Is Work Done by the Sun in a Circular Earth Orbit?

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In a perfectly circular orbit, the gravitational force from the sun acts as a centripetal force, which does not perform work on the Earth during its displacement since the force is always perpendicular to the direction of motion. The discussion clarifies that there is no tangential component of the gravitational force in this scenario, meaning no work is done by the sun on the Earth. Even when considering small displacements, the lack of a tangential force component confirms that the work done remains zero. The conclusion emphasizes that in a circular orbit, the gravitational force does not contribute to work done on the orbiting body. Understanding this concept is crucial for grasping the dynamics of celestial mechanics.
Saraharris38
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Okay, this is not a homework question, but a conceptual question already answered in the book.

Problem: Suppose the Earth's orbit around the sun is perfectly circular. What is the work done by the gravitational force of the sun on the Earth through a small displacement over a small time interval?

Answer: No work is done

My problem:

I understand that the centripetal portion of the force does not do any work on the Earth through a small displacement because it is perpendicular and thus the cos factor is zero. However, doesn't the tangential portion of the force act (for very small displacements) essentially parallel to the displacement, and thus do a bit of work?

Thanks!
 
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Saraharris38 said:
However, doesn't the tangential portion of the force act (for very small displacements) essentially parallel to the displacement, and thus do a bit of work?
But there's no tangential component of the force--you are assuming a perfectly circular orbit.
 
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