Question

A planet orbits a star in an elliptical orbit. At a particular instant the momentum of the planet is $\left(-2.6\times {10}^{29},-1.0\times {10}^{29},0\right)\mathit{k}\mathit{g}\mathbf{.}\mathit{m}\mathbf{/}\mathit{s}$, and the force on the planet by the star is $\left(-2.5\times {10}^{22},-1.4\times {10}^{23},0\right)\mathit{N}$. Find ${\mathit{F}}_{\mathbf{\parallel }}$and ${\mathit{F}}_{\mathbf{\perp }}$.

Short answer

$\left(-6.87\times {10}^{22},-2.64\times {10}^{22},0\right)N$and$\left(4.37\times {10}^{22},-1.13\times {10}^{23},0\right)N$

Step-by-step solution

Identification of the given data 

The given data is listed below as,

• The planet’s momentum is, $p=\left(-2.6\times {10}^{29},-1.0\times {10}^{29},0\right)kg.m/s$.

• The force exerted on the planet is, $F=\left(-2.5\times {10}^{22},-1.4\times {10}^{23},0\right)N$.

Significance of the parallel force

The parallel force mainly acts in the same or the opposite direction at some points of an object.

From the momentum principle, the equation of the parallel component of the force is expressed as,

$F=\left|F\right|\cos \theta \hat{p}$ …(1)

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

Determination of the parallel force of the planet

The magnitude of the momentum of the planet can be expressed as,

$p=\sqrt{{p^2}_x+{p^2}_y+{p^2}_z}$

Here, $p_x$,$p_y$, and $p_z$are the momentum at the $x$, $y$and $z$direction respectively.

For $p_x=-2.6\times {10}^{29}kgm/s$, $p_y=-1.0\times {10}^{29}kgm/s$and $p_z=0$,

$$\begin{gathered} p=\sqrt{{\left(-2.6\times {10}^{29}kgm/s\right)}^2+{\left(-1.0\times {10}^{29}kgm/s\right)}^2+0} \\ =2.785\times {10}^{29}kgm/s \end{gathered}$$

Write the expression for the unit vector $\hat{p}$.

$\hat{p}=\frac{\overline{p}}{p}$

Here, $\overline{p}$is the momentum of the planet and $p$is the magnitude of the momentum.

For $\overline{p}=\left(-2.6\times {10}^{29},-1.0\times {10}^{29},0\right)kg.m/s$and$p=2.785\times {10}^{29}kg.m/s$ .

$$\begin{gathered} \hat{p}=\frac{\left(-2.6\times {10}^{29},-1.0\times {10}^{29},0\right)kg.m/s}{2.785\times {10}^{29}kg.m/s} \\ =\left(-0.9335,-0.359,0\right) \end{gathered}$$

Rewrite equation (1)

$\overline{F_{\parallel }}=\left(\overline{F}.\hat{p}\right)\hat{p}$

For $\hat{p}=\left(-0.9335,-0.359,0\right)$and $\overline{F}=\left(-2.5\times {10}^{22},-1.4\times {10}^{23},0\right)N$$$\begin{gathered} \overline{F}=\left(\left(-2.5\times {10}^{22},-1.4\times {10}^{23},0\right)N\times \left(-0.9335,-0.359,0\right)\right)\times \left(-0.9335,-0.359,0\right) \\ =\left(2.33375\times {10}^{22}+5.026\times {10}^{22}\right)N\times \left(-0.9335,-0.359,0\right) \\ =7.3635\times {10}^{22}N\times \left(\left(-0.9335,-0.359,0\right)\right) \\ =\left(-6.87\times {10}^{22},-2.64\times {10}^{22},0\right)N \end{gathered}$$

Determination of the perpendicular force of the planet

The equation of force for the planet can be expressed as,

$$\begin{gathered} F_{net}=\overline{F_{\parallel }}+\overline{F} \\ \overline{F_{\perp }}=F_{net}-\overline{F_{\parallel }} \end{gathered}$$

For $F_{net}=\left(-2.5\times {10}^{22},-1.4\times {10}^{23},0\right)N$, and $\overline{F_{\parallel }}=\left(-6.87\times {10}^{22},-2.64\times {10}^{22},0\right)N$.

$$\begin{gathered} \overline{F_{\perp }}=\left(\left(-2.5\times {10}^{22},-1.4\times {10}^{23},0\right)N-\left(-6.87\times {10}^{22},-2.64\times {10}^{22},0\right)N\right) \\ =\left(4.37\times {10}^{22},-1.13\times {10}^{23},0\right)N \end{gathered}$$

Thus, the values of $\overline{F_{\parallel }}$and $F_{\perp }$are $\left(-6.87\times {10}^{22},-2.64\times {10}^{22},0\right)N$and $\left(4.37\times {10}^{22},-1.13\times {10}^{23},0\right)N$ respectively.