Answer to Question #169719 in Physics for Pushpo

Question

Obtain the escape velocity of an atmospheric particle 100 km above the earth’s surface. Given that ME = 5.98 × 1024 kg, RE = 6370 km, G = 6.67 × 10-11 N.m2/kg2. 

Answer 1

The escape velocity can be found as follows:

$$v_{esc}=\sqrt{\dfrac{2GM}{R+h}},$$ $$v_{esc}=\sqrt{\dfrac{2\cdot6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot5.98\cdot10^{24}\ kg}{6370\cdot10^3\ m+100\cdot10^3\ m}}=11.1\ \dfrac{km}{s}.$$