The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fve, where ve is its escape velocity from the surface of the earth. The value of f is :

Question

Two identical particles each of mass \( m \) go around a circle of radius \( a \) under the action of their mutual gravitational attraction. The angular speed of each particle will be:

Answer 1

To solve this problem, we need to find the escape velocity of a body from a platform which is at a height equal to the radius of the Earth \( R \). First, let's recall the formula for escape velocity from the surface of the Earth:

v_e = \sqrt{\frac{2GM}{R}}

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

Now, the platform is at a height \( R \) from the surface, so the total distance from the center of the Earth is \( 2R \). The escape velocity from this point is:

v_{2R} = \sqrt{\frac{2GM}{2R}} = \sqrt{\frac{GM}{R}}

To find the factor \( f \), such that the escape velocity from the platform is \( f \times v_e \), we need to express \( v_{2R} \) in terms of \( v_e \):

v_{2R} = \sqrt{\frac{GM}{R}} = \frac{1}{\sqrt{2}} \times \sqrt{\frac{2GM}{R}} = \frac{1}{\sqrt{2}} \times v_e

Thus, the factor \( f \) is:

f = \frac{1}{\sqrt{2}}

Therefore, the value of \( f \) is \(\frac{1}{\sqrt{2}}\), which is the correct answer.

Answer 2