Suppose we project the particle the particle with speed veve, so that it just reaches infinity (r→∞)(r→∞). 
Applying energy conservation principle 
Ki+Ul=Kf+UfKi+Ul=Kf+Uf 
12m0v2e+m0[−GMe2R3{3R2−(R2)2}]=012m0ve2+m0[-GMe2R3{3R2-(R2)2}]=0 
⇒ve=11GMe4R−−−−−√⇒ve=11GMe4R.

Problem:
A narrow tunnel is dug along the diameter of the earth, and a particle of mass m is placed at a distance of R/2 from the center. Find the escape speed of the particle from that place?

Solution:
To find the escape speed of a particle at a certain distance from the center of the Earth, we need to consider the gravitational force acting on the particle and the concept of gravitational potential energy.

Gravitational Potential Energy:
The gravitational potential energy (U) of a particle of mass m at a distance r from the center of the Earth can be calculated using the equation:

U = -G * (M * m) / 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.

Escape Speed:
The escape speed is the minimum speed required for a particle to escape the gravitational pull of a celestial body. Mathematically, the escape speed (v) can be calculated using the equation:

v = sqrt(2 * |U| / m)

where |U| is the absolute value of the gravitational potential energy and m is the mass of the particle.

Calculations:
In this problem, the particle is placed at a distance of R/2 from the center of the Earth. The mass of the particle is given as m.

Let's calculate the gravitational potential energy of the particle at this distance.

U = -G * (M * m) / (R/2)
U = -2 * G * (M * m) / R

Now, let's calculate the escape speed using the formula mentioned above.

v = sqrt(2 * |U| / m)
v = sqrt(2 * |-2 * G * (M * m) / R| / m)
v = sqrt(4 * G * (M * m) / R)

Final Answer:
Therefore, the escape speed of the particle at a distance of R/2 from the center of the Earth is sqrt(4 * G * (M * m) / R).