A satellite of mass `1000 kg` is supposed to orbit the earth at a height of `2000 km` above the earth's surface. Find `a.` its speed in the orbit `b`. its kinetic energy. `c`. The potential energy of the earth-satellite system and `d`. its time period. Mass of the earth `=6xx10^24kg`.

Explanation

To solve the problem step by step, we will calculate the speed of the satellite in orbit, its kinetic energy, the potential energy of the Earth-satellite system, and its time period.

### Given Data:
- Mass of the satellite, \( m = 1000 \, \text{kg} \)
- Height of the satellite above the Earth's surface, \( h = 2000 \, \text{km} = 2000 \times 10^3 \, \text{m} \)
- Mass of the Earth, \( M = 6 \times 10^{24} \, \text{kg} \)
- Radius of the Earth, \( R_e = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)
- Gravitational constant, \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)

### Step 1: Calculate the Radius of the Orbit
The radius of the orbit \( r \) is the sum of the Earth's radius and the height of the satellite above the Earth's surface.

\[ r = R_e + h = (6400 \times 10^3) + (2000 \times 10^3) = 8400 \times 10^3 \, \text{m} \]

### Step 2: Calculate the Speed of the Satellite
The speed \( v \) of a satellite in orbit is given by the formula:

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

Substituting the values:

\[ v = \sqrt{\frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24})}{8400 \times 10^3}} \]

Calculating this gives:

\[ v \approx 6902.38 \, \text{m/s} \]

### Step 3: Calculate the Kinetic Energy of the Satellite
The kinetic energy \( KE \) of the satellite is given by:

\[ KE = \frac{1}{2} mv^2 \]

Substituting the values:

\[ KE = \frac{1}{2} \cdot 1000 \cdot (6902.38)^2 \]

Calculating this gives:

\[ KE \approx 2.38 \times 10^{10} \, \text{J} \]

### Step 4: Calculate the Potential Energy of the Earth-Satellite System
The potential energy \( PE \) of the Earth-satellite system is given by:

\[ PE = -\frac{GMm}{r} \]

Substituting the values:

\[ PE = -\frac{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24}) \cdot (1000)}{8400 \times 10^3} \]

Calculating this gives:

\[ PE \approx -4.76 \times 10^{10} \, \text{J} \]

### Step 5: Calculate the Time Period of the Satellite
The time period \( T \) of the satellite is given by:

\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

Substituting the values:

\[ T = 2\pi \sqrt{\frac{(8400 \times 10^3)^3}{(6.67 \times 10^{-11}) \cdot (6 \times 10^{24})}} \]

Calculating this gives:

\[ T \approx 7642 \, \text{s} \approx 2.12 \, \text{hours} \]

### Summary of Results:
- a. Speed in the orbit: \( \approx 6902.38 \, \text{m/s} \)
- b. Kinetic energy: \( \approx 2.38 \times 10^{10} \, \text{J} \)
- c. Potential energy: \( \approx -4.76 \times 10^{10} \, \text{J} \)
- d. Time period: \( \approx 7642 \, \text{s} \) or \( \approx 2.12 \, \text{hours} \)