6.4 increasing earth’s density (5pts)\nassuming its mass stays the same, how much smaller does earth’s…

Explanation:

Step1: Recall escape - velocity formula

The escape - velocity formula is $v_{e}=\sqrt{\frac{2GM}{R}}$, where $G$ is the gravitational constant, $M$ is the mass of the planet or star, and $R$ is its radius.

Let $M_{E}$ be the mass of the Earth, $R_{E}$ be the radius of the Earth, $M_{S}$ be the mass of the Sun, and $R_{S}$ be the radius of the Sun. The escape - velocity of the Earth is $v_{eE}=\sqrt{\frac{2GM_{E}}{R_{E}}}$, and the escape - velocity of the Sun is $v_{eS}=\sqrt{\frac{2GM_{S}}{R_{S}}}$.

We want $v_{eE}=v_{eS}$, and we are given that $M_{E}$ remains the same. So, $\sqrt{\frac{2GM_{E}}{R_{E}}}=\sqrt{\frac{2GM_{S}}{R_{S}}}$.

Squaring both sides gives $\frac{2GM_{E}}{R_{E}}=\frac{2GM_{S}}{R_{S}}$. Since $G$ cancels out, we have $\frac{M_{E}}{R_{E}}=\frac{M_{S}}{R_{S}}$.

We know that the mass of the Sun $M_{S}\approx1.989\times 10^{30}\text{ kg}$ and the mass of the Earth $M_{E}\approx5.972\times 10^{24}\text{ kg}$. Let the new radius of the Earth be $R_{E}'$ such that $\frac{M_{E}}{R_{E}'}=\frac{M_{S}}{R_{S}}$.

We can rewrite it as $R_{E}'=\frac{M_{E}}{M_{S}}R_{S}$.

The radius of the Sun $R_{S}\approx6.96\times 10^{8}\text{ m}$.

Step2: Calculate the new radius of the Earth

$R_{E}'=\frac{5.972\times 10^{24}}{1.989\times 10^{30}}\times6.96\times 10^{8}$

First, calculate $\frac{5.972\times 10^{24}}{1.989\times 10^{30}}=\frac{5.972}{1.989}\times10^{24 - 30}\approx3.0\times10^{-6}$.

Then, $R_{E}'=3.0\times10^{-6}\times6.96\times 10^{8}=3.0\times6.96\times10^{-6 + 8}=208.8\text{ m}$.

The actual radius of the Earth $R_{E}\approx6.371\times 10^{6}\text{ m}$.

To find out how many times smaller, calculate the ratio $\frac{R_{E}'}{R_{E}}=\frac{208.8}{6.371\times 10^{6}}\approx3.3\times10^{-5}$.

Answer:

The Earth's radius has to be approximately $3.3\times 10^{-5}$ times its original radius.