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`

Answer

To solve the problem, we need to analyze the motion of the satellite before and after the change in direction of its velocity vector.

### Step-by-Step Solution:

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

2. **Change in Velocity:**
- At a certain point in its orbit, the satellite's velocity vector is suddenly turned by an angle \( \theta = \cos^{-1}(3/5) \).
- The new speed remains constant at \( v_0 \), but the direction changes.

3. **Components of Velocity:**
- After the change in direction, the new velocity components can be expressed as:
- \( v_{x} = v_0 \cos(\theta) \)
- \( v_{y} = v_0 \sin(\theta) \)
- Since \( \cos(\theta) = \frac{3}{5} \), we have \( v_{x} = \frac{3}{5} v_0 \).
- To find \( \sin(\theta) \), we use the Pythagorean identity:
\[ \sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
- Thus, \( v_{y} = v_0 \sin(\theta) = \frac{4}{5} v_0 \).

4. **Angular Momentum Conservation:**
- The angular momentum before the change is given by:
\[ L_i = m v_0 r_0 \]
- After the change, the angular momentum is:
\[ L_f = m v_{p} r_{p} = m \left(\frac{3}{5} v_0\right) r_{p} \quad \text{(at perigee)} \]
\[ L_f = m v_{a} r_{a} = m \left(\frac{4}{5} v_0\right) r_{a} \quad \text{(at apogee)} \]

5. **Setting Angular Momenta Equal:**
- Since angular momentum is conserved:
\[ m v_0 r_0 = m \left(\frac{3}{5} v_0\right) r_{p} = m \left(\frac{4}{5} v_0\right) r_{a} \]
- Canceling \( m v_0 \) from both sides gives:
\[ r_0 = \frac{3}{5} r_{p} \quad \text{and} \quad r_0 = \frac{4}{5} r_{a} \]

6. **Finding Ratios:**
- From the above equations, we can express \( r_{p} \) and \( r_{a} \):
\[ r_{p} = \frac{5}{3} r_0 \quad \text{and} \quad r_{a} = \frac{5}{4} r_0 \]
- Now, we need to find the ratio of speeds at perigee and apogee:
\[ \frac{v_{p}}{v_{a}} = \frac{r_{a}}{r_{p}} = \frac{\frac{5}{4} r_0}{\frac{5}{3} r_0} = \frac{3}{4} \]

7. **Final Result:**
- Therefore, the ratio of speed at perigee to speed at apogee is:
\[ \frac{v_{p}}{v_{a}} = \frac{3}{4} \]