A sky lab of mass `2 xx 10^(3)kg` is first launched from the surface of earth in a circular orbit of radius `2R` and them it is shifted from this circular orbit to another circular orbit of radius `3R`. Calculate the energy required
(a) to place the lab in the first orbit,
(b) to shift the lab from first orbit to the second orbit. `(R = 6400 km`, `g = 10 m//s^(2))`

Explanation

To solve the problem, we need to calculate the energy required to place the Skylab in the first orbit and then the energy required to shift it from the first orbit to the second orbit.

### Given Data:
- Mass of Skylab, \( m = 2 \times 10^3 \, \text{kg} \)
- Radius of Earth, \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)

### Step 1: Calculate the energy required to place the lab in the first orbit (radius = \( 2R \))

1. **Total Energy at the Surface of the Earth (TE1)**:
\[ TE_1 = PE_1 + KE_1 \]
Where:
- Potential Energy (PE) at the surface:
\[ PE_1 = -\frac{GMm}{R} \]
- Kinetic Energy (KE) at the surface is zero since it is at rest:
\[ KE_1 = 0 \]
Thus,
\[ TE_1 = -\frac{GMm}{R} \]

2. **Total Energy in the First Orbit (TE2)**:
For the first orbit at radius \( 2R \):
\[ PE_2 = -\frac{GMm}{2R} \]
The orbital velocity \( v \) at this radius is given by:
\[ v = \sqrt{\frac{GM}{2R}} \]
Therefore, the Kinetic Energy (KE) is:
\[ KE_2 = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{GM}{2R}\right) = \frac{GMm}{4R} \]
Thus,
\[ TE_2 = PE_2 + KE_2 = -\frac{GMm}{2R} + \frac{GMm}{4R} = -\frac{GMm}{4R} \]

3. **Work Done to Place the Skylab in the First Orbit**:
The work done (energy required) to place the Skylab in the first orbit is:
\[ W_1 = TE_2 - TE_1 \]
Substituting the values:
\[ W_1 = \left(-\frac{GMm}{4R}\right) - \left(-\frac{GMm}{R}\right) = -\frac{GMm}{4R} + \frac{4GMm}{4R} = \frac{3GMm}{4R} \]

### Step 2: Calculate the energy required to shift the lab from the first orbit to the second orbit (radius = \( 3R \))

1. **Total Energy in the Second Orbit (TE3)**:
For the second orbit at radius \( 3R \):
\[ PE_3 = -\frac{GMm}{3R} \]
The orbital velocity \( v \) at this radius is:
\[ v = \sqrt{\frac{GM}{3R}} \]
Therefore, the Kinetic Energy (KE) is:
\[ KE_3 = \frac{1}{2} mv^2 = \frac{1}{2} m \left(\frac{GM}{3R}\right) = \frac{GMm}{6R} \]
Thus,
\[ TE_3 = PE_3 + KE_3 = -\frac{GMm}{3R} + \frac{GMm}{6R} = -\frac{2GMm}{6R} + \frac{GMm}{6R} = -\frac{GMm}{6R} \]

2. **Work Done to Move from First Orbit to Second Orbit**:
The work done (energy required) to shift the Skylab from the first orbit to the second orbit is:
\[ W_2 = TE_3 - TE_2 \]
Substituting the values:
\[ W_2 = \left(-\frac{GMm}{6R}\right) - \left(-\frac{GMm}{4R}\right) = -\frac{GMm}{6R} + \frac{3GMm}{12R} = \frac{GMm}{12R} \]

### Final Calculation:
1. **Substituting Values**:
- Gravitational constant \( G \) can be related to \( g \) as \( G = g \cdot R^2/M \), but we will use \( GM \) directly.
- Using \( g = 10 \, \text{m/s}^2 \) and \( R = 6400 \times 10^3 \, \text{m} \):
\[ W_1 = \frac{3GMm}{4R} \quad \text{and} \quad W_2 = \frac{GMm}{12R} \]

### Final Answers:
- Energy required to place the lab in the first orbit:
\[ W_1 = 9.6 \times 10^{10} \, \text{J} \]
- Energy required to shift the lab from the first orbit to the second orbit:
\[ W_2 = 1.1 \times 10^{10} \, \text{J} \]