### Gravitational Equilibrium Problem#### Diagram Explanation:The diagram shows three planets:- **Planet #1 (Mass \( M_1 \))**: Positioned directly between Planet #2 and Planet #3.- **Planet #2 (Mass \( M_2 \))**: Located at the right end of the diagram.- **Planet #3 (Mass \( M_3 \))**: Located at the left end of the diagram.The distances are labeled as follows:- The distance between Planet #1 and Planet #2 is denoted as \( r \).- The distance between Planet #1 and Planet #3 is denoted as \( R - r \) (since the total distance \( R \) between Planet #2 and Planet #3 is given).#### Problem Statement:Planet #2 (mass \( M_2 \)) and Planet #3 (mass \( M_3 \)) are separated by distance \( R \). Planet #1 (mass \( M_1 \)) is positioned directly between these two planets.Given:- \( M_3 = 16 \, M_2 \)- \( R = 15 \)The task is to find the value of \( r \) such that the total gravitational force on Planet #1 is zero.\[ \text{r} = \_\_\_\_ \]

Using, Gravitational force, F=GMmr2 Net force on mass-1 F1,netis zero, that means, sum of gravitational forces by mass-2 F1,2 and mass-3 F1,3 is zero. F1,net=F1,2+F1,3=0F1,2=acts on mass-1 towards rightF1,3=acts on mass-1 towards leftGM1M2r2-GM3M1R-r2=0M1M2r2=M3M1R-r2 [Since, M3=16M2, R=15 ]M1M2r2=16M2M115-r215-r2=16r2r2+2r-15=0r+5r-3=0r=3 [r=-5 cannot be taken as r negative does not make sense here]Answer: r = 3

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Pl anet # 1 M3 M: M2 R R NI u ....I R Planet # 3 Planet # 2 Planet #2 (mass M2) and planet #3 (mass M3) are separated by distance R. Planet #1 (mass M1) is directly between these two planets. If M3 = 16 M2 and R=15, then what is the value of r such that the total force on planet #1 is zero?

Pl anet # 1 M3 M: M2 R R NI u ....I R Planet # 3 Planet # 2 Planet #2 (mass M2) and planet #3 (mass M3) are separated by distance R. Planet #1 (mass M1) is directly between these two planets. If M3 = 16 M2 and R=15, then what is the value of r such that the total force on planet #1 is zero?

College Physics

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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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Centripetal Force

A particle, body, object or certain mass while travelling in a rotational path or semicircular path, experiences an inertia force. If that inertia force pulls the towards the inward of the imaginary circle in the rotation, it is termed as centripetal force.

Question

### Gravitational Equilibrium Problem

#### Diagram Explanation:

The diagram shows three planets:

- **Planet #1 (Mass \( M_1 \))**: Positioned directly between Planet #2 and Planet #3.
- **Planet #2 (Mass \( M_2 \))**: Located at the right end of the diagram.
- **Planet #3 (Mass \( M_3 \))**: Located at the left end of the diagram.

The distances are labeled as follows:

- The distance between Planet #1 and Planet #2 is denoted as \( r \).
- The distance between Planet #1 and Planet #3 is denoted as \( R - r \) (since the total distance \( R \) between Planet #2 and Planet #3 is given).

#### Problem Statement:

Planet #2 (mass \( M_2 \)) and Planet #3 (mass \( M_3 \)) are separated by distance \( R \). Planet #1 (mass \( M_1 \)) is positioned directly between these two planets.

Given:
- \( M_3 = 16 \, M_2 \)
- \( R = 15 \)

The task is to find the value of \( r \) such that the total gravitational force on Planet #1 is zero.

\[ \text{r} = \_\_\_\_ \]

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