Two hypothetical planets of masses `m_(1) and m_(2)` are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is 'd' ?
(speed of `m_(1) ` is `v_(1)` and that of `m_(2)` is `v_(2)`)
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Options

A. `v_(1) = v_(2)`
B. `v_(1) = m_(2) sqrt((2G)/(d(m_(1) + m_(2)))) " " v_(2) = - m_(1) sqrt((2G)/(d(m_(1) + m_(2))))`
C. `v_(1) = m_(1) sqrt((2G)/(d(m_(1) + m_(2)))) " " v_(2) = m_(2) sqrt((2G)/(d(m_(1) + m_(2))))`
D. `v_(1) = m_(2) sqrt((2G)/(m_(1))) " " v_(2) = m_(1) sqrt((2G)/(m_(2)))`

Answer

To solve the problem of finding the speeds of two hypothetical planets of masses \( m_1 \) and \( m_2 \) when they are at a separation \( d \), we will use the principles of conservation of momentum and conservation of energy.

### Step 1: Conservation of Momentum
Initially, both planets are at rest, so their initial velocities \( v_1 \) and \( v_2 \) are zero. According to the conservation of momentum:

\[ m_1 v_1 + m_2 v_2 = 0 \]

This implies:

\[ v_1 = -\frac{m_2}{m_1} v_2 \quad \text{(Equation 1)} \]

### Step 2: Conservation of Energy
The initial potential energy when the planets are infinitely far apart is zero. When they are at a separation \( d \), the potential energy \( U \) is given by:

\[ U = -\frac{G m_1 m_2}{d} \]

The total mechanical energy is conserved, so we have:

\[ \text{Initial Energy} = \text{Final Energy} \]

Initially, the kinetic energy is zero, so:

\[ 0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 - \frac{G m_1 m_2}{d} \]

Rearranging gives:

\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{G m_1 m_2}{d} \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 1 into Equation 2
Substituting \( v_1 \) from Equation 1 into Equation 2:

\[ \frac{1}{2} m_1 \left(-\frac{m_2}{m_1} v_2\right)^2 + \frac{1}{2} m_2 v_2^2 = \frac{G m_1 m_2}{d} \]

This simplifies to:

\[ \frac{1}{2} m_1 \frac{m_2^2}{m_1^2} v_2^2 + \frac{1}{2} m_2 v_2^2 = \frac{G m_1 m_2}{d} \]

Combining the terms gives:

\[ \frac{1}{2} \left(\frac{m_2^2}{m_1} + m_2\right) v_2^2 = \frac{G m_1 m_2}{d} \]

### Step 4: Solve for \( v_2^2 \)
Multiplying both sides by \( \frac{2d}{m_2} \):

\[ \left(\frac{m_2}{m_1} + 1\right) v_2^2 = \frac{2G m_1}{d} \]

Thus,

\[ v_2^2 = \frac{2G m_1}{d\left(\frac{m_2}{m_1} + 1\right)} \]

### Step 5: Find \( v_1 \)
Now, substituting back to find \( v_1 \):

\[ v_1 = -\frac{m_2}{m_1} v_2 \]

Substituting the expression for \( v_2 \):

\[ v_1 = -\frac{m_2}{m_1} \sqrt{\frac{2G m_1}{d\left(\frac{m_2}{m_1} + 1\right)}} \]

### Final Result
The speeds of the planets when their separation is \( d \) are:

\[ v_1 = \sqrt{\frac{2G m_2}{d(m_1 + m_2)}}, \quad v_2 = \sqrt{\frac{2G m_1}{d(m_1 + m_2)}} \]