91Ó°ÊÓ

The altitude of one of the satellite is $d_1=7500\mathrm{km}$.

The altitude of the other satellite is $d_2=15000\mathrm{km}$.

The orbital speed of a satellite does get affected by the variation in mass of the object and radius of the object.

The relation of speed of a satelliteis given by,

$v=\sqrt{\frac{Gm}{r}}$

Here, m is the mass of the Earth, G is the gravitational constant and r is the distance of the satellite.

The ratio of speed of the satellite can be written as:

$\begin{array}{c}\frac{v_1}{v_2}=\frac{\sqrt{\frac{Gm}{r_1}}}{\sqrt{\frac{Gm}{r_2}}}\\ \frac{v_1}{v_2}=\sqrt{\frac{r_2}{r_1}}\\ \frac{v_1}{v_2}=\sqrt{\frac{r_{\mathrm{E}}+d_2}{r_{\mathrm{E}}+d_1}}\end{array}$

Here, $r_{\mathrm{E}}$is the radius of the Earth.

On plugging the values in the above relation.

$\begin{array}{c}\frac{v_1}{v_2}=\sqrt{\frac{6.38×{10}^6\mathrm{m}+15000\mathrm{km}×\frac{1000\mathrm{m}}{1\mathrm{km}}}{6.38×{10}^6\mathrm{m}+7500\mathrm{km}×\frac{1000\mathrm{m}}{1\mathrm{km}}}}\\ \frac{v_1}{v_2}=1.24\end{array}$

Thus, $1.24$is the required factor by which the speed is faster for close satellite as compared to the far satellite.