6.2 escape velocity of earth (5pts)\nthe sun is 333,000 times the mass and 109 times the radius of the…

6.2 escape velocity of earth (5pts)\nthe 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?
Answer
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:
$11.18\ km/s$