If `d` is the distance between the centre of the earth of mass `M_(1)` and the moon of mass `M_(2)`, then the velocity with which a body should be projected from the mid point of the line joining the earth and the moon, so that it just escape is

Options

A. `sqrt((4G(M_(1)+M_(2)))/(d))`
B. `sqrt((4G)/(d)(M_(1)M_(2))/((M_(1)+M_(2))))`
C. `sqrt((2G)/(d)((M_(1)+M_(2))/(M_(1)M_(2))))`
D. `sqrt((2G)/(d)(M_(1)+M_(2))`

Explanation

To solve the problem of finding the velocity with which a body should be projected from the midpoint between the Earth and the Moon so that it just escapes their gravitational influence, we can follow these steps:

### Step 1: Understand the Concept of Escape Velocity
Escape velocity is the minimum velocity needed for an object to break free from the gravitational attraction of a celestial body without any further propulsion. In this case, we need to consider the gravitational effects of both the Earth and the Moon.

**Hint:** Remember that escape velocity depends on the gravitational potential energy and kinetic energy.

### Step 2: Identify the Forces at the Midpoint
The midpoint between the Earth and the Moon is at a distance of \( \frac{d}{2} \) from both the Earth and the Moon. The gravitational potential energy (U) at this point due to both bodies needs to be considered.

**Hint:** Gravitational potential energy is given by the formula \( U = -\frac{G M m}{r} \), where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, \( m \) is the mass of the object, and \( r \) is the distance from the center of the mass.

### Step 3: Calculate the Gravitational Potential Energy
The total gravitational potential energy at the midpoint due to both the Earth and the Moon can be expressed as:
\[ U = -\frac{G M_1 m}{\frac{d}{2}} - \frac{G M_2 m}{\frac{d}{2}} = -\frac{2G m (M_1 + M_2)}{d} \]
where \( M_1 \) is the mass of the Earth and \( M_2 \) is the mass of the Moon.

**Hint:** Combine the potential energies from both the Earth and the Moon.

### Step 4: Relate Kinetic Energy to Potential Energy
For the body to escape, its kinetic energy (K.E.) must equal the negative of the gravitational potential energy:
\[ K.E. = -U \]
The kinetic energy is given by:
\[ K.E. = \frac{1}{2} m v^2 \]
Setting these equal gives:
\[ \frac{1}{2} m v^2 = \frac{2G m (M_1 + M_2)}{d} \]

**Hint:** Remember to cancel the mass \( m \) of the object since it appears on both sides.

### Step 5: Solve for Escape Velocity
Now, we can solve for \( v \):
\[ \frac{1}{2} v^2 = \frac{2G (M_1 + M_2)}{d} \]
Multiplying both sides by 2:
\[ v^2 = \frac{4G (M_1 + M_2)}{d} \]
Taking the square root gives us the escape velocity:
\[ v = \sqrt{\frac{4G (M_1 + M_2)}{d}} \]

**Hint:** The escape velocity formula involves the sum of the masses and the distance between them.

### Final Answer
Thus, the velocity with which a body should be projected from the midpoint of the line joining the Earth and the Moon so that it just escapes is:
\[ v = \sqrt{\frac{4G (M_1 + M_2)}{d}} \]