Answer to Question #153584 in Quantum Mechanics for rk

Question

Obtain expression for escape velocity?

Answer 1

Solution. The escape velocity is this is the speed that must be given to a physical object so that it, moving parallel to the cosmic body, could not fall on it, but at the same time it would remain in a constant orbit. In this case, the object in orbit will be at rest, since two forces will act on it: centrifugal force and gravity.

$$\frac{mv^2_{esc}}{R}=\frac{GMm}{R^2}$$

where m is body mass; v_esc is escape velocity, R is orbital radius; G=6.67x10^-11Nm²/kg² is gravitational constant; M is cosmic body mass. Therefore, the escape velocity is eqaul to

$$v_{esc}=\sqrt{\frac{GM}{R}}$$

Answer.

$$v_{esc}=\sqrt{\frac{GM}{R}}$$

where R is orbital radius; G=6.67x10^-11Nm²/kg² is gravitational constant; M is cosmic body mass.