A particle is situated at a height `3 R` from the earth surface . The velocity with which it should be projected vertically upward so that it does not return to earth is

Options

A. `sqrt((GM)/(R ))`
B. `sqrt((2GM)/(R ))`
C. `sqrt((GM)/(2R ))`
D. `sqrt((GM)/(4R ))`

Answer

To solve the problem of finding the velocity with which a particle should be projected vertically upward from a height of \(3R\) above the Earth's surface so that it does not return, we can follow these steps:

### Step 1: Determine the distance from the center of the Earth
The height of the particle above the Earth's surface is \(3R\). Therefore, the distance from the center of the Earth to the particle is:
\[ \text{Distance from center} = R + 3R = 4R \]

### Step 2: Write the expression for gravitational potential energy
The gravitational potential energy \(U\) of a mass \(m\) at a distance \(r\) from the center of the Earth (with mass \(M\)) is given by:
\[ U = -\frac{GMm}{r} \]
At a distance of \(4R\), the potential energy \(U_i\) of the particle is:
\[ U_i = -\frac{GMm}{4R} \]

### Step 3: Write the expression for kinetic energy
The kinetic energy \(K\) of the particle when it is projected with velocity \(V\) is given by:
\[ K = \frac{1}{2}mv^2 \]
Thus, the initial total energy \(E_i\) of the system (potential energy + kinetic energy) is:
\[ E_i = K + U_i = \frac{1}{2}mV^2 - \frac{GMm}{4R} \]

### Step 4: Determine the final energy condition
For the particle to not return to Earth, it must reach a point where it is infinitely far away. At this point, both the potential energy \(U_f\) and kinetic energy \(K_f\) will be zero:
\[ E_f = 0 \]

### Step 5: Set the initial energy equal to the final energy
Using the conservation of energy principle, we set the initial energy equal to the final energy:
\[ \frac{1}{2}mV^2 - \frac{GMm}{4R} = 0 \]

### Step 6: Solve for the velocity \(V\)
Rearranging the equation gives:
\[ \frac{1}{2}mV^2 = \frac{GMm}{4R} \]
Dividing both sides by \(m\) (assuming \(m \neq 0\)):
\[ \frac{1}{2}V^2 = \frac{GM}{4R} \]
Multiplying both sides by 2:
\[ V^2 = \frac{2GM}{4R} = \frac{GM}{2R} \]
Taking the square root:
\[ V = \sqrt{\frac{GM}{2R}} \]

### Final Answer
The velocity with which the particle should be projected vertically upward so that it does not return to Earth is:
\[ V = \sqrt{\frac{GM}{2R}} \]