- #36
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
- 16,039
- 2,676
The quantitycloud360 said:as the bracket 2g/gamma(stuff inside bracket), part is positive?
[tex]\frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) [/tex]
is not positive; it's negative. Because r>R, i.e. the particle is above the planet's surface, the first exponential will be smaller than the second exponential, right?
The original inequality actually gives youcloud360 said:so we let
V^2=2g/(gamma*e^(-gamma*R))
[tex]V^2 \ge \frac{2g}{\gamma} e^{-\gamma R}[/tex]
The exponential isn't in the denominator. So now you have
[tex]v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \ge \frac{2g}{\gamma} e^{-\gamma R} + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) [/tex]