Escape velocity from Earth

Problem

Using $v_\text{fuga} = \sqrt{2GM/R}$, calculate the escape velocity from the Earth’s surface ($M_T = 5{,}97\cdot 10^{24}$ kg, $R_T = 6{,}37\cdot 10^6$ m).

Solution

The escape velocity is the minimum velocity needed to move indefinitely away from the planet (zero total mechanical energy). The formula is: $v_\text{fuga} = \sqrt{\frac{2 G M_T}{R_T}}$ Calculating the product $G M_T = 6{,}67\cdot 10^{-11}\cdot 5{,}97\cdot 10^{24} = 3{,}98\cdot 10^{14}$ m$^3$/s$^2$. Substituting: $v_\text{fuga} = \sqrt{\frac{2\cdot 3{,}98\cdot 10^{14}}{6{,}37\cdot 10^6}} = \sqrt{\frac{7{,}96\cdot 10^{14}}{6{,}37\cdot 10^6}} = \sqrt{1{,}25\cdot 10^8}$ $v_\text{fuga} \approx 1{,}12\cdot 10^4~\text{m/s} \approx 11{,}2~\text{km/s}$ $\ev{v_\text{fuga} \approx 11{,}2~\text{km/s}}$

Links

Topics: Gravitation Concepts: Escape velocity Skills: Symbolic setup