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In the notes I received from high school teacher, it says:

How much work is done by the gravitational field of the planet in bringing a satellite from an infinite distance to a position $r$ away?

In general,

$W=\Delta U=\int_{\infty}^{r} F\,dr= \int_{\infty}^{r} \frac{GMm}{r^2}\,dr=-\frac{GMm}{r}$

But since U (at $\infty$) is zero, we can say that the work done in bringing a mass from an infinite distance away to the point $r$ is equal to its potential energy U.

I see some problems with this. I’ve been reading uni textbooks to enhance my understanding, and every single one of them says that $W_{grav}=-\Delta U$, and they also have $W= \int_{\infty}^{r} -\frac{GMm}{r^2}\,dr$ with a negative sign in front of the force. I’ve been trying to decode what my high school teacher has said in his notes. Is he talking about the work done by the gravitational field on the mass? In that case his working out is incorrect right? Or is he trying to figure out the work done by some other force to move the object from infinity towards r?

Also, I don’t quite understand why uni textbooks put a minus sign in front of the force.

Thanks for reading this post, help much appreciated.

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2 Answers 2

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No,the first statement is incorrect. It does not represent the work done by gravity in bringing the satellite closer to the planet; rather, it is a measure of the work done by an external force in bringing the two together, such that each one does not gain any kinetic energy. In simpler terms, it is the work done by an external force, equal in magnitude but opposite in direction to the gravitational force, such that the two objects move towards each other with zero acceleration (constant velocity), so the change in kinetic energy $ΔK = 0$. In many books, you may notice that they say that the objects move towards each other slowly. That is just another way of saying this.

The expression $W_{grav} = -ΔU$ is correct; In fact, for any system, the work done by a conservative force that is internal to the system is given by the expression: $$W_{int} = -ΔU $$ In this case, the system comprises of the planet and the satellite, and the internal force is the gravitational force of attraction, which is also conservative.

Also, I don’t quite understand why uni textbooks put a minus sign in front of the force.

This is because the force is attractive, and hence the direction of force is opposite in direction to the radius vector joining the two objects.

Also be sure to note that the gravitational potential energy of a two particle system is given by: $$U_g(r) = -\frac{G.m_1.m_2}{r}$$ Assuming that at infinity, the potential energy is zero. Notice the negative sign, as a result of $F(r)$ being negative, as correctly stated in the books you have read. The correct way of finding the work done by gravity is then: $$W_{grav} = -ΔU_{g} = -(-\frac{G.m_1.m_2}{r} - 0) = \frac{G.m_1.m_2}{r}$$

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Basically, this problem boils down to perspective.

Your high school teacher told you, how much work 'YOU ( The satellite )' did, in bringing the satellite (itself). The negative sign implies that work was done on it instead of it doing any work.

If the work done by a body is negative, work is being done 'on' the body, instead of 'by' the body.

The physics books are telling you the same story from the perspective of the gravitational field, which is doing positive work on the satellite. Hence the negative sign, that ultimately makes the work done, positive.

So, in the first part, you found work done by satellite on the field, and the second part, work done by gravity on the satellite, and they have opposite signs, because one of them is doing work, and the other has work being done on it.

So, work done by a conservative force is always negative grad U. I hope you understand the difference.

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