Answer:
The radius R of the orbit of the geosynchronous satellite is [tex]4.225x10^{7} m[/tex]
Explanation:
By means of the equation of the orbital speed, the orbital period can be known:
[tex]v = \frac{2 \pi r}{T}[/tex] Â (1)
Where r is the orbital radius and T is the orbital radius
r can be isolated from equation 1:
[tex]T \cdot v= 2 \pi r[/tex] Â
[tex]r = \frac{T \cdot v}{2 \pi}[/tex] (2)
Notice that, from equation 2 is necessary to find the velocity before the orbital radius can be determined. That can be done through the Law of Universal gravity.
[tex]F = G\frac{M \cdot m}{r^{2}}[/tex] (3)
Then, replacing Newton's second law in equation 3 it is gotten:
[tex]m \cdot a= G\frac{M \cdot m}{r^{2}}[/tex] (4)
However, a is the centripetal acceleration since it is a circular motion:
[tex]a = \frac{v^{2}}{r}[/tex] Â (5)
Replacing equation 5 in equation 4 it is gotten:
[tex]m \frac{v^{2}}{r} = G\frac{M \cdot m}{r^{2}}[/tex]
[tex]m \cdot v^{2} = G \frac{M \cdot m}{r^{2}}r[/tex]
[tex]m \cdot v^{2} = G \frac{M \cdot m}{r}[/tex]
[tex]v^{2} = G \frac{M \cdot m}{rm}[/tex]
[tex]v^{2} = G \frac{M}{r}[/tex]
[tex]v = \sqrt{\frac{G M}{r}}[/tex] Â (6)
Where v is the orbital speed, G is the gravitational constant, M is the Earth mass, and r is the Earth radius.
Finally, equation 6 can be replaced in equation 2:
[tex]r = \frac{T \cdot v}{2 \pi}[/tex]
[tex]r = \frac{T \sqrt{\frac{G M}{r}}}{2 \pi}[/tex]
[tex]r^{2} = \frac{T^{2} G M}{4 \pi^{2} r}[/tex]
[tex]r^{3} = \frac{T^{2} G M}{4 \pi^{2}}[/tex]
[tex]r = \sqrt[3]{\frac{T^{2} G M}{4 \pi^{2}}}[/tex] (7)
Where r is the orbital radius, T is the orbital period, G is gravitational constant and M is the Earth mass.
Since it is a geosynchronous satellite, it will have the same orbital period of the Earth (24 hours).
It is necessary to express the period in seconds:
[tex]T = 24 h x \frac{3600 s}{1 h}[/tex] â [tex]86400 s[/tex]
[tex]T = 86400 s[/tex]
Hence, all the values can be replaced in equation 7:
[tex]r = \sqrt[3]{\frac{(86400 s)^{2}(6.67x10^{-11}N.m^{2}/kg^{2})(5.98x10^{24}kg)}{4 \pi^{2}}}[/tex]
[tex]r = \sqrt[3]{7.54x10^{22}} m^{3}[/tex]
[tex]r = 4.225x10^{7} m[/tex]
So the radius R of the orbit of the geosynchronous satellite is [tex]4.225x10^{7} m[/tex]
This question involves the concepts of the time period, orbital radius, and gravitational constant. Â
The radius of the orbit of the geosynchronous satellite is "42250 km".
The theoretical time period of the comet around the earth can be found using the following formula:
[tex]\frac{T^2}{R^3}=\frac{4\pi^2}{GM}[/tex]
where,
T = Time Period of Satellite = 24 h = 86400 s (geosynchronous have same time period as Earth)
R = Orbital Radius = ?
G = Gravitational Constant = 6.67 x 10âťÂšÂš N.m²/kg²
M = Mass of Earth = 5.98 x 10²ⴠkg
Therefore,
[tex]\frac{(86400\ s)^2}{R^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.98\ x\ 10^{24}\ kg)}\\\\R = \sqrt[3]{\frac{(86400\ s)^2(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.98\ x\ 10^{24}\ kg)}{4\pi^2}}[/tex]
R = 4.225 x 10⡠m = 42250 km
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Learn more about the orbital time period here:
brainly.com/question/14494804?referrer=searchResults
The attached picture shows the derivation of the formula for orbital speed.
