A particle of mass `m` is thrown upwards from the surface of the earth, with a velocity `u`. The mass and the radius of the earth are, respectively, `M` and `R`. `G` is gravitational constant `g` is acceleration due to gravity on the surface of earth. The minimum value of `u` so that the particle does not return back to earth is

Options

A. `sqrt((2GM)/(R^(2)))`
B. `sqrt((2GM)/(R))`
C. `sqrt((2gM)/(R^(2)))`
D. `sqrt(2gR^(2))`

Explanation

To determine the minimum velocity \( u \) required for a particle to escape the gravitational pull of the Earth without returning, we can use the principle of conservation of mechanical energy. Here’s a step-by-step solution:

### Step 1: Understand the Energy Conservation Principle
When the particle is thrown upwards, it has kinetic energy due to its initial velocity \( u \) and potential energy due to its position in the gravitational field of the Earth. As it rises, it will lose kinetic energy and gain potential energy until it reaches a point where its velocity becomes zero (the maximum height).

### Step 2: Write the Expression for Total Mechanical Energy
The total mechanical energy \( E \) at the surface of the Earth (where the particle is thrown) can be expressed as:
\[ E = \text{Kinetic Energy} + \text{Potential Energy} \]
At the surface, the kinetic energy is:
\[ \text{K.E.} = \frac{1}{2} m u^2 \]
The potential energy at the surface of the Earth is given by:
\[ \text{P.E.} = -\frac{GMm}{R} \]
Thus, the total energy at the surface is:
\[ E = \frac{1}{2} m u^2 - \frac{GMm}{R} \]

### Step 3: Set the Total Energy at Maximum Height to Zero
For the particle to escape the gravitational field and not return, the total mechanical energy at the maximum height must be zero:
\[ 0 = \frac{1}{2} m u^2 - \frac{GMm}{R} \]

### Step 4: Solve for \( u^2 \)
Rearranging the equation gives:
\[ \frac{1}{2} m u^2 = \frac{GMm}{R} \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} u^2 = \frac{GM}{R} \]
Multiplying both sides by 2:
\[ u^2 = \frac{2GM}{R} \]

### Step 5: Take the Square Root to Find \( u \)
Taking the square root of both sides gives:
\[ u = \sqrt{\frac{2GM}{R}} \]

### Step 6: Substitute \( g \) in Terms of \( G \)
We know that the acceleration due to gravity \( g \) at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
From this, we can express \( GM \) as:
\[ GM = gR^2 \]
Substituting this back into our equation for \( u \):
\[ u = \sqrt{\frac{2gR^2}{R}} = \sqrt{2gR} \]

### Final Answer
Thus, the minimum value of \( u \) so that the particle does not return back to Earth is:
\[ u = \sqrt{2gR} \]

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