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--- 183_notes:escape_speed [2021/04/01 12:35] – [Escape Speed] stumptyl
+++ 183_notes:escape_speed [2021/04/01 12:37] (current) – [Calculating the Escape Speed] stumptyl
@@ Line 26: / Line 26: @@
   * Final: $r=\infty$; $v=0$
-The [[183_notes:energy_cons|Energy Principle tells you that the energy of this system is conserved]]. Moreover, you know from the final state where $r\rightarrow \infty$ and $v\rightarrow 0$ that both the final potential and final kinetic energies are zero. Hence, the total energy of the system must be zero.
+The [[183_notes:energy_cons|Energy Principle tells you that the energy of this system is conserved]]. __Moreover, you know from the final state where $r\rightarrow \infty$ and $v\rightarrow 0$ that both the final potential and final kinetic energies are zero. Hence, the total energy of the system must be zero.__
 $$\Delta E_{sys} = W_{surr} = 0 \longrightarrow E_{sys,f} = E_{sys,i}$$
@@ Line 40: / Line 40: @@
 $$v_{esc} = \sqrt{\dfrac{2GM}{R}}$$
-This speed defines the minimum speed needed to leave the planet and never return under the gravitational interaction between the object and the planet. Notice that this speed doesn't depend on the mass of the launched object (so long as it is much less than the planet).
+This speed defines the minimum speed needed to leave the planet and never return under the gravitational interaction between the object and the planet. **__Reminder that this speed term utilizes the SI unit of m/s.__**  Notice that this speed doesn't depend on the mass of the launched object (so long as it is much less than the planet).