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. `6.4xx10^3` km
B. `1.6xx10^3` km
C. `1.28xx10^4` km
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-by-Step Solution:

1. **Understanding \( E_1 \)**:
The energy \( E_1 \) required to take a satellite from the Earth's surface to a height \( h \) can be expressed using the gravitational potential energy formula. The potential energy at the Earth's surface (point 1) is:
\[ U_1 = -\frac{GMm}{R} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( R \) is the radius of the Earth.

At height \( h \) (point 2), the potential energy is:
\[ U_2 = -\frac{GMm}{R+h} \]

The work done (or energy required) to move the satellite from point 1 to point 2 is:
\[ E_1 = U_2 - U_1 = \left(-\frac{GMm}{R+h}\right) - \left(-\frac{GMm}{R}\right) \]
Simplifying this, we get:
\[ E_1 = GMm \left(\frac{1}{R} - \frac{1}{R+h}\right) \]
\[ E_1 = GMm \left(\frac{(R+h) - R}{R(R+h)}\right) = \frac{GMmh}{R(R+h)} \]

2. **Understanding \( E_2 \)**:
The kinetic energy \( E_2 \) required for the satellite to be in a circular orbit at height \( h \) is given by:
\[ E_2 = \frac{1}{2} mv^2 \]
The orbital velocity \( v \) at height \( h \) is given by:
\[ v = \sqrt{\frac{GM}{R+h}} \]
Therefore, substituting for \( v \):
\[ E_2 = \frac{1}{2} m \left(\sqrt{\frac{GM}{R+h}}\right)^2 = \frac{1}{2} m \frac{GM}{R+h} \]

3. **Setting \( E_1 \) equal to \( E_2 \)**:
Now we set \( E_1 \) equal to \( E_2 \):
\[ \frac{GMmh}{R(R+h)} = \frac{1}{2} m \frac{GM}{R+h} \]
We can cancel \( GMm \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{h}{R(R+h)} = \frac{1}{2(R+h)} \]
Cross-multiplying gives:
\[ 2h = R \]
Thus, we have:
\[ h = \frac{R}{2} \]

4. **Calculating \( h \)**:
Given that the radius of the Earth \( R = 6.4 \times 10^3 \) km:
\[ h = \frac{6.4 \times 10^3}{2} = 3.2 \times 10^3 \text{ km} \]

### Final Answer:
The height \( h \) for which \( E_1 \) and \( E_2 \) are equal is:
\[ h = 3.2 \times 10^3 \text{ km} \]