With what velocity should a particle be projected so that its height 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)`

Answer

To find the velocity with which a particle should be projected so that its height becomes equal to the radius of the Earth, we can use the principle of conservation of energy. Here's a step-by-step solution:

### Step 1: Define the problem
We need to determine the initial velocity \( V \) required to project a particle from the surface of the Earth to a height equal to the radius of the Earth \( r \).

### Step 2: Understand the energy conservation principle
The total mechanical energy (kinetic energy + potential energy) of the particle must remain constant throughout its motion, as there are no external forces doing work on it.

### Step 3: Write the expressions for potential energy
1. At the surface of the Earth (height = 0), the potential energy \( PE_1 \) is given by:
\[ PE_1 = -\frac{GMm}{r} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the particle, and \( r \) is the radius of the Earth.

2. At the maximum height (which is equal to the radius of the Earth), the total distance from the center of the Earth is \( 2r \). The potential energy \( PE_2 \) at this height is:
\[ PE_2 = -\frac{GMm}{2r} \]

### Step 4: Write the expressions for kinetic energy
1. The kinetic energy \( KE_1 \) at the surface of the Earth when the particle is projected with velocity \( V \) is:
\[ KE_1 = \frac{1}{2} m V^2 \]

2. At the maximum height, the kinetic energy \( KE_2 \) is zero because the particle momentarily comes to rest:
\[ KE_2 = 0 \]

### Step 5: Apply conservation of energy
According to the conservation of energy:
\[ KE_1 + PE_1 = KE_2 + PE_2 \]
Substituting the expressions we have:
\[ \frac{1}{2} m V^2 - \frac{GMm}{r} = 0 - \frac{GMm}{2r} \]

### Step 6: Simplify the equation
Rearranging the equation gives:
\[ \frac{1}{2} m V^2 = \frac{GMm}{r} - \frac{GMm}{2r} \]
\[ \frac{1}{2} m V^2 = \frac{GMm}{2r} \]

### Step 7: Cancel out the mass of the particle
Since \( m \) is present in all terms, we can cancel it out:
\[ \frac{1}{2} V^2 = \frac{GM}{2r} \]

### Step 8: Solve for \( V \)
Multiplying both sides by 2 gives:
\[ V^2 = \frac{GM}{r} \]
Taking the square root of both sides results in:
\[ V = \sqrt{\frac{GM}{r}} \]

### Final Answer
The velocity with which the particle should be projected is:
\[ V = \sqrt{\frac{GM}{r}} \]

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