The angle between the gravitational force on a planet by a star and the momentum of the planet is $61°$at a particular instant. At this instant the magnitude of the planet’s momentum isrole="math" localid="1654013162020" $\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{29}}\mathbf{}\mathbf{kgm}\mathbf{/}\mathbf{s}$, and the magnitude of the gravitational force on the planet is role="math" localid="1654013174728" $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{N}$. (a) What is the parallel component of the force on the planet by the star? (b) What will be the magnitude of the planet’s momentum after 8h?

The given data is listed below as,

$\circ$The angle between the planet’s gravitational force and momentum is,$\theta =60°$

$\circ$The magnitude of the planet’s initial momentum is, ${\mathrm{p}}_{\mathrm{i}}=3.1\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}$

$\circ$The magnitude of the planet’s gravitational force is $\overrightarrow{\mathrm{F}}=1.8\times {10}^{23}\mathrm{N}$

The parallel force acts in the opposite or same direction at the different points of a particular object.

The equation of the parallel component of the force is expressed as-

$\overrightarrow{F_{\parallel }}=\left|\overrightarrow{F}\right|\cos \theta p$....$\left(1\right)$

Here, $F_{\parallel }$is the parallel force,$\overrightarrow{\left|F\right|}$ is the absolute value of the gravitational force, $p$is the unit vector and $\theta$is the angle between the momentum and gravitational force

(a) For,$\overrightarrow{\left|F\right|}=1.8\times {10}^{23}N$and $\theta =0$in equation (1).

$$\begin{gathered} F_{\parallel }=\left(1.8\times {10}^{23}N\times \cos 0°\right)p \\ =\left(1.8\times {10}^{23}N\right)p \end{gathered}$$

Thus, the parallel component of the force on the planet by the star is role="math" localid="1654016428235" $\left(1.8\times {10}^{23}N\right)\mathbf{p}$

$$\begin{gathered} {\mathrm{P}}_{\mathrm{f}}={\mathrm{P}}_{\mathrm{i}}+{\mathrm{F}}_{\mathrm{net}}∆\mathrm{t} \\ \mathrm{Here},{\mathrm{P}}_{\mathrm{f}}\mathrm{is}\mathrm{the}\mathrm{final}\mathrm{momentum},{\mathrm{P}}_{\mathrm{i}}\mathrm{is}\mathrm{the}\mathrm{initial}\mathrm{momentum},{\mathrm{F}}_{\mathrm{net}}\mathrm{is}\mathrm{the}\mathrm{net}\mathrm{force}\mathrm{exerted}\mathrm{by}\mathrm{the}\mathrm{planet}\mathrm{and}∆\mathrm{t}\mathrm{is}\mathrm{the}\mathrm{difference}\mathrm{in}\mathrm{the}\mathrm{time}\mathrm{period} \\ \mathrm{For}{\mathrm{p}}_{\mathrm{i}}=3.1\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s},{\mathrm{p}}_{\mathrm{i}}=1.8\times {10}^{23}\mathrm{N}\mathrm{and}∆\mathrm{t}=8\mathrm{h}-0=8\mathrm{h} \\ {\mathrm{p}}_{\mathrm{f}}=\left(3.1\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)+(1.8\times {10}^{23}\mathrm{N})\times \left(8\mathrm{h}\frac{3600\mathrm{s}}{1\mathrm{h}}\right) \\ =\left(3.1\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}\right)+5.184\times {10}^{27}\mathrm{N}.\mathrm{s}\times \frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}} \\ =3.15184\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ \mathrm{Thus},\mathrm{the}\mathrm{magnitude}\mathrm{of}\mathrm{the}\mathrm{planet}’\mathrm{s}\mathrm{momentum}\mathrm{after}8\mathrm{h}\mathrm{is}3.15184\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$The equation of the magnitude of the planet’s momentum after 8h is expressed as,