Use the expression you wrote down in question 4) relating speed (v), period (P), and distance from the Sun (R), to eliminate P from the proportionality form of Kepler's Law (P² x a³). You should be left with an expression only in terms of speed (v) and distance (R). Simplify your answer. 3 2 a R3 ラ v 2 Ra to v?R= Constant Rearrange this proportionality to solve for speed (v) in terms of distance (R).
Use the expression you wrote down in question 4) relating speed (v), period (P), and distance from the Sun (R), to eliminate P from the proportionality form of Kepler's Law (P² x a³). You should be left with an expression only in terms of speed (v) and distance (R). Simplify your answer. 3 2 a R3 ラ v 2 Ra to v?R= Constant Rearrange this proportionality to solve for speed (v) in terms of distance (R).
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#6 please
![## Keplerian Orbits
For non-solid systems, the force of gravity determines orbital motion. We know that the strength of gravity depends on mass and distance. Therefore, we can learn about the mass distribution (how much mass is located in a given area) in astronomical systems by studying rotation curves. We will start with a simple case: our solar system.
### 4)
Let’s consider the planets orbiting the Sun. They actually travel in ellipses, but their orbits are nearly circular. As you did for the ants, write a formula for the speed of a planet around the Sun in terms of its distance from the Sun, \( R \). Use the period, \( P \), for the time required for one orbit. Rearrange your speed formula to solve for \( P \).
\[
v = \frac{2\pi R}{P} \implies P = \frac{2\pi R}{v}
\]
Recall the proportionality relation form of Kepler’s Law of planetary motion, \( P^2 \propto a^3 \). Note that \( a \) is the same as \( R \) when we’re approximating the planetary orbits as circles, so use \( R \) for the distance.
### 5)
Use the expression you wrote down in question 4 relating speed (\( v \)), period (\( P \)), and distance from the Sun (\( R \)), to eliminate \( P \) from the proportionality form of Kepler’s Law (\( P^2 \propto a^3 \)). You should be left with an expression only in terms of speed (\( v \)) and distance (\( R \)). Simplify your answer.
\[
\left( \frac{2\pi R}{v} \right)^2 \propto R^3
\]
\[
\frac{4\pi^2 R^2}{v^2} \propto R^3 \implies 4\pi^2 \frac{R^2}{v^2} \propto R^3 \implies \frac{1}{v^2} \propto \frac{1}{R} \to R \propto \frac{v^2}{4\pi^2} \equiv \text{constant}
\]
### 6)
Rearrange this proportionality to solve for speed (\(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa7c87654-3157-4119-9345-56717eaa1677%2F43b635a3-886f-4da8-9740-0760863e07ae%2Fbbd5c8_processed.png&w=3840&q=75)
Transcribed Image Text:## Keplerian Orbits
For non-solid systems, the force of gravity determines orbital motion. We know that the strength of gravity depends on mass and distance. Therefore, we can learn about the mass distribution (how much mass is located in a given area) in astronomical systems by studying rotation curves. We will start with a simple case: our solar system.
### 4)
Let’s consider the planets orbiting the Sun. They actually travel in ellipses, but their orbits are nearly circular. As you did for the ants, write a formula for the speed of a planet around the Sun in terms of its distance from the Sun, \( R \). Use the period, \( P \), for the time required for one orbit. Rearrange your speed formula to solve for \( P \).
\[
v = \frac{2\pi R}{P} \implies P = \frac{2\pi R}{v}
\]
Recall the proportionality relation form of Kepler’s Law of planetary motion, \( P^2 \propto a^3 \). Note that \( a \) is the same as \( R \) when we’re approximating the planetary orbits as circles, so use \( R \) for the distance.
### 5)
Use the expression you wrote down in question 4 relating speed (\( v \)), period (\( P \)), and distance from the Sun (\( R \)), to eliminate \( P \) from the proportionality form of Kepler’s Law (\( P^2 \propto a^3 \)). You should be left with an expression only in terms of speed (\( v \)) and distance (\( R \)). Simplify your answer.
\[
\left( \frac{2\pi R}{v} \right)^2 \propto R^3
\]
\[
\frac{4\pi^2 R^2}{v^2} \propto R^3 \implies 4\pi^2 \frac{R^2}{v^2} \propto R^3 \implies \frac{1}{v^2} \propto \frac{1}{R} \to R \propto \frac{v^2}{4\pi^2} \equiv \text{constant}
\]
### 6)
Rearrange this proportionality to solve for speed (\(
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