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Q6.6-43PE

Expert-verifiedFound in: Page 224

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication and weather observation because the satellite remains above the same point on Earth (provided it orbits in the equatorial plane in the same direction as Earth’s rotation). Calculate the radius of such an orbit based on the data for the moon in Table 6.2.**

The geosynchronous satellite's orbital radius is $4.23\times {10}^{7}\mathrm{m}$.

**A satellite is an item that is placed into orbit on purpose, and those are known as artificial satellites. There are natural satellites that orbit other planets. For example, the moon orbits around the earth, and the earth orbits the sun.**

Kepler's third law, written in mathematical form in, relates the period, or time, for one orbit to the radius of the orbit,

$\frac{{{T}_{1}}^{2}}{{{T}_{2}}^{2}}=\frac{{{r}_{1}}^{3}}{{{r}_{2}}^{3}}$

For the Moon, we'll use the subscript "m", and for the satellite, we'll use the subscript "s".

- The Orbital radius of the moon from the Table 6.2, ${r}_{m}=3.84\times {10}^{5}\text{\hspace{0.33em}km}=3.84\times {10}^{8}\text{\hspace{0.33em}m}$.
- Time taken by the moon to complete one orbit is, ${T}_{m}=0.07481\text{\hspace{0.33em}y}$.
- The Orbital radius of the geosynchronous earth satellite, ${\text{r}}_{\text{s}}\text{=?}$.
- The Orbital Period of the geosynchronous earth satellite, ${T}_{s}=1\text{\hspace{0.33em}day}=1\text{\hspace{0.33em}day}\times \frac{1\text{\hspace{0.33em}y}}{365\text{\hspace{0.33em}day}}=\frac{1}{365}\text{\hspace{0.33em}y}$.

Applying Kepler’s third law,

$\begin{array}{l}\frac{{{T}_{s}}^{2}}{{{T}_{m}}^{2}}=\frac{{{r}_{s}}^{3}}{{{r}_{m}}^{3}}\\ {{r}_{s}}^{3}=\frac{{{T}_{s}}^{2}}{{{T}_{m}}^{2}}\times {{r}_{m}}^{3}\\ {r}_{s}={\left(\frac{{{T}_{s}}^{2}}{{{T}_{m}}^{2}}\right)}^{\frac{1}{3}}\times {r}_{m}\\ {r}_{s}={\left(\frac{{T}_{s}}{{T}_{m}}\right)}^{\frac{2}{3}}\times {r}_{m}\end{array}$

Putting the values ${\mathrm{T}}_{\mathrm{s}},{\mathrm{T}}_{\mathrm{m}}\mathrm{and}{\mathrm{r}}_{\mathrm{m}}$ in the above expression, and we get,

$\begin{array}{l}{r}_{s}={\left(\frac{\frac{1}{365}y}{0.07481y}\right)}^{\frac{2}{3}}\times 3.84\times {10}^{8}\text{\hspace{0.33em}m}\\ {r}_{s}=4.23\times {10}^{7}\text{\hspace{0.33em}m}\end{array}$

Hence, the orbital radius of the geosynchronous satellite is ${\text{4.23\xd710}}^{\text{7}}\text{\hspace{0.33em}m}$.

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