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
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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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### 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.

\[ \text{r} = \_\_\_\_ \]
Transcribed Image Text:### 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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