Question

At a particular instant the magnitude of the momentum of a planet is $\mathbf{2}\mathbf{.}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{29}}\mathbf{kg}\mathbf{.}\mathbf{m}\mathbf{/}\mathbf{s}$, and the force exerted on it by the star it is orbiting is $\mathbf{8}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{22}}\mathbf{N}$. The angle between the planet's momentum and the gravitational force exerted by the star is $\mathbf{123}\mathbf{°}$.

(a) What is the parallel component of the force on the planet by the star?

(b) What will the magnitude of the planet's momentum be after 9h?

Short answer

The parallel component of the force on the planet by the star is ${\overrightarrow{\mathrm{F}}}_{\left|\right|}=\left(-4.8\times {10}^{22},0,0\right)\mathrm{N}$

The magnitude of the planet's momentum be after 9h is${\overrightarrow{\mathrm{p}}}_{\mathrm{future}}=2.28\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}$

Step-by-step solution

Given data

At a particular instant the magnitude of the momentum of a planet is $2.3\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}$, and the force exerted on it by the star it is orbiting is $8.9\times {10}^{22}\mathrm{N}$. The angle between the planet's momentum and the gravitational force exerted by the star is$123°$ .

The concept of parallel component and perpendicular component

The parallel component of the force $\mathit{F}\mathbf{\left|}\mathbf{\right|}$. The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We separate the force into two parts: a parallel force $\overrightarrow{\mathbf{F}}\mathbf{|}$ to the momentum and a perpendicular force${\mathit{F}}_{\mathbf{⟂}}$to the momentum.

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line

 Computation of forces

We are given the magnitude of the momentum of the planet and let us call this momentum $\left(p_{now}\right)$and it is given by role="math" localid="1656683699039" ${\mathrm{p}}_{\mathrm{now}}=2.3\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}$Also, we are given the force exerted on the planet $\mathrm{F}=8.9\times {10}^{22}\mathrm{N}$and the angle between the planet and the star is $\theta =123°$.

We are asked to find the parallel component of the force $F\left|\right|$. The momentum here is not constant, where the planet moving along a curving path with varying speed where the rate change in momentum and the force may be varying in magnitude and direction. We divide the force here into two parts: a parallel force $\overrightarrow{F}|$to the momentum and a perpendicular force $F_{⟂}$to the momentum.

The parallel force exerted to the momentum will speed or reduce the velocity of the planet and does not change its moving line. Let us apply the direction cosines, we could obtain the parallel force as next

${\overrightarrow{F}}_{\left|\right|}=\left|\overrightarrow{F}\right|\cos \theta \stackrel{⏜}{p}$ ………………………… (1)

Where the parallel force $\overrightarrow{F}$is in the opposite direction of $\stackrel{⏜}{p}$as the angle between them is larger than $90°$. Now we can plug our values for $\theta$and $\left|\overrightarrow{F}\right|$into equation (1) to get the parallel force to the planet.

$$\begin{gathered} {\overrightarrow{F}}_{\left|\right|}=\left|\overrightarrow{F}\right|\cos \theta \stackrel{⏜}{p} \\ =\left(8.9\times {10}^{22}N\right)\cos 123°\stackrel{⏜}{p} \\ =-4.8\times {10}^{22}N\stackrel{⏜}{p} \end{gathered}$$

  Get its vector because all the force is in one direction.

As this force is in one direction, we could get its vector as next

${\overrightarrow{\mathrm{F}}}_{\left|\right|}=\left(-4.8\times {10}^{22},0,0\right)\mathrm{N}$

Or

${\overrightarrow{\mathrm{F}}}_{\left|\right|}\left(0,-4.8\times {10}^{22},0\right)\mathrm{N}$

OR

${\overrightarrow{\mathrm{F}}}_{\left|\right|}\left(0,0,-4.8\times {10}^{22},0\right)\mathrm{N}$

The cosine of $123°$the angle between $\overrightarrow{F}$and $\overrightarrow{p}$, is a negative number, so ${\overrightarrow{F}}_{\left|\right|}$is opposite to $\stackrel{⏜}{p}$. The magnitude of the planet's momentum will decrease.

The parallel component of the force on the planet by the star is${\overrightarrow{\mathrm{F}}}_{\left|\right|}\left(-4.8\times {10}^{22},0,0\right)\mathrm{N}$

Calculation of momentum

Determine the momentum after 9h . Let us name this momentum $p_{future}$. From the result in part (a), the force is in the negative direction of the momentum. Hence, the parallel force is nonzero, and the term $F\left|\right|∆t$ would need to be opposite in direction to $p_{now}$where the term $\overline{)\overline{)F}}∆t$is the impulsive given to the planet during and result in a new momentum to a planet $p_{future}$where the parallel force $F\left|\right|$is associated with changes in the magnitude of the object's momentum.

$P_{future}={\overrightarrow{p}}_{now}+F_{\left|\right|}∆t$ ……………….…… (2)

Now we can plug our values for role="math" localid="1656684825612" ${\overrightarrow{p}}_{now},F_{\left|\right|}and∆t$into equation (2) and get ${\overrightarrow{p}}_{future}$were the time should be in second.

$$\begin{gathered} {\overrightarrow{\mathrm{p}}}_{\mathrm{future}}={\overrightarrow{\mathrm{p}}}_{\mathrm{now}}+{\mathrm{F}}_{\left|\right|}∆\mathrm{t} \\ =2.3\times {10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}+\left(-4.8\times {10}^{22}\mathrm{N}\right)\left(9\mathrm{h}\times 3600\mathrm{s}/\mathrm{h}\right) \\ =2.28\times {10}^{29}\mathrm{kg},\mathrm{m}/\mathrm{s} \end{gathered}$$

The magnitude of the planet's momentum be after 9h is ${\overrightarrow{\mathrm{p}}}_{\mathrm{future}}=2.28\times {10}^{29}\mathrm{kg},\mathrm{m}/\mathrm{s}$