Gravitational Attraction The gravitational attraction F on a body a distance r fr

Gravitational Attraction The gravitational attraction ...

$$\begin{array}{c}{\text { Gravitational Attraction The gravitational attraction } F \text { on a }} \\ {\text { body a distance } r \text { from the center of Earth, where } r \text { is greater }} \\ {\text { than the radius of Earth, is a function of its mass } m \text { and the }} \\ {\text { distance } r \text { as follows: }} \\ {F=\frac{m g R^{2}}{r^{2}}}\end{array}$$ where $R$ is the radius of Earth and $g$ is the force of gravity- about 32 feet per second per second (ft per sec $^{2} ) .$ (a) Find and interpret $F_{m}$ and $F_{r}$ . (b) Show that $F_{m}>0$ and $F_{r}<0 .$ Why is this reasonable?

$$\begin{array}{c}{\text { Gravitational Attraction The gravitational attraction } F \text { on a }} \\ {\text { body a distance } r \text { from the center of Earth, where } r \text { is greater }} \\ {\text { than the radius of Earth, is a function of its mass } m \text { and the }} \\ {\text { distance } r \text { as follows: }} \\ {F=\frac{m g R^{2}}{r^{2}}}\end{array}$$
where $R$ is the radius of Earth and $g$ is the force of gravity-
about 32 feet per second per second (ft per sec $^{2} ) .$
(a) Find and interpret $F_{m}$ and $F_{r}$ .
(b) Show that $F_{m}>0$ and $F_{r}<0 .$ Why is this reasonable?

Key Concepts

Partial Differentiation

Partial differentiation is the process of finding the derivative of a multivariable function with respect to one variable while holding the other variables constant. This concept allows us to understand how a function changes in response to changes in one particular variable, independently of the others.

Interpretation of Partial Derivatives

Interpreting partial derivatives in a physical context involves understanding the rate at which the quantity of interest (in this case, gravitational force) changes with respect to a change in one parameter (such as mass or distance) while keeping the other parameters constant. A positive partial derivative indicates that the function increases as that variable increases, while a negative derivative indicates a decrease.

Sign Analysis of Partial Derivatives

The sign of a partial derivative provides insight into how the function behaves: a positive sign means that an increase in the variable leads to an increase in the function's value, and a negative sign means that an increase in the variable results in a decrease in the function's value. This analysis is crucial in verifying that the mathematical model aligns with physical intuition, such as the gravitational force increasing with mass and decreasing with distance.

Transcript

00:01 The gravitational attraction force f is given by mg, big r square, divided by small r square, where m is the mass of the body, big r is the radius of earth, small r is the distance from body to the center of the earth, which is greater than the radius of the earth, and g is the acceleration of gravity, which is about 32 feet per second square.

00:32 So in part a, we got to find the partial derivatives of the force with respect to the mass and with respect to the distance from the body to the center of the earth.

00:42 So first we have that partial derivative of the force with respect to the mass.

00:52 In that partial derivative, g, big r square divided by small r square is a constant and the derivative of m for respect to m is one.

01:05 So this is the partial derivative of the force with respect to the mass and that represents the rate of changing force per unit changing mass while the distance r is held constant now let's calculate if it is a partial derivative of the force with respect to the distance small r that is we respect to the distance from the body to the center of the earth and that is equal to in this derivative m g big r square is a constant and that's get to be multiplied by the partial derivative respect to the small r of one divided by small r square and this is equal to m g big r square times the partial derivative for respect to a small r of small r to the negative two and this is equal to m g big r square times negative two times small r to the negative two minus one is negative three and finally this is equal to negative two m g big r squared divided by small r to the third and this is the partial derivative of the force with respect to the distance small r and that that represents the rate of change in force per unit changing distance while the mass is held constant.

02:50 Now let's see in part b that the partial derivative of the force with respect to the mass is positive and that the partial derivative of the force will respect to the distance from the body to the center of the earth is negative.

03:11 The first relationship, partial derivative of the force, with respect to the mass is positive is true because in the formula we obtain in par a, we see that all the quantities appear in that formula are positive.

03:34 Big r square, small r square, and g, which is 32, so all is positive.

03:40 So partial derivative of the force with respect to the mass is positive.

03:47 The other one, partial derivative of the fourths, with respect to the small r, is negative.

03:54 We see that expression we obtain here.

03:58 All is positive, r, big r square...

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