Calculating Orbital Radius for Geosynchronous Satellite Using Kepler's Third Law

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In summary, to find the orbital radius of the moon in Earth radii, you use Kepler's Third Law: T^2≈R^3. To find the orbital radius of the satellite in low Earth orbit, you use Kepler's Third Law: T^2≈R^3. To find the orbital radius of the satellite in geosynchronous orbit, you use Kepler's Third Law: T^2≈R^3. To find the orbital radius of the satellite in geosynchronous orbit, you use Kepler's Third Law: T^2≈R^3. To find the orbital radius of the satellite in geosynchronous orbit, you use Kepler's Third Law: T^2
  • #1
tchouhan
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Hello everyone, I'm in a beginner physics class at uni, and after going through this problem over and over, I figured I'd make an account and ask for help since I might be here often.

Homework Statement



The Moon's distance from Earth is approximately 60 Earth radii, and it takes the Moon 27 days to orbit Earth. A satellite just above Earth's atmosphere orbits in 84 min. Show that these data are consistent with Kepler's third law [itex] T^2≈R^3 [/itex]. What radius orbit is required for a satellite to have a period of exactly 1 day? (Such an orbit is called geosynchronous).



Homework Equations



[itex]T^2≈R^3[/itex],

The Attempt at a Solution



I found the orbital radius in astronomical units, it's ##0.17614## AU. I think I'm confused as how to figure out the orbital radius for the satellite. I always get very large numbers, because 1 AU is equal to ~24,000 Earth radii. I don't understand how to show this relationship because I can't even understand it myself! Could someone walk me step by step? I think I am to show the orbital radius of the Moon in Earth radii, but it's ##4135.95## ER, which seems far too large.
 
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  • #2
Bag astronomical units. You are dealing with objects close to the earth.

Set up a ratio using Kepler's Third Law.

You are given for the Moon: T = 27 days, R = 60 Re, where Re = radius of the earth

For the satellite in low Earth orbit: T = 84 min, R is roughly Re, since the atmosphere is relatively thin

Obviously, some units will have to be converted to make a proper ratio.

Then, for the satellite in geosynchronous orbit, T = 1 day, but R = ?

Have a go at this approach.
 
  • #3
I'm mostly confused as to what units go with which. ##27/365.25## would give me the time in years.

But then what do I convert the Earth radius to? Miles? Kilometers?

And for the last part, ##1/365.25## which I then square gives me ##0.00273##. What is this number? Miles? KM?
 
  • #4
For ##T^2 \propto R^3 ##, units do not matter, as long as you use the same units in both cases. That is, you can use, for example, minutes for both the satellite and the Moon, and the Earth's radius for both the satellite and the moon.
 
  • #5
Try using time in days, distance in Earth radii. Then it's just a matter of figuring out what 84 minutes is in units of days and grinding through some simple math. You'll get an answer in Earth radii, which is consistent with the units used in the two given values.
 
  • #6
You're not seeing the big picture. As long as you use consistent units for your ratio, you'll be OK.

For the moon, the period is in days. For the satellite, the period is in minutes. See the mismatch in time units? To be consistent, express all periods in days or minutes, one or the other.

The units of the ratio are immaterial. You want to take a known ratio for one orbital situation so that you can find the unknown component in a second orbital situation.
 
  • #7
Alright, I've got it figured out. Thank you everyone, the conversions were going right over my head.
 

1. What is Kepler's 3rd Law Problem?

Kepler's 3rd Law Problem is a mathematical problem that describes the relationship between the orbital period and distance of a planet or satellite orbiting around a larger body, such as a star or a planet.

2. Who was Johannes Kepler?

Johannes Kepler was a German mathematician, astronomer, and astrologer who is best known for his three laws of planetary motion, including the 3rd Law which is named after him.

3. How is Kepler's 3rd Law Problem solved?

Kepler's 3rd Law Problem is solved by using the equation T^2 = (4π^2/GM) * r^3, where T is the orbital period, G is the gravitational constant, M is the mass of the larger body, and r is the distance between the two bodies.

4. What is the significance of Kepler's 3rd Law Problem?

Kepler's 3rd Law Problem is significant because it allows scientists to calculate the distance between a planet and its star based on its orbital period, or vice versa. This has greatly contributed to our understanding of the solar system and the universe.

5. Are there any limitations to Kepler's 3rd Law Problem?

Yes, there are limitations to Kepler's 3rd Law Problem, as it assumes that the orbiting bodies are in circular orbits and that there are no other external forces acting on them. In reality, most orbits are elliptical and there could be other factors affecting the orbital period and distance.

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