Consider a spacecraft in an elliptical orbit around the earth At the lowest point or perigee of its orbit it is `300km` above the earth's surface at the highest point or apogee it is `3000km` above the earth surface
(a) What is the period of the spacecraft's orbit
(b) using conservation of angular momentum find the ratio of spacecraft's speed at perigee to its speed at apogee
(c) Using conservation of energy find the speed at perigee and the speed at apogee
(d) it is derised to have the spacecraft escape from the earth completrly if the spacecraft's rockets are fired at perigee by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee Which point in the orbit is the most efficient to use ?

Answer

To solve the problem step by step, we will break it down into the four parts as specified in the question.

### Given Data:
- Height at perigee (h1) = 300 km
- Height at apogee (h2) = 3000 km
- Radius of the Earth (R_e) = 6400 km

### Step 1: Calculate the distances from the center of the Earth
1. **Distance at Perigee (R_p)**:
\[ R_p = R_e + h_1 = 6400 \, \text{km} + 300 \, \text{km} = 6700 \, \text{km} \]

2. **Distance at Apogee (R_a)**:
\[ R_a = R_e + h_2 = 6400 \, \text{km} + 3000 \, \text{km} = 9400 \, \text{km} \]

### Step 2: Calculate the semi-major axis (a)
The semi-major axis (a) of the elliptical orbit is given by:
\[ a = \frac{R_p + R_a}{2} = \frac{6700 \, \text{km} + 9400 \, \text{km}}{2} = 8050 \, \text{km} \]

### Step 3: Calculate the period of the spacecraft's orbit (T)
Using Kepler's third law:
\[ T = 2\pi \sqrt{\frac{a^3}{GM}} \]
Where \( G \) is the gravitational constant and \( M \) is the mass of the Earth. We can express \( GM \) in terms of \( g \) (acceleration due to gravity at Earth's surface):
\[ g = \frac{GM}{R_e^2} \implies GM = g R_e^2 \]
Substituting \( g \approx 9.8 \, \text{m/s}^2 \) and \( R_e = 6400 \times 10^3 \, \text{m} \):
\[ GM \approx 9.8 \times (6400 \times 10^3)^2 \]
Now substituting \( a \) in meters:
\[ a = 8050 \times 10^3 \, \text{m} \]
Calculating \( T \):
\[ T = 2\pi \sqrt{\frac{(8050 \times 10^3)^3}{GM}} \]
This will yield \( T \approx 7160 \, \text{s} \).

### Step 4: Ratio of speeds at perigee and apogee (using conservation of angular momentum)
Using the conservation of angular momentum:
\[ m v_p R_p = m v_a R_a \implies \frac{v_p}{v_a} = \frac{R_a}{R_p} \]
Substituting the values:
\[ \frac{v_p}{v_a} = \frac{9400}{6700} \approx 1.4 \]

### Step 5: Calculate speeds at perigee and apogee (using conservation of energy)
Using conservation of energy:
\[ \frac{1}{2} m v_p^2 - \frac{GMm}{R_p} = \frac{1}{2} m v_a^2 - \frac{GMm}{R_a} \]
This simplifies to:
\[ \frac{1}{2} v_p^2 - \frac{GM}{R_p} = \frac{1}{2} v_a^2 - \frac{GM}{R_a} \]
Rearranging gives:
\[ v_p^2 - v_a^2 = 2GM \left( \frac{1}{R_p} - \frac{1}{R_a} \right) \]
Using the earlier derived ratio \( \frac{v_p}{v_a} = 1.4 \), we can find \( v_a \) and \( v_p \):
1. Calculate \( v_a \) using:
\[ v_a = \sqrt{\frac{GM}{R_a}} \approx 5.95 \times 10^3 \, \text{m/s} \]
2. Then calculate \( v_p \):
\[ v_p = 1.4 \times v_a \approx 8.35 \times 10^3 \, \text{m/s} \]

### Step 6: Escape velocity increase at perigee and apogee
1. **At Perigee**:
\[ \Delta v_p = \sqrt{\frac{2GM}{R_p}} - v_p \]
This yields approximately \( \Delta v_p \approx 2596.38 \, \text{m/s} \).

2. **At Apogee**:
\[ \Delta v_a = \sqrt{\frac{2GM}{R_a}} - v_a \]
This yields approximately \( \Delta v_a \approx 3291.53 \, \text{m/s} \).

### Conclusion
The most efficient point to fire the rockets for escape is at apogee, where the required increase in speed is higher.