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Q19P

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Found in: Page 208

### Matter & Interactions

Book edition 4th edition
Author(s) Ruth W. Chabay, Bruce A. Sherwood
Pages 1135 pages
ISBN 9781118875865

# 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 is role="math" localid="1654013162020" ${\mathbf{3}}{\mathbf{.}}{\mathbf{1}}{\mathbf{}}{\mathbf{×}}{{\mathbf{10}}}^{{\mathbf{29}}}{\mathbf{}}{\mathbf{kgm}}{\mathbf{/}}{\mathbf{s}}$ , and the magnitude of the gravitational force on the planet is role="math" localid="1654013174728" $\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{×}{\mathbf{10}}^{\mathbf{23}}\mathbf{}\mathbf{N}$ . (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?

$\left(a\right)8.7×{10}^{22}\mathrm{Np}\phantom{\rule{0ex}{0ex}}\left(\mathrm{b}\right)3.15184×{10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}$

See the step by step solution

## Step 1: 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, ${\mathrm{p}}_{\mathrm{i}}=3.1×{10}^{29}\mathrm{kg}.\mathrm{m}/\mathrm{s}$

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

## Step 2: 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-

$\stackrel{\to }{{F}_{\parallel }}=\left|\stackrel{\to }{F}\right|\mathrm{cos}\theta p$ ....$\left(1\right)$

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

## Step 3: Determination of the parallel component of the force on the planet

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

${F}_{\parallel }=\left(1.8×{10}^{23}N×\mathrm{cos}0°\right)p\phantom{\rule{0ex}{0ex}}=\left(1.8×{10}^{23}N\right)p$

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

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

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