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Let us assume that a gravitational field is created by a mass $M$. An agent is bringing a unit mass from $\infty$ to distance $r < \infty$, both measured from mass $M$.

The agent is always forcing the unit mass with a continuously changing force $\vec F(\vec x)$, $\vec{x}$ being the distance pointing radially out from $M$.

According to classical mechanics, it holds that $\vec F(\vec x) = \frac{GM}{x^2}\hat{x}$, with $G$ being the gravitational constant.

The work is calculated as follows: $$W = \int_\infty^r\vec F(\vec x)\cdot d\vec x$$ $$=\int_\infty^r{{F(x)}\,dx\cdot cos(\pi)}$$ $$=-\int_\infty^r{{\frac{GM}{x^2}}dx}$$ $$=-GM[-\frac{1}{x}]_\infty^r$$ $$=GM[\frac{1}{x}]_\infty^r$$ $$=GM[\frac{1}{r}-\frac{1}{\infty}]$$ $$=\frac{GM}{r}$$

The body moved against the force's direction (the angle between them was always $\pi$). So the work should have been negative. But since $r$ is the scalar distance from $M$, it is positive like $G$ or $M$, yielding the result always positive.

What is wrong here?

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The agent is always forcing the unit mass with a continuously changing Force, $\vec{F}$(x) ... = $\frac{GM}{x^2}\hat{x}$

By your force definition, the agent is not the attractive gravitational force but is something which is restricting the motion to constant velocity because the mass M is pulling in the $-\hat{x}$ direction with a force equal in magnitude to the gravity but opposite in direction. That's okay, but I wanted to state that explicitly. Also, you are calculating only the work done by that agent.

You also have defined the positive direction to be away from $M$, and that's okay, too.

Your work integral calculates the work done by the force of the agent which is holding the mass back from accelerating toward $M$. Notice that, with your symbols, $$W = \int_{\infty}^r \frac{GM}{x^2}\hat{x}\cdot dx(\hat{x})= \int_{\infty}^r \frac{GM}{x^2}~ dx.$$

The $\cos \pi$ fator you have is incorrect. The infinitesimal $dx\hat{x}$ in an integral defines the direction of the positive coordinate change, not the direction of the motion. The direction of motion is contained in the integration limits.

The result of the integral (for a unit mass being moved) is $$W = \left.\frac{-GM}{x}\right|_{\infty}^r= \frac{-GM}{r}.$$

The negative value makes sense because the agent is restraining the motion and acting in the positive $x$ direction while the motion is in the negative $x$ direction. And because the object is moving at constant velocity, the work done by the gravitational field will be the negative of the above so that the net work is zero, in agreement with the work-energy principle: $$\Delta K = W_{net}$$

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  • $\begingroup$ Yes, assuming constant velocity, I agree and I think this is a great answer. But - why constant velocity? Maybe I'm not familiar with agents. $\endgroup$ – zonksoft Apr 29 at 19:02
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    $\begingroup$ I like to give the calc 1 analogy that if we are integrating $f(x)$ from $x=5$ to $x=3$ we do not do $$\int_5^3f(x)\,(-\text dx)$$ we just do $$\int_5^3f(x)\,\text dx$$ The sign (direction) of $\text dx$ is already determined by the limits. $\endgroup$ – BioPhysicist Apr 29 at 19:36
  • $\begingroup$ Thanks for response, I understand that my mistake was to think dx as a tiny displacement of the body. But since u pointed out, that dx is not the displacement vector (instead refers to the increase in x vector), then why should we do dot product with something that is not the displacement vector? @Bill N $\endgroup$ – Imtiaz Kabir Apr 29 at 22:23
  • $\begingroup$ @ImtiazKabir d$\vec{x}$ is an infinitesimal displacement in a positive direction. The actual resulting displacement is determined by the limits of integration. $\endgroup$ – Bill N Apr 30 at 1:15
  • $\begingroup$ @zonksoft There isn't any particular reason to have constant velocity. That's what the OP chose by his force function. The word "agent" simply refers to whatever is providing the force. The force that he defined can't be the gravitational attraction because it's in the wrong direction, but he can define a force anyway he want to. He chose to define it to be equal and opposite to the gravitational force acting on the unit mass. $\endgroup$ – Bill N Apr 30 at 1:19
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You made a mathematical error in trying to prove the result. It arises in many scenarios.

To give you an insight into your mistake I would like to tell you the correct method of integration in physics.

Remember that we always consider an element $dx$ at a distance $x$ from origin in the direction of $x$. What you did was physically correct but while considering the field at $x$ you displaced the unit mass by $dx $ opposite to $x$.

Always remember that for your physics to be mathematically correct consider $dx$ in the direction of $x$. If you are taking $dx$ opposite to $x$ as you did you will have to integrate with a negative sign.

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Your calculation looks like you want to calculate the work done by gravitation. In that case, you're missing a sign. Newton's law of universal gravitation as a vector equation is actually ($m=1$ in your example):

$$\vec F(\vec x) =- G {M m \over {\vert \vec x \vert}^2} {\hat{x}}$$

The missing sign perpetuates through the whole calculation, leading to the correct sign at the end.

In any case, you are calculating the work done by the gravitational field - if you want to take some other force into account (you are talking about "forcing the unit mass with a continuously changing force"), this is not part of your calculation. In that case, you need to define a second force $\vec F_2$ (e.g. a propulsion engine of a rocket) and then both those forces do work. To conclude, forces doing work add up.

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    $\begingroup$ It is correct because he is talking about a force which is in opposite direction of gravitational force $\endgroup$ – Shreyansh Pathak Apr 29 at 17:56
  • $\begingroup$ Thank you Sarcasm, I did some more clarification! $\endgroup$ – zonksoft Apr 29 at 18:10
  • $\begingroup$ OP didn't say the agent was gravity, so it is an attempt to calculate work done by the agent. OP is not consistent with vector directions. $\endgroup$ – Bill N Apr 29 at 18:29
  • $\begingroup$ Yes, I noticed the vector direction thing... ironically, the rest of the calculation after the Newtonian Gravity sign error checks out imho. If there is another force of some kind (e.g. an agent), one would define an $\vec F_2(\vec x)$, add it to the gravitational field and then do the line integral using the sum of the forces. $\endgroup$ – zonksoft Apr 29 at 18:36
  • $\begingroup$ The $\cos \pi$ factor is wrong. $\endgroup$ – Bill N Apr 29 at 18:49

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