Earth escape velocity using $μ$ $mu$

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Last modified by

MichaelBartmess

on

Jul 24, 2020, 6:28:07 PM

Created by

MichaelBartmess

on

Jan 8, 2014, 1:40:11 AM

v_{e} = \sqrt{(\frac{2 \cdot G \cdot M_{\oplus}}{R_{\oplus} + h})}

$v_e = sqrt(((2*G*M_(o+))/(R_(o+)+ h )))$

Altitude

$"Altitude"$

INPUTS

The input value is the altitude above the earth, h.

NOTES

This equation uses the following constants:

( $G$ $G$ ) Universal Gravity constant
( $M_{\oplus}$ $M_(o+)$ ) Mass of the Earth
( $R_{\oplus}$ $R_(o+)$ ) Mean Equatorial Radius

As a fundamental example, escape velocity from the planet Earth is around 11.2 km/sec.

This equation is based on a spherical earth model.

The energy of a parabolic trajectory is has the minimum energy required to escape Earth's gravitational attraction. We can compute the escape velocity of any object at a distance r = $(R_{\oplus}$ $(R_(o+)$ + h) by making energy equal to zero and solving for velocity.

This gives us $V_{esc}$ $V_"esc"$ = ${(2 \frac{μ}{r})}^{\frac{1}{2}}$ $(2mu/r)^(1/2)$

This equation, Earth escape velocity using

μ

$mu$ , references 3 pages

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μ

$mu$ , is used in 1 page

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Earth escape velocity using μμmu

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NOTES

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Earth escape velocity using $μ$ $mu$