Question

(a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth's rotation. (b) If the satellite is directly above the North Pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth = 6 × 10²⁴ kg.

Answer 1

(a) The angular speed of the Earth and the satellite will be the same.

$$\begin{gathered} \mathrm{i}.\mathrm{e}.,\frac{2\mathrm{\pi }}{T_e}=\frac{2\mathrm{\pi }}{T_s} \\ \Rightarrow \frac{1}{24\times 3600}=\frac{1}{2\mathrm{\pi }\sqrt{{\left(R+h\right)}^3/g{h}^2}} \\ \Rightarrow 12\times 3600=3.14\sqrt{\frac{{\left(R+h\right)}^3}{g{R}^2}} \\ \Rightarrow \frac{{\left(\mathrm{R}+h\right)}^3}{g{\mathrm{R}}^2}=\frac{{\left(12\times 3600\right)}^2}{{\left(3.14\right)}^2} \\ \Rightarrow \frac{\left(6400+h^3\right)\times {10}^9}{9.8\times {\left(6400\right)}^2\times {10}^6}=\frac{{\left(12\times 3600\right)}^2}{{\left(3.14\right)}^2} \\ \Rightarrow \frac{\left(6400+h\right)\times {10}^9}{6272\times {10}^9}=432\times {10}^4 \\ \Rightarrow {\left(6400+h\right)}^3=6272\times 432\times {10}^4 \\ \Rightarrow 6400+h={\left(6272\times 432\times {10}^4\right)}^{1/3}-6400 \\ \Rightarrow h=42300\mathrm{km} \end{gathered}$$

(b) Time taken from the North Pole to the equatorial plane is given by
$$\begin{gathered} \frac{1}{4}\mathrm{T} \\ =\frac{1}{4}\times 6.28\sqrt{\frac{{\left(42300+6400\right)}^3}{10\times {\left(6400\right)}^2\times {10}^6}} \\ =3.14\sqrt{\frac{{\left(479\right)}^3\times {10}^6}{{\left(64\right)}^2\times {10}^{11}}} \\ =3.14\sqrt{\frac{497\times 497\times 497}{64\times 64\times {10}^5}} \\ =6\mathrm{h} \end{gathered}$$