Question

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?

Short answer

$$\begin{gathered} \left(a\right)8.7\times {10}^{22}\mathrm{Np} \\ \left(\mathrm{b}\right)3.15184\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

Step-by-step solution

Identification of the given data

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 $\overrightarrow{\mathrm{F}}=1.8\times {10}^{23}\mathrm{N}$

Significance of the parallel force

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

Determination of the parallel component of the force on the planet

(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" $\left(1.8\times {10}^{23}N\right)\mathbf{p}$

 Step 4: Determination of the magnitude of the planet’s final momentum

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