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 solve this problem, we need to determine the minimum speed required for an object to escape from the gravitational pull of the Earth when it is initially placed at a distance of \(3R\) above the Earth's surface. Here, \(R\) is the Earth's radius, \(M\) is the mass of the Earth, and \(G\) is the gravitational constant.

The total energy of the object at the distance of \(3R\) from the Earth's surface (a total of \(4R\) from the Earth's center) is given by the sum of its gravitational potential energy and kinetic energy. For the object to not return to Earth, its velocity must be such that the total mechanical energy at any point is zero or greater.

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

The condition for the object to escape is that the total mechanical energy should be \(\geq 0\): \(K.E. + P.E. \geq 0\)

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

Solving for the kinetic energy term: \(\frac{1}{2}mv^2 \geq \frac{GMm}{4R}\)

Cancelling \(m\) from both sides and solving for \(v\) gives:

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

 

Therefore, the minimum speed required to project the object so that it does not return to Earth is: \(\sqrt{\frac{GM}{2R}}\).

Hence, the correct answer is: \(\sqrt{\frac{GM}{2R}}\).

Answer 2

The minimum speed required for an object to escape the gravitational field of Earth is given by the escape velocity formula: \[ v_{\text{escape}} = \sqrt{\frac{2GM}{R}} \] 
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the distance from the center of the Earth. 
However, in this case, the object is placed at a distance of \( 3R \) from the Earth’s surface. 
The total distance from the center of the Earth is \( 4R \). 
The escape velocity at this distance is: \[ v_{\text{escape}} = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}} \] 
Thus, the minimum speed with which the object must be projected is \( \sqrt{\frac{GM}{2R}} \), and the correct answer is (1).