Geosynchronous Orbit and Radius Calculation for Satellite | 8 Earth Radii

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In summary: I think I've got it now.In summary, a satellite in geosynchronous orbit remains above the Earth's equator and has an average radius of approximately 7 times the Earth's radius. However, the more accurate way to calculate the radius would be to use Kepler's Third Law by setting the orbital period equal to the Earth's rotational period and solving for the radius. The equation for this is GM/r^2 = 4rπ^2/T^2.
  • #1
UrbanXrisis
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A satellite in geosynchronous orbit remains above the Earth's equator as the planet rotates on its axis.

a) calculate the radius of its orbit.

does this mean from the center of the earth?
I looked online and the distance is about 7 Earth radii, which would make the radius of the orbit 8 Earth radii. Earth's radius is 6.37x10^6m.. makeing the radius 5.096x10^7m.

what does it mean by calculate the orbit?

b) the satellite relays a radio signal from a transmitter near the noth pole to a receiver, also near the north pole. Traveling at the speed of light, how long is the radio wave in trasit?

d/v=t
[tex]d=\frac{\sqrt{r_{earth}^2+r_{satellite}^2}}{c}[/tex]
 
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  • #2
UrbanXrisis said:
A satellite in geosynchronous orbit remains above the Earth's equator as the planet rotates on its axis.

a) calculate the radius of its orbit.

does this mean from the center of the earth?
I looked online and the distance is about 7 Earth radii, which would make the radius of the orbit 8 Earth radii. Earth's radius is 6.37x10^6m.. makeing the radius 5.096x10^7m.

what does it mean by calculate the orbit?

b) the satellite relays a radio signal from a transmitter near the noth pole to a receiver, also near the north pole. Traveling at the speed of light, how long is the radio wave in trasit?



d/v=t
[tex]d=\frac{\sqrt{r_{earth}^2+r_{satellite}^2}}{c}[/tex]
a) For use in formulas relating to gravitational forces and geosynchronous orbits, the orbit radius is that measured from the Earth's CENTER.
Hint: At what orbit radius will the gravitational force between Earth and satellite provide the exact centripetal force required for the satellite's rotational velocity "ω" (in radians/sec) to equal Earth's rotational "ω" (in radians/sec)??

b) Remember that the radio signal path is 2-way ("uplink" and "downlink").


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Last edited:
  • #3
UrbanXrisis said:
A satellite in geosynchronous orbit remains above the Earth's equator as the planet rotates on its axis.

a) calculate the radius of its orbit.

does this mean from the center of the earth?
I looked online and the distance is about 7 Earth radii, which would make the radius of the orbit 8 Earth radii. Earth's radius is 6.37x10^6m.. makeing the radius 5.096x10^7m.

what does it mean by calculate the orbit?

b) the satellite relays a radio signal from a transmitter near the noth pole to a receiver, also near the north pole. Traveling at the speed of light, how long is the radio wave in trasit?

d/v=t
[tex]d=\frac{\sqrt{r_{earth}^2+r_{satellite}^2}}{c}[/tex]
You may have misunderstood what they meant by the 'distance' for a geosynchronous satellite. A geosynchronous satellite has an average radius of 42,164 km, or approximately 7 times the Earth's radius. In other words, they didn't mean for you to add that distance to the Earth's radius.

For your problem, multiplying the Earth's radius by 7 gets you an approximate answer, provided that's the way they want you to calculate the radius.

However, I'm not sure that's the way they wanted you to calculate the radius. Since you have thread about Kepler's Second Law, does that mean you learned Kepler's Third Law, as well? If so, in order for a satellite to stay over the same spot on the Earth, it must complete one orbit in the same amount of time the Earth completes one complete rotation (86164 seconds). The orbital period is known, so you can rearrange the equation for orbital period to solve for the radius, instead, and will come much closer to an accurate answer than the rough approximation you found online.
 
  • #4
thx with the help...
[tex]\frac{GM}{r^2}=\frac{4r \pi^2}{T^2}[/tex]

that's the equation I was trying to figure out
 

1. What is a geosynchronous orbit?

A geosynchronous orbit is a type of orbit in which a satellite orbits the Earth at the same speed and direction as the Earth's rotation, so it appears to stay in the same position relative to a fixed point on Earth's surface.

2. How is a geosynchronous orbit different from a geostationary orbit?

A geostationary orbit is a specific type of geosynchronous orbit in which the satellite's orbital plane is directly above the Earth's equator, making it appear to stay in the same position above the Earth's surface at all times. A geosynchronous orbit, on the other hand, can have an orbital plane at any inclination, as long as it maintains the same orbital period as the Earth's rotation.

3. What is the formula for calculating the radius of a satellite's geosynchronous orbit?

The formula for calculating the radius of a geosynchronous orbit is R = 8 * Earth's radius, where R is the distance from the center of the Earth to the satellite in kilometers.

4. How is the radius of a geosynchronous orbit related to the satellite's orbital period?

The radius of a geosynchronous orbit is directly proportional to the satellite's orbital period. This means that as the radius increases, so does the orbital period, and vice versa.

5. What factors can affect the accuracy of calculating the radius of a geosynchronous orbit?

The accuracy of calculating the radius of a geosynchronous orbit can be affected by factors such as atmospheric drag, solar radiation pressure, and gravitational pull from other celestial bodies. Additionally, small variations in the Earth's rotation and shape can also impact the accuracy of the calculation.

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