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## anonymous one year ago a

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1. Michele_Laino

If 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

2. Michele_Laino

now, 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|

3. Michele_Laino

substituting W, we have: $\Large \frac{1}{{16}}\frac{{GMm}}{{{R^2}}} = \frac{{GMm}}{{{{\left( {R + h} \right)}^2}}}$

4. Michele_Laino

and 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

5. Michele_Laino

that's right! it is h=3*R=3*6400= 19200 Km

6. Michele_Laino

ok!

7. Michele_Laino

thanks! :)

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