Question

6.2 escape velocity of earth (5pts)
the sun is 333,000 times the mass and 109 times the radius of the earth. what velocity (in km/s) is required to leave earths surface?

Explanation:

Step1: Recall escape - velocity formula

The escape - velocity formula is $v = \sqrt{\frac{2GM}{r}}$, where $G$ is the gravitational constant ($G=6.67\times 10^{- 11}\ m^{3}\ kg^{-1}\ s^{-2}$), $M$ is the mass of the planet, and $r$ is the radius of the planet.

Step2: Use standard values for Earth

The mass of the Earth $M_E$ and radius of the Earth $r_E$. The standard mass of the Earth $M_E = 5.97\times 10^{24}\ kg$ and the standard radius of the Earth $r_E=6371\times 10^{3}\ m$.

Step3: Calculate escape - velocity

Substitute the values into the formula:
\[

$$\begin{align*} v&=\sqrt{\frac{2\times6.67\times 10^{-11}\times5.97\times 10^{24}}{6371\times 10^{3}}}\\ &=\sqrt{\frac{2\times6.67\times5.97\times 10^{-11 + 24}}{6371\times 10^{3}}}\\ &=\sqrt{\frac{2\times6.67\times5.97\times 10^{13}}{6371\times 10^{3}}}\\ &=\sqrt{\frac{79.5798\times 10^{13}}{6371\times 10^{3}}}\\ &=\sqrt{1.25\times10^{10}}\\ & = 11180.34\ m/s \end{align*}$$

\]

Step4: Convert to km/s

To convert from m/s to km/s, divide by 1000: $v=\frac{11180.34}{1000}=11.18\ km/s$

Answer:

$11.18\ km/s$