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] (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]
[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] (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] (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] ⇒ [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.
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 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.98 x 10²⁴ kg
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 10⁷ m = 42250 km
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.
