# Altitude of Geostationary Orbit

• duchuy
In summary, the conversation discusses how to determine the value of R for an object in geostationary orbit using the formula T = 2piR / v, with T being the revolution period of the satellite, R being the distance between the center of masses, and v being the velocity. The participants also mention Newton's gravitation law and circular movement, as well as the relationship between centripetal force and gravitational force. They suggest equating these forces and using an expression for T to find the value of R.
duchuy
Homework Statement
Deduce at what distance from the center of the Earth are positioned the geostationary satellites which, observed from the terrestrial frame of reference, are motionless in the sky.
Relevant Equations
T = 2piR / v
Hi,
They gave me this formula T = 2piR / v, with T the revolution period of the satellite, R the distance between the center of masses and v the velocity.
They gave me the value of G and the eath's mass and asked to determine the value of R.
I don't even see fromwhere I should start...

Have you studied what Mr. Newton said about gravitation?

berkeman
Gordianus said:
Have you studied what Mr. Newton said about gravitation?
Probably I have, but I'm not quite sure how I would use those formulas for an object in geostationary orbit... and I'm taking physics as a minor so I don't really remember much

O.K. Can you make a search about Newton's gravitation law and circular movement?

And, do you know what a 'geostationary' orbit is.

If a satellite appears to be stationary above the earth, what must its period T be? Can you also write an expression for v in terms of T and R? Then you'd be left with an expression with R as the only unknown.

BUT

Usually to tackle a Q like this, rather than using that equation you've quoted, I'd start from the fact that the gravitational force on the satellite is what provides the centripetal force. So if you can write an expression for each of these (I usually prefer the mrω² version for centripetal force), and equate them, you can do it from there - though you still need to know T of course.
In the end, it will get you to the same place, but doing it this way helps you to understand and therefore remember how this works.

Keith_McClary

## 1. What is the altitude of a geostationary orbit?

The altitude of a geostationary orbit is approximately 35,786 kilometers (22,236 miles) above the Earth's equator.

## 2. Why is the altitude of a geostationary orbit important?

The altitude of a geostationary orbit is important because it allows satellites to match the Earth's rotation and remain in a fixed position above a specific location on the Earth's surface.

## 3. How does the altitude of a geostationary orbit affect satellite communication?

The altitude of a geostationary orbit allows satellites to maintain a constant line of sight with ground-based communication devices, making it ideal for satellite communication.

## 4. Can the altitude of a geostationary orbit change?

Yes, the altitude of a geostationary orbit can change due to factors such as atmospheric drag or gravitational forces from other celestial bodies. However, adjustments are typically made to maintain the orbit's position above a specific location on Earth.

## 5. What is the difference between the altitude of a geostationary orbit and a low Earth orbit?

The altitude of a geostationary orbit is much higher than a low Earth orbit, which typically ranges from 160 to 2,000 kilometers (99 to 1,243 miles) above the Earth's surface. Geostationary orbits also have a much longer orbital period, taking 24 hours to complete one orbit compared to approximately 90 minutes for a low Earth orbit.

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