Question

How far (in meters) above the earth's surface will the acceleration of gravity be 21.0 % of what it is on the surface?

Answer 1

Answer:

7532m

Explanation:

Gravity on the surface of the earth with radius R is given by:

$F_1=\frac{Gm_{earth}m_{object}}{R^2_{earth}}$

Gravity a distance r above the surface:

$F_2=\frac{Gm_{earth}m_{object}}{(R+r)^2}$

How big is r if:

$F_2=0.21\times F_1$

You get the following equation:

$R^2=0.21(R+r)^2=0.21R^2+0.41Rr+0.21r^2$

Solve for r:

$r^2+2Rr-\frac{79}{21} R^2=0$

$r=-R+\sqrt{R^2+\frac{79}{21}R^2}=-R+\sqrt{\frac{100}{21}R^2}=(\frac{10}{\sqrt{21}}-1)R$