91Ó°ÊÓ

The given data is listed below as,

• The angle between the planet’s gravitational force and momentum is, $\mathrm{θ}={61}^0$
  

• The magnitude of the planet’s initial momentum is,$p_i=3.1×{10}^{29}kgm/s$

• The magnitude of the planet’s gravitational force is,$F=1.8×{10}^{23}N$

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-

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

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

(a)

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

$$\begin{gathered} F=\left(1.8×{10}^{23}N×\cos 0^{∘}\right)\hat{p} \\ =\left(1.8×{10}^{23}N\right)\hat{p} \end{gathered}$$

Thus, the parallel component of the force on the planet by the star is $\left(1.8×{10}^{23}N\right)\hat{p}$.

The equation of the magnitude of the planet’s momentum after $8h$is expressed as,

$p_f=p_i+F_{net}∆t$

Here, $p_f$is the final momentum, $p_i$is the initial momentum, $F_{net}$is the net force exerted by the planet and $∆t$is the difference in the time period.

For$p_i=3.1×{10}^{29}kgm/s$, $p_i=1.8×{10}^{23}N$and $∆t=8h-0=8h$.

$$\begin{gathered} p_f=\left(3.1×{10}^{29}kgm/s\right)+\left(1.8×{10}^{23}N\right)×\left(8h×\frac{3600s}{1h}\right) \\ =\left(3.1×{10}^{29}kgm/s\right)+5.184×{10}^{27}N.s×\frac{1kg.m/s^2}{1N} \\ =3.15184×{10}^{29}kgm/s \end{gathered}$$

Thus, the magnitude of the planet’s momentum after $8h$is $3.15184×{10}^{29}kgm/s$.