# Why does gravity becomes negative when finding the negative derivative of potential? (I know $F=-dU/dx$) Why is $F$ itself negative?

By definition of the gravitational energy:

$$-\frac{dU}{dx}=-\frac{GMm}{x^2}=F$$

Then by integrating we get:

$$U(x)=\int\frac{GMm}{x^2}=-\frac{GMm}{x}$$

(I am not confused why force is negative derivative of the potential function)

I am confused about the negative sign in front of the force here in the first equation.I know that in force vector there should be a negative sign to indicate the opposite direction to unit vector. But isn't the whole equation just in scalar form?

(See both function $U$ and variable $x$ are scalars)

So why do we put minus sign here? Even if it is in vector form,how do we define the direction $dx$ and how does it relate to the unit vector in $F$ (which makes a minus sign in F's vector equation).

I know that physically work done equals to negative change in potential function, and it makes sense in this equation. But I wish to get a mathematical explanation.

And, I have an another question, when using line integral to find work done by the field:

$$w=\int_c \mathbf{F}\cdot\mathrm{d}\mathbf{r}$$

Where $\mathrm{d}\mathbf{r}$ here is the element for position vector, if we goes from a larger radius to a smaller radius, the force vector should have same direction to change in position vector, therefore the work done:

$$w=\int_c \mathbf{F}\cdot\mathrm{d}\mathbf{r}=\int_{R_2}^{R_1}\frac{GMm}{R^2}\mathrm{d}R=\frac{GMm}{R_2}-\frac{GMm}{R_1}$$

When I goes from smaller radius to a lager radius, the formula becomes:

$$w=\int_c \mathbf{F}\cdot\mathrm{d}\mathbf{r}=\int_{R_1}^{R_2}-\frac{GMm}{R^2}\mathrm{d}R=\frac{GMm}{R_2}-\frac{GMm}{R_1}$$

Where $\mathrm{d}\mathbf{r}$ is positional vector, $\mathrm{d}R$ is the radius element and $R_2>R_1$. The position vector should be reversible to each other because of the direction we choose. See the work done here becomes same, but we know that gravity is conservative, so the two outcomes here should just be reversible to each other.

What goes wrong with my deductions here? Hope that you can solve my problem, and please forgive my ignorance as I am just a high school student.

• – Qmechanic Jun 10 '18 at 15:07
• Here’s the MathJax tutorial so you can embed and edit mathematical type without importing images. – gen-z ready to perish Jun 10 '18 at 15:18
• So your title question (v7) is effectively: Why ${\bf F}$ and ${\bf r}$ have opposite directions? Answer: Because gravity is attractive. – Qmechanic Jun 10 '18 at 15:31

Negative sign in the expression of force of gravitation simply implies attractive nature of this force. It indicates that the force tends to decrease the separation between to particles. This is answer of your first question.

When we use a minus sign to write, as a scalar, the gravitational force on a 'test' mass, m, we are really referring to the component of the force along the line $\vec{\text{Mm}}$, a line from the 'source' mass, M, going through m. Because the force on m is in the direction $\vec{\text{mM}}$, we have$$F_r=-\frac{GMm}{r^2}$$You may well have met the same sort of thing for a body performing shm along a line, in which case$$F_x=-kx$$

[If we write $F=\frac{GMm}{r^2}$, then we're talking about the magnitude of the force. Magnitudes of vectors are, by definition, always positive. Components can be positive or negative.]

As for your second question, the loss in PE of the system as m goes from $R_1$ to $R_2$ is the work done by the gravitational field so$$U(R_2)-U(R_1)=-\int_{R_{1}}^{R_{2}}-\frac{GMm}{r^2}dr=GMm \left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

So if $R_2 >R_1$ then $U(R_2)>U(R_1)$. This makes sense: the further m is from M, the greater the potential energy of the system. [We can get work out of the system as m "falls back" towards M: we might have some crazy arrangement for generating electricity as m falls.]

You don't need to do the integration differently for the two cases $R_2 >R_1$ and $R_2 < R_1$.

For convenience it is usual to assign zero potential energy to the system when the separation of the masses is infinite. So putting $R_1=\infty$ and $R_2=R$ and $U(R_1)=0$, we have$$U(R)= -\frac{GMm}{R}$$

I will lay down the basics first,

• Gravitational potential of an object is the difference between the potential energy at infinity(which is zero) and the potential energy at the point it is in.
• Change in potential energy between two points is negative of the work done by conservative force in moving the object from one point to other.
• According to the above statement gravitational potential energy is negative of the work done by gravitational force in moving object O2 from B to infinity .

Now coming to your question, Force is negative because object is moved towards right ,as in the image but gravitational force is acting towards left.So work done by gravitational force is -F.dr