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in my revision guide it defines gravitational potential difference as:

The gravitational potential difference is work done in moving a unit mass.

It then goes on to explain the gravitational potential difference between two points, etc. However, this left me a bit confused. Shouldn't the definition be:

The gravitational potential difference between two points is the work done in moving a unit mass from one point to the other.

(Could we also add: Where we take the positive direction as that of increasing potential?)

Please clarify, thanks!

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2 Answers 2

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I have highlighted some key word lacking in your revision. Also, work has a very specific definition. The difference in gravitational potential difference between $\vec{r}_1$ and $\vec{r}_2$ is the negative of the work done on a unit mass by the external gravitational field as the unit mass moves from $\vec{r}_1$ to $\vec{r}_2$.

As an example, consider a mass $M$ located at x=0 and producing the gravitational field in a space. A unit mass moving from $x_1 > 0$ to $x_2 > x_1$ will have work done on it by the field: $$W = \int \vec{F}\cdot d\vec{r} =\int_{x_1}^{x_2} \frac{-GM(1)}{x^2} dx= GM\left(\frac{1}{x_2}-\frac{1}{x_1}\right).$$

The difference of gravitational potential is the negative of this. This makes sense because the field is doing negative work on the unit mass as it moves away from $M$. The gravitation potential is getting higher which agrees with moving opposite the force.

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Words are imprecise. Your wording is much less imprecise than the one from the textbook. As the work is positive or negative (depending on the direction taken), there is no need to define a "positive direction", but the potential difference depends on the order of $\vec r_1$ and $\vec r_2$ (that is: if the potential difference moving from 1 to 2 is positive, the one for moving from 2 to 1 will be its opposite).

My try of a precise definition:

The graviational potential difference $V(\vec r_2) - V(\vec r_1)$ between the points $\vec r_1$ and $\vec r_2$ is given by: $$V(\vec r_2) - V(\vec r_1) = -\int_{\vec r_1}^{\vec r_2} d\vec r \cdot \vec F_G(\vec r)$$ where $\vec F_G(\vec r)$ denotes the gravitational force on the unit mass at the point $\vec r$. It is a remarkable property of the gravitational force, that the gravitational potential differnce does not depend on the path taken from $\vec r_1$ to $\vec r_2$.

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  • $\begingroup$ You've got a sign error. You need a negative sign in front of the integral to account for the fact that moving against the force of gravity takes you to a higher gravitational potential. $\endgroup$
    – Mark H
    Commented May 7, 2015 at 22:02

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