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

Explanation

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

### Step 1: Understand the scenario
We want to project a particle such that its maximum height (h) equals the radius of the Earth (R). This means that when the particle reaches its maximum height, it will have zero kinetic energy and will only have gravitational potential energy.

### Step 2: Write the conservation of energy equation
At the point of projection (point 1), the total mechanical energy is the sum of kinetic energy (KE) and gravitational potential energy (PE):
- KE at point 1: \( KE_1 = \frac{1}{2} m u^2 \)
- PE at point 1: \( PE_1 = -\frac{GMm}{R} \)

At the maximum height (point 2), the kinetic energy becomes zero and the potential energy is:
- KE at point 2: \( KE_2 = 0 \)
- PE at point 2: \( PE_2 = -\frac{GMm}{2R} \)

### Step 3: Set up the conservation of energy equation
According to the conservation of energy:
\[ KE_1 + PE_1 = KE_2 + PE_2 \]
Substituting the expressions for KE and PE:
\[ \frac{1}{2} m u^2 - \frac{GMm}{R} = 0 - \frac{GMm}{2R} \]

### Step 4: Simplify the equation
Rearranging the equation gives:
\[ \frac{1}{2} m u^2 = \frac{GMm}{R} - \frac{GMm}{2R} \]
Factoring out \( GMm \):
\[ \frac{1}{2} m u^2 = GMm \left(\frac{1}{R} - \frac{1}{2R}\right) \]
\[ \frac{1}{2} m u^2 = GMm \left(\frac{2}{2R} - \frac{1}{2R}\right) \]
\[ \frac{1}{2} m u^2 = \frac{GMm}{2R} \]

### Step 5: Cancel out the mass (m)
Since mass (m) appears on both sides, we can cancel it out:
\[ \frac{1}{2} u^2 = \frac{GM}{2R} \]

### Step 6: Solve for u
Multiplying both sides by 2 gives:
\[ u^2 = \frac{GM}{R} \]
Taking the square root:
\[ u = \sqrt{\frac{GM}{R}} \]

### Conclusion
The velocity with which the particle should be projected is:
\[ u = \sqrt{\frac{GM}{R}} \]