1. Given data

• Mass of satellite: $m = 1000\,\text{kg}$
• Mass of earth: $M = 6 \times 10^{24}\,\text{kg}$
• Radius of earth: $R = 6.4 \times 10^6\,\text{m}$
• Height above earth: $h = 270\,\text{km} = 2.7 \times 10^5\,\text{m}$
• Gravitational constant: $G = 6.67 \times 10^{-11}\,\text{N m}^2\text{kg}^{-2}$

2. Orbital radius

For the satellite, $r = R + h = 6.4 \times 10^6 + 2.7 \times 10^5 = 6.67 \times 10^6\,\text{m}$

3. Kinetic energy in circular orbit

For a satellite in circular orbit, $K = \frac{GMm}{2r}$

4. Substitute values

$K = \frac{(6.67 \times 10^{-11})(6 \times 10^{24})(1000)}{2(6.67 \times 10^6)}$

First simplify numerator: $6.67 \times 6 \times 10^{-11+24+3} = 40.02 \times 10^{16}$

Denominator: $2 \times 6.67 \times 10^6 = 13.34 \times 10^6$

So, $K = \frac{40.02 \times 10^{16}}{13.34 \times 10^6}$

$K = 3.0 \times 10^{10}\,\text{J}$

5. Required integer

The kinetic energy is $3 \times 10^{10}\,\text{J}$

So the blank is $3$.