91Ó°ÊÓ

The given data is listed below as,

$∘$The angle between the planet’s gravitational force and momentum is,$\mathrm{θ}=60°$

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

$∘$The magnitude of the planet’s gravitational force is $\stackrel{→}{\mathrm{F}}=1.8×{10}^{23}\mathrm{N}$

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-

$\stackrel{→}{F_{âˆ\yen }}=\left|\stackrel{→}{F}\right|\cos \mathrm{θ}p$....$\left(1\right)$

Here, $F_{âˆ\yen }$is the parallel force,$\stackrel{→}{\left|F\right|}$ is the absolute value of the gravitational force, $p$is the unit vector and $\mathrm{θ}$is the angle between the momentum and gravitational force

(a) For,$\stackrel{→}{\left|F\right|}=1.8×{10}^{23}N$and $\mathrm{θ}=0$in equation (1).

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

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

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