Apparent Weight and Geosynchronous Orbit

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SUMMARY

Geosynchronous orbits, positioned at an altitude of 3.58 x 107 m, allow communications satellites to remain fixed over a point on the equator as the Earth rotates. The orbital period for such satellites is 24 hours, and the local gravitational acceleration (g) at this altitude is approximately 0.223 m/s². The apparent weight of a 2000 kg satellite in this orbit is not directly calculable using standard weight equations, as the perceived weight is influenced by the centripetal force required for circular motion rather than the gravitational force alone.

PREREQUISITES
  • Understanding of geosynchronous orbits and their characteristics
  • Familiarity with gravitational acceleration and its calculation
  • Knowledge of centripetal force and its role in circular motion
  • Basic physics concepts related to weight and normal force
NEXT STEPS
  • Study the relationship between gravitational force and centripetal force in orbital mechanics
  • Learn how to calculate apparent weight in non-inertial reference frames
  • Explore the physics of satellite motion and the effects of altitude on gravitational acceleration
  • Investigate the implications of weight perception in varying gravitational fields
USEFUL FOR

Students of physics, aerospace engineers, and professionals involved in satellite technology and orbital dynamics will benefit from this discussion.

JeYo
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Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the Earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58 x 10^7m. What is the period of the motion? What is the value of g at this orbit? What is the apparent weight of a 2000kg satellite at this g?




Now, I found the period to be 24hr, and the value of g to be 0.223m/s/s. But I cannot seem to find a formula to correctly calculate the apparent weight of the satellite with this information.
 
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I have already tried these equations: W = mω^2r and W = ma, with a being 0.223 m/s/s
 
JeYo said:
Now, I found the period to be 24hr, and the value of g to be 0.223m/s/s. But I cannot seem to find a formula to correctly calculate the apparent weight of the satellite with this information.

Trick question! I also get your answer for the local gravitational acceleration of 0.224 m/(sec^2). However, keep in mind that the local "weight force", which is the magnitude of the gravitational force acting on the satellite, is what is supplying the centripetal force keeping it on its circular orbit.

Think about how weight is perceived. When you stand on a scale, what you read off is not your weight, but the "normal" force applied upward by the spring inside the scale supporting your weight. What you perceive as your weight when you are standing on a floor or the ground is the effect of the normal force from that surface pushing up against the soles of your feet; it's not actually your "weight" that you feel, but a force of equal magnitude.

If you jumped out of a high window (don't do it!), you would indeed have a weight force acting on you, accelerating you toward the Earth (while the Earth also immeasurably accelerates toward you), but what would you feel? By the same token, what normal force is pushing back on the satellite? If you were in that orbit (inside a spacesuit, presumably), what would you feel?
 

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