A planet of mass 4 x 1024 kg is at location (5 x 10¹¹, -5 × 101¹1, 0) m. A star of mass 5 × 1030 kg is at location (-6 × 10¹1, 5 x 10¹¹, 0) m. It will be useful to draw a diagram of the situation, including the relevant vectors. (d) What is the magnitude of the force exerted on the planet by the star?Fon planet!N(e) What is the magnitude of the force exerted on the star by the planet?Fon starlN(f) What is the force (vector) exerted on the planet by the star? (Note the change in units.)Fon planetx 1020 N(g) What is the force (vector) exerted on the star by the planet? (Note the change in units.)x 1020 Non star=

Mass of the planet m=4×1024 kg Mass of the star M=5×1030 kg Position of the planet…

Answered: A planet of mass 4 x 1024 kg is at…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Transcribed Image Text:A planet of mass 4 x 1024 kg is at location (5 x 10¹¹, -5 × 101¹1, 0) m. A star of mass 5 × 1030 kg is at location (-6 × 10¹1, 5 x 10¹¹, 0) m. It will be useful to draw a diagram of the situation, including the relevant vectors.

(d) What is the magnitude of the force exerted on the planet by the star?
Fon planet!
N
(e) What is the magnitude of the force exerted on the star by the planet?
Fon starl
N
(f) What is the force (vector) exerted on the planet by the star? (Note the change in units.)
Fon planet
x 1020 N
(g) What is the force (vector) exerted on the star by the planet? (Note the change in units.)
x 1020 N
on star
=

Expert Solution

Step 1

Mass of the planet $m = 4 \times 10^{24} k g$

Mass of the star $M = 5 \times 10^{30} k g$

Position of the planet $\vec{r_{1}} = (5 \times 10^{11} \hat{i} - 5 \times 10^{11} \hat{j}) m$

Position of the star ${\vec{r}}_{2} = (- 6 \times 10^{11} \hat{i} + 5 \times 10^{11} \hat{j})$

Position vector $\vec{r}$ pointing from planet to star is given by

$\vec{r} = (\vec{r_{2}} - \vec{r_{1}})$

$= {(- 6 \times 10^{11} \hat{i} + 5 \times 10^{11} \hat{j}) - (5 \times 10^{11} \hat{i} - 5 \times 10^{11} \hat{j})} = (- 11 \times 10^{11} \hat{i} + 10 \times 10^{11} \hat{j})$

$|\vec{r}| = \sqrt{{(- 11 \times 10^{11})}^{2} + {(10 \times 10^{11})}^{2}} = 1.486 \times 10^{12} m$

Unit vector $\hat{r}$ in the direction of $\vec{r}$ is

$\hat{r} = \frac{\vec{r}}{|\vec{r}|}$

$= \frac{(- 11 \times 10^{11} \hat{i} + 10 \times 10^{11} \hat{j})}{1.486 \times 10^{12}} = (- 0.74 \hat{i} + 0.67 \hat{j}) m$

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