With what velocity should a particle be projected so that its maximum height attained becomes equal to radius of earth ?

Options

A. `((GM)/(R))^(1//2)`
B. `((8GM)/(R))^(1//2)`
C. `((2GM)/(R))^(1//2)`
D. `((4GM)/(R))^(1//2)`

Explanation

To solve the problem of determining the velocity with which a particle should be projected so that its maximum height equals the radius of the Earth, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential energy
When a particle is projected upwards, it will lose kinetic energy as it gains potential energy until it reaches its maximum height. At maximum height, all the kinetic energy will have been converted into potential energy.

### Step 2: Write the equations for kinetic and potential energy
The decrease in kinetic energy (KE) can be expressed as:
\[ \text{KE} = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is the initial velocity.

The increase in potential energy (PE) as the particle rises to height \( h \) is given by:
\[ \text{PE} = mgh \]
where \( g \) is the acceleration due to gravity and \( h \) is the height attained.

### Step 3: Set the decrease in kinetic energy equal to the increase in potential energy
At maximum height, we have:
\[ \frac{1}{2} mv^2 = mgh \]
Since \( h \) is equal to the radius of the Earth (denote it as \( R \)), we can substitute \( h \) with \( R \):
\[ \frac{1}{2} mv^2 = mgR \]

### Step 4: Simplify the equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = gR \]

### Step 5: Solve for \( v^2 \)
Multiplying both sides by 2 gives:
\[ v^2 = 2gR \]

### Step 6: Take the square root to find \( v \)
Taking the square root of both sides, we find:
\[ v = \sqrt{2gR} \]

### Step 7: Substitute the value of \( g \)
We know that the gravitational acceleration \( g \) can also be expressed as:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth. Substituting this into our equation gives:
\[ v = \sqrt{2 \cdot \frac{GM}{R^2} \cdot R} = \sqrt{\frac{2GM}{R}} \]

### Final Answer
Thus, the velocity with which the particle should be projected is:
\[ v = \sqrt{2gR} \]
or equivalently,
\[ v = \sqrt{\frac{2GM}{R}} \]