A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ______ \( \times 10^{10}\text{ J} \).
(Mass of earth \( = 6 \times 10^{24}\text{ kg} \), Radius of earth \( = 6.4 \times 10^6\text{ m} \), Gravitational constant \( = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2} \))

Explanation:
Let us calculate the kinetic energy of the satellite step-by-step using gravitational principles:

1. Identify the given parameters:
Mass of the satellite \( (m) = 1000\text{ kg} = 10^3\text{ kg} \)
Mass of the Earth \( (M_e) = 6 \times 10^{24}\text{ kg} \)
Radius of the Earth \( (R_E) = 6.4 \times 10^6\text{ m} \)
Height of the orbit \( (h) = 270\text{ km} = 2.7 \times 10^5\text{ m} \)

2. Calculate the orbital radius \( (r) \):
The total distance of the satellite from the center of the Earth is:
\( r = R_E + h = 6.4 \times 10^6\text{ m} + 0.27 \times 10^6\text{ m} = 6.67 \times 10^6\text{ m} \)

3. Apply the Kinetic Energy formula:
The kinetic energy of a satellite revolving in a stable circular orbit is given by:
\( \text{KE} = \frac{G M_e m}{2r} = \frac{G M_e m}{2(R_E + h)} \)

Substituting the correct values into the expression:
\( \text{KE} = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 10^3}{2 \times 6.67 \times 10^6} \)

Canceling out the common term \( 6.67 \) from both the numerator and denominator simplifies the math:
\( \text{KE} = \frac{6 \times 10^{16}}{2 \times 10^6} \)
\( \text{KE} = 3 \times 10^{10}\text{ J} \)

Hence, Option C is the correct choice.