Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, $$\frac{K_A}{K_B}$$ is:

Let the mass of the earth be denoted by $$M_E$$ and the universal gravitational constant by $$G$$. For a satellite moving in a circular orbit, the necessary centripetal force comes entirely from the gravitational attraction of the earth. Hence we write the force-centripetal balance equation:

$$\frac{G\,M_E\,m_s}{r^2} = \frac{m_s\,v^2}{r}.$$

Here $$m_s$$ is the mass of the satellite, $$r$$ is the orbital radius and $$v$$ is the orbital speed. Cancelling the common factor $$m_s$$ and one power of $$r$$ we get

$$v^2 = \frac{G\,M_E}{r}.$$

The kinetic energy of a body of mass $$m_s$$ moving with speed $$v$$ is given by the standard formula $$K = \tfrac12 m_s v^2$$. Substituting the expression for $$v^2$$ obtained above, we have

$$K = \tfrac12 m_s \left(\frac{G\,M_E}{r}\right) = \frac{G\,M_E\,m_s}{2\,r}.$$

This shows that for satellites around the same planet, the kinetic energy is directly proportional to the satellite mass and inversely proportional to the orbital radius:

$$K \propto \frac{m_s}{r}.$$

Now we apply this result to the two satellites mentioned in the question.

Satellite A has mass $$m_A = m$$ and orbital radius $$r_A = R$$. Therefore,

$$K_A = \frac{G\,M_E\,m}{2\,R}.$$

Satellite B has mass $$m_B = 2m$$ and orbital radius $$r_B = 2R$$. Therefore,

$$K_B = \frac{G\,M_E\,(2m)}{2\,(2R)} = \frac{2\,G\,M_E\,m}{4\,R} = \frac{G\,M_E\,m}{2\,R}.$$

We observe that the factors $$G$$, $$M_E$$ and $$2R$$ neatly cancel to give the same numerator and denominator as in the expression for $$K_A$$. Hence

$$K_A = K_B.$$

Taking the ratio,

$$\frac{K_A}{K_B} = \frac{K_B}{K_B} = 1.$$

Hence, the correct answer is Option C.