1. Given

   • Radius of Earth: $R = 6.4 \times 10^3\ \text{km}$
   • Satellite is taken to height $h$ above Earth, so orbital radius is $r = R + h$

2. Energy required to take the satellite from Earth surface to height $h$

   The increase in gravitational potential energy is $E_1 = GMm\left(\frac{1}{R} - \frac{1}{r}\right)$

3. Kinetic energy of satellite in circular orbit at radius $r$

   For circular orbit, $\frac{mv^2}{r} = \frac{GMm}{r^2} \Rightarrow v^2 = \frac{GM}{r}$

   Hence kinetic energy is $E_2 = \frac{1}{2}mv^2 = \frac{GMm}{2r}$

4. Condition given: $E_1 = E_2$

   So, $GMm\left(\frac{1}{R} - \frac{1}{r}\right) = \frac{GMm}{2r}$

   Cancel $GMm$: $\frac{1}{R} - \frac{1}{r} = \frac{1}{2r}$

   Rearranging, $\frac{1}{R} = \frac{1}{r} + \frac{1}{2r} = \frac{3}{2r}$

   Therefore, $r = \frac{3R}{2}$

5. Find height $h$

   Since $r = R + h = \frac{3R}{2}$ we get $h = \frac{3R}{2} - R = \frac{R}{2}$

   Now substitute $R = 6.4 \times 10^3\ \text{km}$: $h = \frac{6.4 \times 10^3}{2} = 3.2 \times 10^3\ \text{km}$

6. Option check

   • A: $1.6 \times 10^3$ km ✗
   • B: $3.2 \times 10^3$ km ✓
   • C: $6.4 \times 10^3$ km ✗
   • D: $1.28 \times 10^4$ km ✗

Therefore, the correct answer is Option B.