Calculating the Force Opposing Gravity in Orbital Simulations

AI Thread Summary
To calculate the force opposing gravity in orbital simulations, consider the scenario of the Earth and the Sun. The Earth's movement at approximately 1.47 kilometers per second generates a centrifugal force that counteracts the Sun's gravitational pull. The formula v^2/r can be used to determine this opposing force, where v is the velocity and r is the radius of the orbit. A quick method is to reference the Sun's gravitational force on the Earth, as both forces are in equilibrium and roughly equal. Understanding these forces is crucial for accurate orbital simulation programming.
bkelly
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I am writing a program to do some orbital simulations. For ease of conversation, let's start with the sun and the earth. If the Earth were standing still, the sun would pull it in quite rapidly. As the Earth is moving at about 1.47 kilometers per second, that movement creates a force opposing the suns gravity. How do I calculate that force?
 
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v^2/r ?
 
Quick and dirty way: look up the Sun's gravitational force on the Earth (the two forces are in equilibrium, so they are roughly the same amount of "force").
 
comparing a flat solar panel of area 2π r² and a hemisphere of the same area, the hemispherical solar panel would only occupy the area π r² of while the flat panel would occupy an entire 2π r² of land. wouldn't the hemispherical version have the same area of panel exposed to the sun, occupy less land space and can therefore increase the number of panels one land can have fitted? this would increase the power output proportionally as well. when I searched it up I wasn't satisfied with...

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