Answer
The speed of the satellite which is closer to the Earth is faster by a factor of 1.24.
Work Step by Step
Let's use 6380 km as the radius of the Earth. The force of gravity provides the centripetal force to keep the satellites moving around in a circle.
We can calculate the speed $v_1$ of the the satellite that is closer to the Earth. The distance above the Earth's surface is 7500 km, so the orbital radius is 7500 km + 6380 km which is 13,880 km. Therefore,
$\frac{m_s~v_1^2}{r} = \frac{Gm_E~m_s}{r^2}$
$v_1^2 = \frac{G~m_E}{r}$
$v_1 = \sqrt{\frac{G~m_E}{r}}$
$v_1 = \sqrt{\frac{(6.67\times 10^{-11}~N\cdot m^2/kg^2)(5.98\times 10^{24}~kg)}{(1.388\times 10^7~m)}}$
$v_1 = 5.36\times 10^3~m/s$
We can calculate the speed $v_2$ of the the satellite that is farther from the Earth. The distance above the Earth's surface is 15,000 km, so the orbital radius is 15,000 km + 6380 km which is 21,380 km.
$\frac{m_s~v_2^2}{r} = \frac{Gm_E~m_s}{r^2}$
$v_2^2 = \frac{G~m_E}{r}$
$v_2 = \sqrt{\frac{G~m_E}{r}}$
$v_2 = \sqrt{\frac{(6.67\times 10^{-11}~N\cdot m^2/kg^2)(5.98\times 10^{24}~kg)}{(2.138\times 10^7~m)}}$
$v_2 = 4.32\times 10^3~m/s$
We can find the ratio of $v_1/v_2$.
$\frac{v_1}{v_2} = \frac{5.36\times 10^3~m/s}{4.32\times 10^3~m/s} = 1.24$
The speed of the closer satellite is faster by a factor of 1.24
