Question

A comet orbits a star in an elliptical orbit, as shown in Figure 5.41. The momentum of the comet at locationis shown in the diagram. At the instant the comet passes each location labeled A, B, C, D, E, and F, answer the following questions about the net force on the comet and the rate of change of the momentum of the comet:

(a) Draw an arrow representing the direction and relative magnitude of the gravitational force on the comet by the star.

(b) Is${\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\perp$zero to nonzero?

(c) Is${\overrightarrow{\mathbf{F}}}_{\mathbf{net}}$zero or nonzero?

(d) Is$\mathbf{d}\mathbf{\left|}\overrightarrow{\mathbf{p}}\mathbf{\right|}\mathbf{/}\mathbf{ds}$positive, negative, or zero?

(e) Is$\mathbf{d}\overline{\mathbf{p}}\mathbf{/}\mathbf{dr}$zero or nonzero?

Short answer

(a) The relative magnitudes are fully explained according to Newton's law of universal gravitation.

(b)We have ${\overrightarrow{\mathrm{F}}}_{\mathrm{net}}┴\ne 0$for all the points

(c)The results will be ${\overrightarrow{\mathrm{F}}}_{\mathrm{net}\parallel }$Is zero for points and. Is nonzero for points B, C, E and.

(d)The results are, $\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}$is positive for points B and C . $\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}$Is zero for points A and D . $\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}$Is negative for points E and F.

(e) $\mathrm{d}\hat{\mathrm{p}}/\mathrm{dt}$will be nonzero for all the points.

Step-by-step solution

Given

A comet orbits a star in an elliptical orbit, as shown in Figure

The formula of Newton's law of universal gravitation used to determine the relative magnitudes.

First, it is important to consider that the force exerted by the star on the comet is defined by Newton's law of universal gravitation:

$\mathrm{f}=\frac{\mathrm{GMm}}{{\mathrm{r}}^2}$

Applying the Newton's law of universal gravitation to determine the relative magnitude.

The distance between the point A and the star $\left({\mathrm{d}}_{\mathrm{A}}\right)$is approximately fourth of the distance between point D and the star $\left(d_D\right)$. This implies that the force on the comet at point $A\left(f_A\right)$is one sixteenth of the force at point $D\left(f_D\right)$.

$$\begin{gathered} {\mathrm{d}}_{\mathrm{A}}=4{\mathrm{d}}_{\mathrm{D}} \\ {\mathrm{f}}_{\mathrm{A}}=\frac{\mathrm{GMm}}{{\left(4{\mathrm{d}}_{\mathrm{D}}\right)}^2} \\ =\frac{1}{16}\frac{{\displaystyle \stackrel{{\mathrm{d}}_{\mathrm{D}}}{\overbrace{\mathrm{GMm}}}}}{{\mathrm{f}}_{\mathrm{D}}} \\ {\mathrm{f}}_{\mathrm{A}}=\frac{1}{16}{\mathrm{f}}_{\mathrm{D}} \end{gathered}$$

Similarly, we have the following relations as well:

$$\begin{gathered} {\mathrm{d}}_{\mathbf{B}}={\mathrm{d}}_{\mathrm{F}} \\ =3{\mathrm{d}}_{\mathrm{D}} \\ {\mathrm{f}}_{\mathrm{B}}={\mathrm{f}}_{\mathrm{F}} \\ =\frac{1}{9}{\mathrm{f}}_{\mathrm{D}} \\ \mathrm{Solve}\mathrm{further}, \\ {\mathrm{d}}_{\mathrm{C}}={\mathrm{d}}_{\mathrm{E}} \\ =2{\mathrm{d}}_{\mathrm{D}} \\ {\mathrm{f}}_{\mathrm{C}}={\mathrm{f}}_{\mathrm{E}} \\ =\frac{1}{4}{\mathrm{f}}_{\mathrm{D}} \end{gathered}$$

Proceeding with diagram with the obtained relations.

With the relations obtained in the previous step, we proceed to draw a diagram in which these proportions are approximately the same.

The relative magnitudes are defined according to Newton's law of universal gravitation.

Defining the perpendicular net force is nonzero for all the points.

Note that there is always a component of the gravitational force that acts on the comet that points inwards the curve. Therefore the perpendicular net force is nonzero for all of the points.

${\overrightarrow{F}}_{net┴}\ne 0.$

The component of net force changes the direction of the momentum.

In other words, there is always a component of the net force that changes the direction of the momentum $\left(\overrightarrow{P}\right)$. As a result there isn't any point in which ${\overrightarrow{F}}_{net┴}\ne 0$is zero.

For all the points we have:${\overrightarrow{F}}_{net┴}\ne 0$

Identify the points whose tangential component is zero to nonzero.

Added some tangential lines to the image drawn in part (a), to help us identify in which points the tangential component is zero or nonzero. We conclude the following:

-${\overrightarrow{F}}_{net║}$is zero for points A and D.

-${\overrightarrow{F}}_{net║}$is nonzero for points B, C, E and F.

Added some tangential lines to the image.

Note that only in points and the gravitational attraction force doesn't have a component in the tangential (parallel) direction.

${\overrightarrow{F}}_{net║}$Is zero for points A and D.

${\overrightarrow{F}}_{net║}$Is nonzero for points B, C, E and F.

Applying Newton's second law.

$$\begin{gathered} \mathrm{Note}\mathrm{that}\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}\mathrm{is}\mathrm{equal}\mathrm{to}{\mathrm{F}}_{\mathrm{net}║},\mathrm{the}\mathrm{component}\mathrm{of}\mathrm{the}\mathrm{force}\mathrm{that}\mathrm{increases}\mathrm{the} \\ \mathrm{tangential}\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{comet}(\mathrm{according}\mathrm{to}\mathrm{Newton}\text{'}\mathrm{s}\mathrm{second}\mathrm{law}). \\ \frac{\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|}{\mathrm{dt}}={\mathrm{F}}_{\mathrm{net}║} \\ \mathrm{And}\mathrm{to}\mathrm{the}\mathrm{right}\mathrm{in}\mathrm{the}\mathrm{lower}\mathrm{part}),\mathrm{otherwise}\mathrm{it}\mathrm{will}\mathrm{be}\mathrm{negative}.\mathrm{Therefore},\mathrm{we}\mathrm{conclude} \\ \mathrm{the}\mathrm{following}: \\ -\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}\mathrm{is}\mathrm{positive}\mathrm{for}\mathrm{pointis}\mathrm{zero}\mathrm{for}\mathrm{points}\mathrm{B}\mathrm{and}\mathrm{C}. \\ -\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}\mathrm{is}\mathrm{zero}\mathrm{for}\mathrm{points}\mathrm{A}\mathrm{and}\mathrm{D}. \\ -\left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}\mathrm{is}\mathrm{negative}\mathrm{for}\mathrm{points}\mathrm{E}\mathrm{and}\mathrm{F} \end{gathered}$$

Drawing the diagram with the values of the points.          

We have the ones that are positive (they are in the counterclockwise direction), and in red the negative ones.

$$\begin{gathered} \left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}\mathrm{Is}\mathrm{positive}\mathrm{for}\mathrm{points}\mathrm{B}\mathrm{and}\mathrm{C}. \\ \left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}\mathrm{Is}\mathrm{zero}\mathrm{for}\mathrm{points}\mathrm{A}\mathrm{and}\mathrm{D}. \\ \left|\mathrm{d}\overrightarrow{\mathrm{p}}\right|/\mathrm{dt}\mathrm{Is}\mathrm{negative}\mathrm{for}\mathrm{points}\mathrm{E}\mathrm{and}\mathrm{F}. \end{gathered}$$

Explaining the force is that acts perpendicular to the trajectory of the comet.

$$\begin{gathered} \mathrm{Note}\mathrm{that}\mathrm{d}\overrightarrow{\mathrm{p}}/\mathrm{dt}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{force}{\mathrm{F}}_{\mathrm{net}}\mathrm{that}\mathrm{acts}\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{trajectory}\mathrm{of}\mathrm{the} \\ \mathrm{comet} \\ {\mathrm{F}}_{\mathrm{net}║}=\frac{\mathrm{d}\overrightarrow{\mathrm{p}}}{\mathrm{dt}} \end{gathered}$$

Observing the diagram.

As it can be observed in the diagram, $F_{net┴}$is nonzero for all the points, hence $d\hat{p}/dt$is nonzero for all the points. Note that for points A and D this force is equal to the total force, however for points B,C

E And F, it is a component of the total force. We have drawn $F_{net┴}$in green, whereas the total force is drawn in orange.

$\mathrm{d}\hat{\mathrm{p}}/\mathrm{dt}$is nonzero for all the points.