Apparent Weight and Geosynchronous Orbit

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Communications satellites in geosynchronous orbits maintain a fixed position over the equator, with an altitude of 3.58 x 10^7 meters and a motion period of 24 hours. The gravitational acceleration at this altitude is approximately 0.223 m/s². The apparent weight of a satellite, however, is influenced by the centripetal force required to maintain its circular orbit. The discussion highlights that the perceived weight is not the actual weight but rather the normal force acting on the satellite. Understanding these forces is crucial for calculating the satellite's apparent weight in orbit.
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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