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 _________ $$\times 10^{10}$$ J.
(Mass of earth = $$6 \times 10^{24}$$ kg, Radius of earth = $$6.4 \times 10^6$$ m, Gravitational constant = $$6.67 \times 10^{-11}$$ Nm$$^2$$ kg$$^{-2}$$)

Solution :

For a satellite revolving around the earth in circular orbit :

$$K = \frac{GMm}{2r}$$

Given :

$$m = 1000\text{ kg}$$

$$M = 6 \times 10^{24}\text{ kg}$$

$$G = 6.67 \times 10^{-11}\text{ Nm}^2\text{kg}^{-2}$$

Radius of earth :

$$R = 6.4 \times 10^6\text{ m}$$

Height of satellite :

$$h = 270\text{ km}$$

$$= 2.7 \times 10^5\text{ m}$$

Orbital radius :

$$r = R+h$$

$$= 6.4 \times 10^6 + 2.7 \times 10^5$$

$$= 6.67 \times 10^6\text{ m}$$

Substituting values :

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

$$= \frac{6 \times 10^{16}}{2 \times 10^6}$$

$$= 3 \times 10^{10}\text{ J}$$

Final Answer :

$$3$$