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anonymous
 one year ago
a
anonymous
 one year ago
a

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1If I call with m the mass of an object upon the earth surface, then its weight W is: \[\Large W = \frac{{GMm}}{{{R^2}}}\] where G is the Newton's constant, M is the earth mass, and R is the earth radius

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, if I call with h, the requested height then I can write: \[\Large \frac{W}{{16}} = \frac{{GMm}}{{{{\left( {R + h} \right)}^2}}}\] dw:1437246469242:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1substituting W, we have: \[\Large \frac{1}{{16}}\frac{{GMm}}{{{R^2}}} = \frac{{GMm}}{{{{\left( {R + h} \right)}^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and after a simplification, we can write: \[\Large \frac{1}{{16}} = \frac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}\] or: \[\Large \frac{1}{{16}} = {\left( {\frac{R}{{R + h}}} \right)^2}\] and finally: \[\Large \frac{1}{4} = \frac{R}{{R + h}}\] please solve the last equation for h

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that's right! it is h=3*R=3*6400= 19200 Km
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