## Related questions with answers

A satellite that always looks down at the same spot on the Earth’s surface is called a geosynchronous satellite. a. Find the distance of this satellite from the surface of the Earth. b. Could the satellite be looking down at any point on the surface of the Earth?

Solution

Verified$\textbf{.a)}$ Since the satellite always looks down at the same spot on the Earth, it must be having the same period as the Earth, namely (24 hours). The period of the satellite can be written as follows

$T^{2}=\frac{4\pi^{2} r^{3}}{GM}$

Rearrange the relation to isolate $(r)$

$\begin{align*} r&=\left(\frac{T^{2} GM}{4\pi^{2}}\right)^{1/3}\\ &=\left( \frac{(6.67 \times 10^{-11} \mathrm{~ m^{3}/kg \cdot s^{2}}) \times (5.98\times 10^{24} \mathrm{~ kg}) \times (24\times 3600 \mathrm{~ s})^{2}}{4\pi^{2}}\right)^{1/3}\\ &=4.22\times 10^{7}\mathrm{~ m} \end{align*}$

Notice that this is the distance from the center of the Earth, so the distance from the surface of the Earth can be determined as follows

$r=4.22\times 10^{7}\mathrm{~ m} - 6.38\times 10^{6} \mathrm{~ m}$

$r=3.6 \times 10^{7} \mathrm{~ m}$

$\textbf{.b)}$ The satellite can't be looking down at any point, it has to be above the equator so it has the same period as the Earth.

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