## Algebra: A Combined Approach (4th Edition)

Given: $v=\sqrt{\dfrac{2Gm}{r}}$ $G=6.67\times10^{-11}m^3/kg\bullet s^2$. mass of Earth is $5.97\times10^{24}$ kg radius of Earth is $6.37\times10^6$ m Find escape velocity for Earth in meters per second. Plug in given information into the formula and solve: $v=\sqrt{\dfrac{2(6.67\times10^{-11}m^3/kg\bullet s^2)(5.97\times10^{24}kg)}{6.37\times10^6m}}$ $v=\sqrt{\dfrac{7.96398\times10^{14}m^3/s^2}{6.37\times10^6m}}$ $v=\sqrt{125023233.9m^2/s^2}$ $v=11181.37889$ m/s Round answer to the nearest whole number: The escape velocity for Earth is 11181 meters per second.