91Ó°ÊÓ

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

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 $D\left(f_D\right)$.

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

Similarly, we have the following relations as well:

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

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.

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.

${\stackrel{→}{F}}_{net┴}≠0.$

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

For all the points we have:${\stackrel{→}{F}}_{net┴}≠0$

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:

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

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

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

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

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

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

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

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

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

As it can be observed in the diagram, $F_{netâ”´}$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 $F_{netâ”´}$in green, whereas the total force is drawn in orange.

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