The satellite of mass $m$ is moving in a circular orbit of radius $r,$

$\therefore $ Kinetic energy of the satellite, $K = \frac{{GMm}}{{2r}}\,\,\,\,\,\,…\left( i \right)$

Potential energy of the satellite, $U = \frac{{ – GMm}}{r}\,\,\,\,\,…\left( {ii} \right)$

Orbital speed of satellite, $v = \sqrt {\frac{{GM}}{r}} \,\,\,\,\,…\left( {iii} \right)$

$Time – period\,of\,satellite,$

$\,T = {\left[ {\left( {\frac{{4{\pi ^2}}}{{GM}}} \right){r^3}} \right]^{1/2}}\,\,\,\,……\left( {iv} \right)$

$Given\,{m_{{s_1}}} = 4{m_{{s_2}}}$

Since $M,\, r$ is same for both the satellites ${S_1}\,and\,{S_2}$

$\therefore $ From equation $\left( {ii} \right)$ , we get $U \propto m$

$\therefore \frac{{{U_{{s_1}}}}}{{{U_{{s_2}}}}} = \frac{{{m_{{s_1}}}}}{{{m_{{s_2}}}}} = 4\,\,\,or,\,\,\,{U_{{s_1}}} = 4{U_{{s_2}}}$

Option $\left( a \right)$ is wrong.

From $\left( {iii} \right)$, since  $v$ is independent of the mass of a satellite, the orbital speed is same for both satellites ${S_1}\,and\,{S_2}$.

Hence option $(b)$ is correct

From $(i)$, we get $K\, \propto m$

$\therefore \frac{{{K_{{s_1}}}}}{{{K_{{s_2}}}}} = \frac{{{m_{{s_1}}}}}{{{m_{{s_2}}}}} = 4\,\,or,\,\,{K_{{s_1}}} = 4{K_{{s_2}'}}$

Hence option $(c)$ is wrong.

From $(iv)$ , since $T$ is independent of the mass of a satellite, time period is same for both the satellites ${S_1}\,and\,{S_2}$ . Hence option $(d)$ is wrong.