A satellite is to be put into an orbit `600 km` above the surface of the earth. If its vertical velocity after launching is `2400 m//s` at this height, calculate the magnitude and direction in the impulse required to put the satellite directly increase. The mass of the satellite is `60 kg` and the radius of the earth is `6400 km`. Take `g=10 ms^(-2)`

Explanation

To solve the problem, we need to calculate the impulse required to change the satellite's velocity from its vertical launch velocity to the required orbital velocity. Let's break down the solution step by step.

### Step 1: Determine the radius of the satellite's orbit
The radius of the Earth (R) is given as 6400 km, and the satellite is at an altitude of 600 km above the Earth's surface. Therefore, the total radius (r) from the center of the Earth to the satellite is:

\[ r = R + \text{altitude} = 6400 \, \text{km} + 600 \, \text{km} = 7000 \, \text{km} = 7000 \times 10^3 \, \text{m} \]

### Step 2: Calculate the orbital speed
The orbital speed (V) of a satellite can be calculated using the formula:

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

Where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. However, we can also use the approximation:

\[ V = \sqrt{g \cdot r} \]

Given \( g = 10 \, \text{m/s}^2 \):

\[ V = \sqrt{10 \, \text{m/s}^2 \cdot 7000 \times 10^3 \, \text{m}} = \sqrt{70000 \times 10^3} = \sqrt{7 \times 10^6} \approx 2645.75 \, \text{m/s} \]

### Step 3: Calculate the change in velocity
The satellite's vertical launch velocity is given as \( V_y = 2400 \, \text{m/s} \). The required horizontal orbital velocity is \( V_x = 2645.75 \, \text{m/s} \).

The change in momentum (impulse) can be calculated using the vector components of the velocities. The total velocity vector after the impulse will have components:

- Vertical component: \( V_y' = 0 \, \text{m/s} \) (since we want to stop the vertical motion)
- Horizontal component: \( V_x' = 2645.75 \, \text{m/s} \)

### Step 4: Calculate the impulse required
The impulse (I) is defined as the change in momentum, which can be calculated as:

\[ I = m \cdot (V_f - V_i) \]

Where \( V_f \) is the final velocity and \( V_i \) is the initial velocity. We can break this down into components:

1. **Vertical Impulse**:
\[ I_y = m \cdot (0 - 2400) = 60 \cdot (-2400) = -144000 \, \text{kg m/s} \]

2. **Horizontal Impulse**:
\[ I_x = m \cdot (2645.75 - 0) = 60 \cdot 2645.75 = 158745 \, \text{kg m/s} \]

### Step 5: Calculate the magnitude of the total impulse
The total impulse can be calculated using the Pythagorean theorem:

\[ I = \sqrt{I_x^2 + I_y^2} = \sqrt{(158745)^2 + (-144000)^2} \]

Calculating this gives:

\[ I = \sqrt{25200000000 + 20736000000} = \sqrt{45936000000} \approx 6780.5 \, \text{kg m/s} \]

### Step 6: Calculate the direction of the impulse
The direction (angle θ) of the impulse can be calculated using:

\[ \tan(\theta) = \frac{I_y}{I_x} \]

Thus,

\[ \theta = \tan^{-1}\left(\frac{-144000}{158745}\right) \]

Calculating this gives:

\[ \theta \approx -41.5^\circ \]

This indicates that the impulse is directed downward and to the left (in the negative vertical direction).

### Final Answer
The magnitude of the impulse required is approximately **6780.5 kg m/s**, and the direction is approximately **-41.5 degrees** from the horizontal axis.

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