What is the speed of a synchronous satellite in orbit around the Earth?

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SUMMARY

The speed of a synchronous satellite in orbit around the Earth can be calculated using the formula v = √(G*M/radius), where G is the gravitational constant (6.67x10^-11), M is the mass of the Earth (5.98x10^24 kg), and radius is the distance from the center of the Earth. To determine the radius, one must add the Earth's radius (6.38x10^6 m) to the altitude of the satellite. Additionally, the relationship between linear speed and angular frequency is expressed as v = ωr, where ω (omega) represents the angular frequency, defined as ω = 2π/T, with T being the orbital period.

PREREQUISITES
  • Understanding of gravitational constant (G) and its value (6.67x10^-11)
  • Knowledge of Earth's mass (5.98x10^24 kg)
  • Familiarity with the concept of orbital radius and altitude
  • Basic understanding of angular frequency (ω) and its formula (ω = 2π/T)
NEXT STEPS
  • Research the calculation of orbital speed for different types of satellites
  • Learn about the relationship between orbital period and altitude for synchronous satellites
  • Explore the implications of angular frequency in satellite motion
  • Study the effects of gravitational forces on satellite trajectories
USEFUL FOR

Aerospace engineers, astrophysicists, students studying orbital mechanics, and anyone interested in satellite technology and physics.

dxlogan187
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A rocket is used to place a synchronous satellite in orbit about the earth. What is the speed of the satellite in orbit?

I know that G= 6.67x10^-11 and the Mass of Earth is 5.98x10^24 kg.

So I'm to assume that I use the equation

v= Square root of([G*M]/radius)

But I don't know how high off the ground the satellite is. I know the radius of the Earth is 6.38X10^6 m, so I'd add that to how high above the Earth it is to get the radius.

Thanks for your time!
 
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The distance from the center of the Earth is the relevant one. However, you just need to recognize that \omega r = v which you could use to eliminate r from your equation.
 
Tide said:
The distance from the center of the Earth is the relevant one. However, you just need to recognize that \omega r = v which you could use to eliminate r from your equation.

What does the w symbol mean? Never seen that before :confused:
 
It's the Greek letter "omega" and it stands for the angular frequency. You may be more familiar with it in this form:

\omega = \frac {2\pi}{T}

where T is the orbital period.
 

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