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I know that the work done in a closed system is $0$ for this particular case. Also, I know that the work done is independent on the path traced but just the initial and final position.

With these conditions, if a planet while revolving around the sun due to gravitational force goes from point $A$ to point $B$, then is the distance taken as the circular path traveled or just the displacement from $A$ to $B$ for a calculation of work done?

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  • $\begingroup$ Since gravity is a conservative force, it shouldn't matter what path you choose to calculate work. i.e. using either $|{\vec{B} - \vec{A}}|$ or the circular path should give the same result. $\endgroup$
    – hsnee
    Commented Dec 27, 2016 at 18:35

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Because of the radial symmetry of the conservative gravitational field of the sun, the only distance that matters with respect to the work is the radial distance. The work done to move the planet on an equipotential surface is always zero because the field is conservative (it has a potential).

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  • $\begingroup$ Do you mean to say work done=Radial distance*Gravitational field in this case?Sorry if this is stupid I am a highschooler. $\endgroup$
    – Suraz Basnet
    Commented Dec 28, 2016 at 4:20
  • $\begingroup$ Gravitational force* $\endgroup$
    – Suraz Basnet
    Commented Dec 28, 2016 at 4:21
  • $\begingroup$ @SurazBasnet Well that would be true if the field was homogenous, but it's not because the gravitational force is proportional to $\frac{1}{r^2}$... Thus, Work = Difference of Potential Energies ($U(\vec B)-U(\vec A)$) = the line integral from point $\vec A$ to point $\vec B$ along any path. That is wriiten in formulas: $\endgroup$
    – Sciencetastic
    Commented Dec 29, 2016 at 12:52
  • $\begingroup$ $W =\int_{\text{path}} \vec{F} \cdot \mathrm{d}\vec{x} = U(\vec B)-U(\vec A)$ where the path is the trajectory taken from $\vec A$ to $\vec B$. Because the gravitational field is conservative, the path you take from $\vec A$ to $\vec B$ doesn't matter at all, but remember that only the radial distance matters in this case because $W= U(\vec B)-U(\vec A)= U(\vec r_B)-U(\vec r_A)={\frac {GMm}{r_B}}-{\frac {GMm}{r_A}}$ where $U(\vec r)$ is the potential energy at point $\vec r$. $\endgroup$
    – Sciencetastic
    Commented Dec 29, 2016 at 13:08
  • $\begingroup$ But @SurazBasnet, your assumption that $W=(B-A)\cdot F_G$ is correct if your body is moving near the surface of the Earth, where the gravitational force can be considered constant $F_G=-mg$. $\endgroup$
    – Sciencetastic
    Commented Dec 29, 2016 at 13:17

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