Question

6.4 increasing earth’s density (5pts) assuming its mass stays the same, how much smaller does earth’s radius have to be for it to have the same escape velocity as the sun?

Explanation:

Step1: Recall escape - velocity formula

The escape - velocity formula is $v_{e}=\sqrt{\frac{2GM}{R}}$, where $v_{e}$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of the planet, and $R$ is the radius of the planet. 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. We want $v_{eE}=v_{eS}$, so $\sqrt{\frac{2GM_{E}}{R_{E}}}=\sqrt{\frac{2GM_{S}}{R_{S}}}$.

Step2: Square both sides and solve for $R_{E}$

Squaring both sides of the equation $\frac{2GM_{E}}{R_{E}}=\frac{2GM_{S}}{R_{S}}$. Canceling out $2G$ from both sides, we get $\frac{M_{E}}{R_{E}}=\frac{M_{S}}{R_{S}}$. Then, $R_{E}=\frac{M_{E}}{M_{S}}R_{S}$. The mass of the Sun $M_{S}\approx1.989\times 10^{30}\ kg$ and the mass of the Earth $M_{E}\approx5.972\times 10^{24}\ kg$.

Step3: Calculate the ratio of radii

$\frac{R_{E}}{R_{S}}=\frac{M_{E}}{M_{S}}=\frac{5.972\times 10^{24}}{1.989\times 10^{30}}\approx3.00\times 10^{-6}$.

Answer:

The radius of the Earth would have to be approximately $3.00\times 10^{-6}$ times the radius of the Sun for the Earth to have the same escape - velocity as the Sun, assuming the Earth's mass remains the same.