Question

With what velocity should a particle be projected so that its height becomes equal to radius of earth?

• A. $\bigg( \frac{GM}{R} \bigg) ^{1/2}$
• B. $\bigg( \frac{8GM}{R} \bigg) ^{1/2}$
• C. $\bigg( \frac{2GM}{R} \bigg) ^{1/2}$
• D. $\bigg( \frac{4GM}{R} \bigg) ^{1/2}$

Answer 1

The Correct Option is A

To solve the problem, we need to determine the velocity with which a particle should be projected from the surface of the Earth so that it reaches a height equal to the radius of the Earth.

Let the radius of the Earth be R and the gravitational constant be G. The mass of the Earth is M.

When a particle is projected with an initial velocity v from the Earth's surface and reaches a height equal to R, its potential energy and kinetic energy at the maximum height can be used to derive this velocity.

At the surface of the Earth:

• Initial kinetic energy = \frac{1}{2}mv^2
• Initial potential energy = -\frac{GMm}{R}

At the height R:

• Potential energy = -\frac{GMm}{2R}
• Kinetic energy = 0 (since it reaches max height)

By conservation of energy, the total initial energy is equal to the total energy at the maximum height:

\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{2R}

Simplifying this equation:

• Add \frac{GMm}{R} to both sides:
• \frac{1}{2}mv^2 = -\frac{GMm}{2R} + \frac{GMm}{R}
• \frac{1}{2}mv^2 = \frac{GMm}{2R}

From this, we can simplify to find the initial velocity v:

v^2 = \frac{GM}{R}

Thus, the required initial velocity v is:

v = \bigg( \frac{GM}{R} \bigg)^{1/2}

This corresponds to the correct option: \bigg( \frac{GM}{R} \bigg)^{1/2}.