91Ó°ÊÓ

The given data can be listed below.

• The period of 1st satellite = 12.0 hours
• The period of 2nd or another satellite = 12.0 hours

Kepler has given three important laws to describe the orbital motions of the planet, satellite, comet, and other celestial bodies. The 3rd law out of these three laws describes the period of the orbital path, that is,

The orbital period of an orbiting celestial body around any heavenly body will be directly proportional to the three-by-two power of the distance between the celestial body and the heavenly body.

$$\begin{gathered} \mathrm{T}=\frac{2{\mathrm{Ï€°ù}}^{3/2}}{\sqrt{\mathrm{Gm}}} \\ \mathrm{r}={\left(\frac{\mathrm{T}\sqrt{\mathrm{Gm}}}{2\mathrm{Ï€}}\right)}^{2/3} \\ \mathrm{r}={\left[\frac{\left(12×3600\mathrm{s}×\sqrt{6.67×{10}^{-11}{\mathrm{m}}^3{\mathrm{Kg}}^{-1}{\mathrm{s}}^{-2}×5.9×{10}^{24}\mathrm{Kg}}\right)}{2×22/7}\right]}^{2/3} \\ \overline{)\mathrm{r}}=2649.40×{10}^4\mathrm{m} \end{gathered}$$

Here, Tis the period, r is the distance between the earth and the satellite, G is the gravitational constant, and m is the mass of the earth.

Hence, the distance between the earth’s surface and 1st satellite is$2649.40×{10}^4\mathrm{m}$

$$\begin{gathered} \mathrm{T}=\frac{2{\mathrm{Ï€°ù}}^{3/2}}{\sqrt{\mathrm{Gm}}} \\ \mathrm{r}={\left(\frac{\mathrm{T}\sqrt{\mathrm{Gm}}}{2\mathrm{Ï€}}\right)}^{2/3} \\ \mathrm{r}={\left[\frac{\left(12×3600\mathrm{s}×\sqrt{6.67×{10}^{-11}{\mathrm{m}}^3{\mathrm{Kg}}^{-1}{\mathrm{s}}^{-2}×5.9×{10}^{24}\mathrm{Kg}}\right)}{2×22/7}\right]}^{2/3} \\ \overline{)\mathrm{r}}=2649.40×{10}^4\mathrm{m} \end{gathered}$$

Here, Tis the period, r is the distance between the earth and the satellite, G is the gravitational constant, and m is the mass of the earth.

Hence, The distance between the earth’s surface and the 2nd satellite is$2649.40×{10}^4\mathrm{m}$.