Question

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 ______ x $10^{10}$ 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$^{-2}$)

Hint:

The formula for the kinetic energy of a satellite in orbit can be derived from the gravitational potential energy, where the radius includes both the radius of the Earth and the height of the satellite above the surface.

Answer 1

Correct Answer: 3

\[ KE = \frac{1}{2} mv^2 = \frac{GMm}{2r} \] \[ KE = \frac{GMm}{2(r_e + h)} \] \[ KE = \frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 1000 \times 6.4 \times 10^6}{2(6.4 \times 10^6 + 2.7 \times 10^5)} \] \[ KE = 3 \times 10^{10} \, \text{J} \]