Question

Two satellites orbit Earth at altitudes of 7500 km and 15,000 km above the Earth’s surface. Which satellite is faster, and by what factor?

Short answer

The factor by which the satellite is faster is $1.24$.

Step-by-step solution

Step 1. Given Data

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

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

Step 2. Understanding the orbital speed

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

Step 3. Evaluating the factor by which the satellite’s speed increases

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.

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

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