A cosmic body `A` moves to the sun with velocity `v_(0)`(when far from the sun) and aiming parameter `l`, the arm of the vector `v_(0)`, relative to the centre of the sun. find the minimum distance by which this body will get to the sun. mass of the sum is `M`.

Answer

To find the minimum distance \( r \) at which a cosmic body \( A \) will approach the Sun, we can use the principles of conservation of angular momentum and energy. Here’s a step-by-step solution:

### Step 1: Understanding the Problem
The cosmic body \( A \) is moving towards the Sun with an initial velocity \( v_0 \) and has an aiming parameter \( l \). The goal is to find the minimum distance \( r \) from the center of the Sun that the body will reach.

### Step 2: Conservation of Angular Momentum
The angular momentum \( L \) of the body at point \( A \) (when it is far from the Sun) is given by:
\[ L_A = m v_0 l \]
where \( m \) is the mass of the cosmic body.

At the minimum distance \( r \) from the Sun, the angular momentum \( L_B \) is given by:
\[ L_B = m v r \]
where \( v \) is the velocity of the body at distance \( r \).

By conservation of angular momentum:
\[ m v_0 l = m v r \]
Thus, we can simplify to:
\[ v_0 l = v r \quad \text{(1)} \]

### Step 3: Conservation of Energy
The total mechanical energy at point \( A \) (far from the Sun) is purely kinetic:
\[ E_A = \frac{1}{2} m v_0^2 \]
At point \( B \) (at distance \( r \)), the total energy is the sum of kinetic and potential energy:
\[ E_B = \frac{1}{2} m v^2 - \frac{G M m}{r} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Sun.

By conservation of energy:
\[ \frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 - \frac{G M m}{r} \]
Cancelling \( m \) from both sides gives:
\[ \frac{1}{2} v_0^2 = \frac{1}{2} v^2 - \frac{G M}{r} \quad \text{(2)} \]

### Step 4: Expressing \( v \) in terms of \( r \)
From equation (1), we can express \( v \):
\[ v = \frac{v_0 l}{r} \]
Substituting this into equation (2):
\[ \frac{1}{2} v_0^2 = \frac{1}{2} \left(\frac{v_0 l}{r}\right)^2 - \frac{G M}{r} \]
This simplifies to:
\[ \frac{1}{2} v_0^2 = \frac{1}{2} \frac{v_0^2 l^2}{r^2} - \frac{G M}{r} \]

### Step 5: Rearranging the Equation
Multiplying through by \( 2r^2 \) to eliminate the fractions:
\[ v_0^2 r^2 = v_0^2 l^2 - 2 G M r \]
Rearranging gives:
\[ v_0^2 r^2 + 2 G M r - v_0^2 l^2 = 0 \]

### Step 6: Solving the Quadratic Equation
This is a quadratic equation in \( r \):
\[ a = v_0^2, \quad b = 2 G M, \quad c = -v_0^2 l^2 \]
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{-2 G M \pm \sqrt{(2 G M)^2 - 4 v_0^2 (-v_0^2 l^2)}}{2 v_0^2} \]
\[ r = \frac{-2 G M \pm \sqrt{4 G^2 M^2 + 4 v_0^4 l^2}}{2 v_0^2} \]
\[ r = \frac{-G M \pm \sqrt{G^2 M^2 + v_0^4 l^2}}{v_0^2} \]

### Step 7: Choosing the Positive Root
Since \( r \) must be positive, we take the positive root:
\[ r = \frac{-G M + \sqrt{G^2 M^2 + v_0^4 l^2}}{v_0^2} \]

### Final Answer
The minimum distance \( r \) at which the cosmic body \( A \) will approach the Sun is:
\[ r = \frac{-G M + \sqrt{G^2 M^2 + v_0^4 l^2}}{v_0^2} \]