Given two points in 3-space, $A = (9, 1, 5)$ and $B=(2,8,7)$, the work done by the gravitational field $\bf{F}$ when an object is moved from point $A$ to point $B$ comes out positive, even though it is work done against gravity! What am I missing here?

$\bf{F}$$=-\nabla V$, $V=-\frac{GMm}{r}$, $r=\sqrt{x^2+y^2+z^2}$. Using the Second Fundamental Theorem for Line Integrals, we have: $V(2,8,7)-V(9,1,5)=-\frac{GMm}{\sqrt{117}}+\frac{GMm}{\sqrt{107}}$, which is positive! But the distance $\|B\|$ is bigger than $\|A\|$. Something is not right, please help.

  • $\begingroup$ So? What's the problem in that? Work done will always come out positive if the direction that the force is exerted in is the same as the direction of displacement of the object. $\endgroup$ – Gummy bears Oct 16 '15 at 7:26
  • $\begingroup$ @Gummy bears No, the work is done against the force in this case. $\endgroup$ – raul Oct 16 '15 at 21:05
  • $\begingroup$ @raul Never checked that. Might be. $\endgroup$ – Gummy bears Oct 17 '15 at 4:47

When you write the formula $$V=-\frac{GMm}{r}$$ you have assumed already that $V$ is 0 at $r= \infty$.

So as you approach the source of the field, the field is doing work on you and hence your potential is decreasing or rather increasing in the negative sense.

So the direction in which you attain stability in this force is the direction in which your potential is decreasing i.e. the force is pulling you. Now, as in the above case, you are trying to oppose the force and move in the opposite direction. So you have to do work against the field.

Hence you are increasing your potential or rather decreasing it in the negative sense.

And you know, $\mathbf{work \,\ done = - (change \,\ in \,\ potential \,\ energy) = -[(more -ve) - (less -ve)] = -[something (-ve)] = positive}$
(Your initial potential is more -ve than your final potential after displacement)

Hope this helps you.


Actually, is simpler than that.

The Second Fundamental Theorem for Line Integrals states:

Suppose that $C$ is a smooth curve (in your case, this could the the line linking the points) given by $\vec r(t)$, where $a<t<b$. Also suppose that $f$ is a function whose gradient vector, $\nabla f$, is continuous on $C$. Then,

\begin{equation} \displaystyle\int_C\nabla f\cdot d\vec r=f(\vec r(b))-f(\vec r(a)). \end{equation}

So there you go. You haven't taken into account the minus sign in $\textbf F=-\nabla V$



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