A small satellite of mass m is revolving around earth in a circular orbit of radius `r_(0)` with speed `v_(0)` . At certain point of its orbit, the direction of motion of satellite is suddenly changed by angle `theta = cos^(-1) `(3/5) by turning its velocity vector , such that speed remains constant. The satellite consequently goes to elliptical orbit around earth. the ratio of speed at perigee to speed at apogee is

Options

A. 3
B. 9
C. `1//3`
D. `1//9`

Explanation

To solve the problem of finding the ratio of speed at perigee to speed at apogee for a satellite that has changed its velocity vector while maintaining a constant speed, we can follow these steps:

### Step 1: Understand the Initial Conditions
The satellite is initially in a circular orbit with radius \( r_0 \) and speed \( v_0 \). The gravitational force provides the necessary centripetal force for circular motion.

### Step 2: Apply the Conservation of Angular Momentum
When the satellite changes its velocity direction by an angle \( \theta = \cos^{-1}(3/5) \), we can use the conservation of angular momentum. The angular momentum before the change is given by:
\[ L = m v_0 r_0 \]
After the change, the new velocity can be broken down into components. The component of the velocity perpendicular to the radius vector is:
\[ v_{\perp} = v_0 \sin(\theta) \]
And the component along the radius is:
\[ v_{\parallel} = v_0 \cos(\theta) \]

### Step 3: Determine the Ratio of Speeds at Perigee and Apogee
Using the conservation of angular momentum, we have:
\[ m v_p r_p = m v_a r_a \]
Where \( v_p \) and \( v_a \) are the speeds at perigee and apogee, and \( r_p \) and \( r_a \) are the respective distances from the center of the Earth.

From the conservation of energy, we can write:
\[ \frac{1}{2} m v_p^2 - \frac{GMm}{r_p} = \frac{1}{2} m v_a^2 - \frac{GMm}{r_a} \]

### Step 4: Use the Geometry of the Situation
Using the angle \( \theta \), we can find the distances at perigee and apogee:
\[ \cos(\theta) = \frac{3}{5} \implies \sin(\theta) = \frac{4}{5} \]

### Step 5: Solve for the Speeds
From the conservation of angular momentum:
\[ v_p r_p = v_a r_a \]
And from energy conservation:
\[ \frac{1}{2} v_p^2 - \frac{GM}{r_p} = \frac{1}{2} v_a^2 - \frac{GM}{r_a} \]

### Step 6: Find the Ratio \( \frac{v_p}{v_a} \)
Rearranging the equations and substituting for \( r_p \) and \( r_a \) gives:
\[ \frac{v_p}{v_a} = \sqrt{\frac{r_a}{r_p}} \]

Using the relationship from the angle and the distances:
\[ \frac{r_a}{r_p} = \frac{5}{9} \]

Thus, we can conclude:
\[ \frac{v_p}{v_a} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3} \]

### Final Result
The ratio of speed at perigee to speed at apogee is:
\[ \frac{v_p}{v_a} = \frac{3}{5} \]