To solve the problem step by step, we will break it down into parts as per the question.
### Given Data:
- Mass of the satellite, \( m = 220 \, \text{kg} \)
- Altitude above the Earth's surface, \( h = 640 \, \text{km} = 640,000 \, \text{m} \)
- Radius of the Earth, \( R_E = 6400 \, \text{km} = 6,400,000 \, \text{m} \)
- Rate of mechanical energy loss per revolution, \( E_{\text{loss}} = 1.4 \times 10^5 \, \text{J} \)
- Number of revolutions, \( N = 1500 \)
### Step 1: Calculate the orbital radius \( R \)
The orbital radius \( R \) is the sum of the Earth's radius and the altitude of the satellite:
\[
R = R_E + h = 6,400,000 \, \text{m} + 640,000 \, \text{m} = 7,040,000 \, \text{m}
\]
**Hint:** Remember to convert all units to meters for consistency.
### Step 2: Calculate the speed \( V \) of the satellite
Using the formula for orbital speed:
\[
V = \sqrt{\frac{G M_E}{R}}
\]
Where:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) (gravitational constant)
- \( M_E = 6 \times 10^{24} \, \text{kg} \) (mass of the Earth)
Substituting the values:
\[
V = \sqrt{\frac{(6.67 \times 10^{-11}) (6 \times 10^{24})}{7,040,000}} \approx 7.53 \, \text{km/s}
\]
**Hint:** The orbital speed formula derives from balancing gravitational force and centripetal force.
### Step 3: Calculate the period \( T \) of the satellite
Using Kepler's third law:
\[
T = 2\pi \sqrt{\frac{R^3}{G M_E}}
\]
Substituting the values:
\[
T = 2\pi \sqrt{\frac{(7,040,000)^3}{(6.67 \times 10^{-11})(6 \times 10^{24})}} \approx 1.63 \, \text{hours}
\]
**Hint:** The period is related to the radius of the orbit and the mass of the Earth.
### Step 4: Calculate the total mechanical energy loss after 1500 revolutions
\[
\text{Total Energy Loss} = E_{\text{loss}} \times N = (1.4 \times 10^5) \times 1500 = 2.1 \times 10^8 \, \text{J}
\]
**Hint:** Mechanical energy loss accumulates over multiple revolutions.
### Step 5: Calculate the final mechanical energy
The initial mechanical energy \( E_i \) is given by:
\[
E_i = -\frac{G M_E m}{2R}
\]
After the energy loss, the final mechanical energy \( E_f \) is:
\[
E_f = E_i - \text{Total Energy Loss}
\]
### Step 6: Calculate the new radius \( R_f \)
Using the final mechanical energy:
\[
E_f = -\frac{G M_E m}{2R_f}
\]
Setting \( E_f = E_i - 2.1 \times 10^8 \) and solving for \( R_f \) gives:
\[
R_f \approx 6812 \, \text{km}
\]
### Step 7: Calculate the new altitude \( h_f \)
\[
h_f = R_f - R_E = 6812 \, \text{km} - 6400 \, \text{km} = 412 \, \text{km}
\]
**Hint:** The altitude is the difference between the final radius and the Earth's radius.
### Step 8: Calculate the new speed \( V_f \) at the final radius
\[
V_f = \sqrt{\frac{G M_E}{R_f}} \approx 7.67 \, \text{km/s}
\]
**Hint:** Speed increases as the satellite moves closer to the Earth.
### Step 9: Calculate the new period \( T_f \)
\[
T_f = 2\pi \sqrt{\frac{R_f^3}{G M_E}} \approx 1.55 \, \text{hours}
\]
**Hint:** The period decreases as the satellite moves closer to the Earth.
### Step 10: Determine if angular momentum is conserved
Angular momentum is not conserved due to air resistance, which creates a net torque on the satellite.
**Hint:** Conservation of angular momentum applies only in the absence of external torques.
### Summary of Answers:
(a) Speed: \( 7.53 \, \text{km/s} \)
(b) Period: \( 1.63 \, \text{hours} \)
(c) Altitude after 1500 revolutions: \( 412 \, \text{km} \)
(d) Speed after 1500 revolutions: \( 7.67 \, \text{km/s} \)
(e) Period after 1500 revolutions: \( 1.55 \, \text{hours} \)
(f) Angular momentum is not conserved.
