - k^2R) = R^2(1-k^2)\)

Finally, solve for \( h \):

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

Thus, the correct answer is:

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