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fizixfan

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I was doing some calculations using the escape velocities from Earth, Moon and Mars. Then by chance I calculated the velocities attained when an object was "dropped" from a height of the radius on each of these bodies, assuming the acceleration due to gravity remained constant during the fall. Escape velocity is the minimum velocity needed to escape a gravitational field.

For example, escape velocity on earth,

Using the simple formula

V2 = velocity after falling distance r; a = acceleration due to gravity; r = radius of Earth (initial velocity = 0), and

a = 9.80665 m/s^2

r = 6,378,000 m

This value of

I thought perhaps this was a coincidence, so I calculated V2 for the Moon and Mars and compared them to the known escape velocities for each body.

Moon:

Escape velocity =

a = 1.62 m/s^2

r = 1,737,150 m

Again, a very close fit.

Mars:

Escape velocity = Vesc =

a = 3.72761 m/s^2

r = 3,389,500 m

V2 = (2*3.72761*3389500)^0.5 = 5026.875 m/s =

Another very close approximation.

I haven't done the calculations for the other planets, moons, or the sun, but I expect I would get similar results. From these results I speculate that Vesc = V2. This seems to be a quick and dirty formula for calculating escape velocity, and might be handy for a stranded astronaut. But am I really just reiterating Vesc = (2Gm/r)^0.5? (G = gravitational constant; m = mass, e.g., of earth; r = radius, e.g., of earth).

I any case, using this method avoids a lot of the very intricate calculations necessary for determining Vesc the standard way, as long as the acceleration due to gravity and radius are known.

Sources:

http://www.livescience.com/50312-how-long-to-fall-through-earth.html

[PLAIN]http://keisan.casio.com/exec/system/1360310353[/PLAIN]

http://www.wolframalpha.com/input/?i=escape+velocity+Mars

For example, escape velocity on earth,

**Vesc = 40,270 km/h**(given).Using the simple formula

**V2 = (2ar)^0.5**, whereV2 = velocity after falling distance r; a = acceleration due to gravity; r = radius of Earth (initial velocity = 0), and

a = 9.80665 m/s^2

r = 6,378,000 m

**V2 =**(2*9.80665*6378000)^0.5 = 11,184.53 m/s =**40,264.29 km/h**This value of

**40,264 km/h**compares very closely to the escape velocity of**40,270 km/h**I thought perhaps this was a coincidence, so I calculated V2 for the Moon and Mars and compared them to the known escape velocities for each body.

Moon:

Escape velocity =

**Vesc =****8,533.6 km/h**a = 1.62 m/s^2

r = 1,737,150 m

**V2**= (2*1.62*1737150)^0.5 = 2,372.418 m/s =**8,540.7 km/h**Again, a very close fit.

Mars:

Escape velocity = Vesc =

**18,108 km/h**a = 3.72761 m/s^2

r = 3,389,500 m

V2 = (2*3.72761*3389500)^0.5 = 5026.875 m/s =

**18,097 km/h**Another very close approximation.

I haven't done the calculations for the other planets, moons, or the sun, but I expect I would get similar results. From these results I speculate that Vesc = V2. This seems to be a quick and dirty formula for calculating escape velocity, and might be handy for a stranded astronaut. But am I really just reiterating Vesc = (2Gm/r)^0.5? (G = gravitational constant; m = mass, e.g., of earth; r = radius, e.g., of earth).

I any case, using this method avoids a lot of the very intricate calculations necessary for determining Vesc the standard way, as long as the acceleration due to gravity and radius are known.

Sources:

http://www.livescience.com/50312-how-long-to-fall-through-earth.html

[PLAIN]http://keisan.casio.com/exec/system/1360310353[/PLAIN]

http://www.wolframalpha.com/input/?i=escape+velocity+Mars

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