The energy required to take a satelite 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 (in km) for which `E_(1)` and `E_(2)` are equal is

Explanation

To solve the problem, we need to find the height \( h \) above the Earth's surface where 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{GMm}{R + h}
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