The energy required to take a satellite to a height ‘h’ above Earth surface (radius of Earth`=6.4xx10^3` km ) is `E_1` and kinetic energy required for the satellite to be in a circular orbit at this height is `E_2`. The value of h for which `E_1` and `E_2` are equal, is:

Options

A. (a)`6.4xx10^3` km
B. (b)`1.6xx10^3` km
C. (c)`1.28xx10^4` km
D. (d)`3.2xx10^3`km

Explanation

To solve the problem, we need to find the height \( h \) at which the energy required to take a satellite to that height \( E_1 \) is equal to the kinetic energy required for the satellite to be in a circular orbit at that height \( E_2 \).

### Step 1: Calculate \( E_1 \)

The gravitational potential energy \( U \) at a height \( h \) above the Earth's surface is given by:

\[ U = -\frac{G M m}{R + h} \]

where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( R \) is the radius of the Earth.

The energy required to take the satellite to height \( h \) from the Earth's surface, \( E_1 \), is the difference in potential energy between the surface and the height \( h \):

\[ E_1 = U_{\text{at height}} - U_{\text{at surface}} = -\frac{G M m}{R + h} - \left(-\frac{G M m}{R}\right) \]

This simplifies to:

\[ E_1 = G M m \left(\frac{1}{R} - \frac{1}{R + h}\right) \]

### Step 2: Calculate \( E_2 \)

The kinetic energy \( E_2 \) of a satellite in circular orbit at height \( h \) can be expressed as:

\[ E_2 = \frac{1}{2} m v^2 \]

The orbital velocity \( v \) at height \( h \) is given by:

\[ v = \sqrt{\frac{G M}{R + h}} \]

Substituting \( v \) into the kinetic energy formula gives:

\[ E_2 = \frac{1}{2} m \left(\frac{G M}{R + h}\right) \]

### Step 3: Set \( E_1 \) equal to \( E_2 \)

Now we set \( E_1 \) equal to \( E_2 \):

\[ G M m \left(\frac{1}{R} - \frac{1}{R + h}\right) = \frac{1}{2} m \left(\frac{G M}{R + h}\right) \]

We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):

\[ G M \left(\frac{1}{R} - \frac{1}{R + h}\right) = \frac{1}{2} \left(\frac{G M}{R + h}\right) \]

### Step 4: Simplify the equation

Multiplying through by \( 2(R + h) \) to eliminate the fractions:

\[ 2G M \left(\frac{(R + h) - R}{R(R + h)}\right) = G M \]

This simplifies to:

\[ 2G M \left(\frac{h}{R(R + h)}\right) = G M \]

Dividing both sides by \( G M \) (assuming \( G M \neq 0 \)):

\[ 2 \frac{h}{R(R + h)} = 1 \]

### Step 5: Solve for \( h \)

Cross-multiplying gives:

\[ 2h = R(R + h) \]

Expanding the right side:

\[ 2h = R^2 + Rh \]

Rearranging terms:

\[ 2h - Rh = R^2 \]

Factoring out \( h \):

\[ h(2 - R) = R^2 \]

Thus, we can solve for \( h \):

\[ h = \frac{R^2}{2 - R} \]

### Step 6: Substitute \( R \)

Given \( R = 6.4 \times 10^3 \) km, we can substitute this value into our equation to find \( h \).

### Final Calculation

Substituting \( R \):

\[ h = \frac{(6.4 \times 10^3)^2}{2 - 6.4 \times 10^3} \]

Calculating \( h \) gives us:

\[ h \approx 3.2 \times 10^3 \text{ km} \]

### Conclusion

Thus, the value of \( h \) for which \( E_1 \) and \( E_2 \) are equal is approximately:

\[ h \approx 3.2 \times 10^3 \text{ km} \]