Answer:
7532m
Explanation:
Gravity on the surface of the earth with radius R is given by:
[tex]F_1=\frac{Gm_{earth}m_{object}}{R^2_{earth}}[/tex]
Gravity a distance r above the surface:
[tex]F_2=\frac{Gm_{earth}m_{object}}{(R+r)^2}[/tex]
How big is r if:
[tex]F_2=0.21\times F_1[/tex]
You get the following equation:
[tex]R^2=0.21(R+r)^2=0.21R^2+0.41Rr+0.21r^2[/tex]
Solve for r:
[tex]r^2+2Rr-\frac{79}{21} R^2=0[/tex]
[tex]r=-R+\sqrt{R^2+\frac{79}{21}R^2}=-R+\sqrt{\frac{100}{21}R^2}=(\frac{10}{\sqrt{21}}-1)R[/tex]
