Earth orbit -- geosynchronous orbit calculation

In summary, the conversation is discussing the correct approach for finding the correct route to solving a circular orbit problem. The equation ε= r_max - r_min / r_max + r_min is mentioned, along with the formula for finding angular velocity. The conversation then delves into using Newton's 2nd law and the force of gravity in order to find the expression for centripetal acceleration. It is suggested to use the formula F= GmM/r^2 and to remember that the acceleration in a circular orbit is ω^2r. The correct expression for a satellite in a circular orbit is mω^2r, not m(r'' + rω).
  • #1
nysnacc
184
3

Homework Statement


upload_2016-9-21_22-36-1.png


Homework Equations



ε= r_max - r_min / r_max + r_min
= r*v2 / gRE

The Attempt at a Solution


I first find that 1 revolution is 2π rad , the angular velocity ω = 2π rad / 24h = π/43200 rad /s
so v = rω where r is the unknown
ε = 0 (circular orbit)

r*(rω)2 / g RE = 0Correct route?
 
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  • #2
Your approach is incorrect. You need to use Newton's 2nd law to find a relation between speed and radius in a circular orbit.
 
  • #3
st
kuruman said:
Your approach is incorrect. You need to use Newton's 2nd law to find a relation between speed and radius in a circular orbit.
start with F= ma, then integrate?
 
  • #4
No need for integration. Remember that the acceleration here is centripetal. What is an expression for F?
 
  • #5
F=ma then a is gravity?
 
  • #6
No, a is the centripetal acceleration. What is an expression for it? F is the force of gravitational attraction between two masses according to Newton's law of gravitational attraction.
 
  • #7
GmM/ r2
 
  • #8
That's F. What about a on the other side of the equation?
 
  • #9
m (r'' + rω) because it is the angular movement ?
 
  • #10
Centripetal acceleration is ω2r, not ωr. Also, what is r'' for a satellite in a circular orbit?
 

1. How is geosynchronous orbit calculated?

Geosynchronous orbit is calculated using a mathematical formula known as Kepler's third law, which relates an object's orbital period to its distance from the central body. The equation is T = 2π√(a³/GM), where T is the orbital period, a is the semi-major axis, G is the gravitational constant, and M is the mass of the central body.

2. What is the difference between geosynchronous and geostationary orbit?

Geosynchronous orbit refers to any orbit around Earth with an orbital period of 24 hours, while geostationary orbit specifically refers to a circular geosynchronous orbit located directly above the equator. This means that an object in geostationary orbit appears stationary from a fixed point on Earth.

3. How high above Earth's surface is geosynchronous orbit?

Geosynchronous orbit is located at an altitude of approximately 35,786 kilometers (22,236 miles) above Earth's surface. This distance is known as the geosynchronous altitude or geostationary altitude.

4. Are all satellites in geosynchronous orbit?

No, not all satellites are in geosynchronous orbit. In fact, most satellites are in much lower orbits, such as low Earth orbit (LEO) or medium Earth orbit (MEO). Geosynchronous orbit is typically reserved for communication and weather satellites due to its stationary position above Earth.

5. How does geosynchronous orbit affect satellite communication?

Geosynchronous orbit is ideal for satellite communication because it allows for a constant line of sight between the satellite and a fixed point on Earth. This means that communication signals do not have to be constantly redirected as the satellite moves, resulting in more reliable and efficient communication. However, geosynchronous orbit also has limitations, such as increased signal delay due to the distance from Earth.

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