Question

An object is kept at rest at a distance of 3R above the earth’s surface where $ R $ is earth’s radius. The minimum speed with which it must be projected so that it does not return to earth is: (Assume $ M $ = mass of earth, $ G $ = Universal gravitational constant)

• A. \( \sqrt{\frac{GM}{2R}} \)
• B. \( \sqrt{\frac{GM}{R}} \)
• C. \( \sqrt{\frac{3GM}{R}} \)
• D. \( \sqrt{\frac{2GM}{R}} \)

Hint:

The escape velocity depends on the distance from the center of the Earth. For a distance greater than Earth's radius, adjust the escape velocity formula accordingly.

Answer 1

The Correct Option is A

To determine the minimum escape velocity, we must find the required initial speed for an object at \(3R\) above Earth's surface to overcome its gravitational pull. \(R\) denotes Earth's radius, \(M\) its mass, and \(G\) the gravitational constant.

The object's total energy at \(3R\) above the surface (a distance of \(4R\) from Earth's center) is the sum of its gravitational potential and kinetic energies. For escape, the total mechanical energy must be non-negative (\(\geq 0\)).

The gravitational potential energy at \(r = 4R\) from Earth's center is: \(-\frac{GMm}{r} = -\frac{GMm}{4R}\).

The escape condition is expressed as: \(K.E. + P.E. \geq 0\)

Substituting the values yields: \(\frac{1}{2}mv^2 - \frac{GMm}{4R} \geq 0\),

Rearranging for kinetic energy gives: \(\frac{1}{2}mv^2 \geq \frac{GMm}{4R}\)

After cancelling \(m\) and solving for \(v\):

• \(\frac{1}{2}v^2 \geq \frac{GM}{4R}\)
• \(v^2 \geq \frac{2GM}{4R}\)
• \(v \geq \sqrt{\frac{GM}{2R}}\)

 

Consequently, the minimum projection speed for the object to escape Earth's gravity is: \(\sqrt{\frac{GM}{2R}}\).

Thus, the correct escape velocity is: \(\sqrt{\frac{GM}{2R}}\).