One of the most impressive achievements in the history of mankind is, undoubtedly, the man on the Moon, our natural satellite that is to 384,000 km .
Tuesday, April 20, 2010
How Long Microlight Batteries Last
From the Earth to the Moon
One of the most impressive achievements in the history of mankind is, undoubtedly, the man on the Moon, our natural satellite that is to 384,000 km .
One of the most impressive achievements in the history of mankind is, undoubtedly, the man on the Moon, our natural satellite that is to 384,000 km .
is not an easy journey, since among many other things, it is necessary to achieve victory, at least in part, the Earth's gravitational attraction. Clear that the Moon helps a little, then, could ask, account
Let cool listening to Frank Sinatra's interpretation of
But then, let's put aside the details and evaluate the orders of magnitude! Let
\\ [F_ \\ Oplus = G \\ frac {M_ \\ Oplus m} {r ^ 2} \\ ]
To reach the moon, the spacecraft must be able to
\\ [F_ \\ Oplus = F_L \\]
\\ [G \\ frac {M_ \\ Oplus m} {r ^ 2} = G \\ frac {m} {M_L (dr) ^ 2} \\]
\\ [\\ frac {(dr) ^ 2} {r ^ 2} = \\ frac {M_L} {M_ \\ Oplus} \\] \\ [\\ frac {d} {r} -1 = \\ pm \\ sqrt {\\ frac {M_L } {M_ \\ Oplus}} \\] 
As in every square, we have two possible solutions. Which of the two we want? In this case we care about the solution associated with positive sign (the other is the other side of the moon!). Putting the values \u200b\u200bof the masses, we obtain the desired result :
No wonder they have had to use a rocket like the Saturn V
Let cool listening to Frank Sinatra's interpretation of
Fly me to the moon 
.
. But then, let's put aside the details and evaluate the orders of magnitude! Let
lunar probe then mass $ m $ on the line joining the centers of the Earth ($ \\ Oplus $) and Moon ($ L $), at a distance $ r $ from the center of the Earth . In this case, Sir Isaac tells us that on the tube are two forces acting in the same direction, opposite directions and magnitudes
\\ [F_ \\ Oplus = G \\ frac {M_ \\ Oplus m} {r ^ 2} \\ ]
\\ [F_L = G \\ frac {m} {M_L (dr) ^ 2} \\]
being $ G $ the universal gravitational constant and $ d $ the distance from Earth to the moon.
To reach the moon, the spacecraft must be able to
overcome the pull of gravity on Earth, but gets a little help: the moon's gravity that attracts it. The point we want to find is one where both forces of attraction are equal, ie there is a distance $ r_0 $ for which both forces $ F_ \\ Oplus F_L $ $ $ and are equal in magnitude. Then:
\\ [G \\ frac {M_ \\ Oplus m} {r ^ 2} = G \\ frac {m} {M_L (dr) ^ 2} \\]
simplifying and grouping a few things:
\\ [\\ frac {(dr) ^ 2} {r ^ 2} = \\ frac {M_L} {M_ \\ Oplus} \\]
\\ [\\ left ( \\ frac {dr} {r} \\ right) ^ 2 = \\ frac {M_L} {M_ \\ Oplus} \\]
\\ [r = \\ frac {d} {1 \\ pm \\ sqrt {\\ frac {M_L} {M_ \\ Oplus}}} \\]
No wonder they have had to use a rocket like the Saturn
: 10 meters in diameter, 110 feet high, and capable of placing 45,000 kg of payload into lunar orbit.
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