The masses and radii of the earth an moon are `M_(1) and R_(1) and M_(2), R_(2)` respectively. Their centres are at a distacne d apart. Find the minimum speed with which the particle of mass m should be projected from a point mid-way between the two centres so as to escape to infinity.

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)))`

Answer

To find the minimum speed with which a particle of mass \( m \) should be projected from a point midway between the centers of the Earth and the Moon, we can follow these steps:

### Step 1: Understand the Setup
We have two bodies: the Earth with mass \( M_1 \) and radius \( R_1 \), and the Moon with mass \( M_2 \) and radius \( R_2 \). The distance between their centers is \( d \). The midpoint between the two centers is at a distance of \( \frac{d}{2} \) from each body.

### Step 2: Calculate the Gravitational Potential Energy
At the midpoint, we need to calculate the gravitational potential energy due to both the Earth and the Moon. The gravitational potential \( V \) at a distance \( r \) from a mass \( M \) is given by:

\[ V = -\frac{GM}{r} \]

Thus, the potential energy \( U \) of the particle of mass \( m \) at the midpoint due to the Earth is:

\[ U_1 = -\frac{G M_1 m}{\frac{d}{2}} = -\frac{2 G M_1 m}{d} \]

And the potential energy due to the Moon is:

\[ U_2 = -\frac{G M_2 m}{\frac{d}{2}} = -\frac{2 G M_2 m}{d} \]

### Step 3: Total Gravitational Potential Energy
The total gravitational potential energy \( U \) at the midpoint is the sum of the potential energies due to both bodies:

\[ U = U_1 + U_2 = -\frac{2 G M_1 m}{d} - \frac{2 G M_2 m}{d} = -\frac{2 G (M_1 + M_2) m}{d} \]

### Step 4: Set Up the Energy Conservation Equation
To escape to infinity, the particle must have enough kinetic energy to overcome the gravitational potential energy. At infinity, the potential energy is zero, and thus the kinetic energy \( K \) at the midpoint must equal the absolute value of the total potential energy:

\[ K = -U \]

The kinetic energy \( K \) is given by:

\[ K = \frac{1}{2} m v^2 \]

Setting the kinetic energy equal to the absolute value of the potential energy, we have:

\[ \frac{1}{2} m v^2 = \frac{2 G (M_1 + M_2) m}{d} \]

### Step 5: Solve for the Minimum Speed \( v \)
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ \frac{1}{2} v^2 = \frac{2 G (M_1 + M_2)}{d} \]

Multiplying both sides by 2 gives:

\[ v^2 = \frac{4 G (M_1 + M_2)}{d} \]

Taking the square root of both sides, we find the minimum speed:

\[ v = \sqrt{\frac{4 G (M_1 + M_2)}{d}} \]

### Final Answer
The minimum speed with which the particle of mass \( m \) should be projected from the midpoint to escape to infinity is:

\[ v = \sqrt{\frac{4 G (M_1 + M_2)}{d}} \]

---