In the above question, escape speed from the centre of earth is :

Options

A. `sqrt(2gR)`
B. `sqrt(gR)`
C. `sqrt(3gR)`
D. `sqrt((3gR)/2)`

Explanation

To find the escape speed from the center of the Earth, we can follow these steps:

### Step 1: Understand Escape Velocity
Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without any additional propulsion.

### Step 2: Set Up the Energy Equation
When an object is at the center of the Earth and we want to throw it to escape the gravitational field, we need to consider both the potential energy (PE) and kinetic energy (KE) of the object. The total energy (E) at the center must equal the total energy at infinity (which is zero).

### Step 3: Write the Potential Energy at the Center
The potential energy (PE) at the center of the Earth is given by the formula:
\[ PE = -\frac{3}{2} \frac{G M_e m}{R_e} \]
where:
- \( G \) is the gravitational constant,
- \( M_e \) is the mass of the Earth,
- \( m \) is the mass of the object,
- \( R_e \) is the radius of the Earth.

### Step 4: Write the Kinetic Energy
If we give the object a velocity \( V \), its kinetic energy (KE) will be:
\[ KE = \frac{1}{2} m V^2 \]

### Step 5: Set Up the Total Energy Equation
The total energy at the center of the Earth can be expressed as:
\[ E = PE + KE = -\frac{3}{2} \frac{G M_e m}{R_e} + \frac{1}{2} m V^2 \]
To escape the gravitational field, this total energy must equal zero:
\[ -\frac{3}{2} \frac{G M_e m}{R_e} + \frac{1}{2} m V^2 = 0 \]

### Step 6: Simplify the Equation
We can cancel the mass \( m \) from both sides since it is not zero:
\[ -\frac{3}{2} \frac{G M_e}{R_e} + \frac{1}{2} V^2 = 0 \]
Rearranging gives:
\[ \frac{1}{2} V^2 = \frac{3}{2} \frac{G M_e}{R_e} \]

### Step 7: Solve for Escape Velocity \( V \)
Multiplying both sides by 2:
\[ V^2 = 3 \frac{G M_e}{R_e} \]
Taking the square root:
\[ V = \sqrt{3 \frac{G M_e}{R_e}} \]

### Step 8: Express in Terms of Gravitational Acceleration \( g \)
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M_e}{R_e^2} \]
Thus, \( G M_e = g R_e^2 \). Substituting this into our equation for \( V \):
\[ V = \sqrt{3 \frac{g R_e^2}{R_e}} = \sqrt{3 g R_e} \]

### Final Answer
The escape speed from the center of the Earth is:
\[ V = \sqrt{3 g R_e} \]