Question

How high above the earth's surface is g reduced to 8.80m/^2?

Answer 1

Gravity changes as the altitude change. The gravitational force is proportional to 1/R2,  where R is the distance from the center of the Earth the radius of earth where gravity is 9.8 m/s^2 is 6400 km this will serve as the zero mark.

g1/(g2) = R2^2/(R1)^2

so we set the constant values to R1 and the unknown distance as x

(9.8)/(8.80) = (6400-x)2/(6400)^2

solving for x we will get 

x = 345.85 km above the earths surface

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Answer 2

Answer:

$355.8km$

Explanation:

We use the equation to calculate gravitational acceleration $g$

$g=\frac{GM}{r^2}$

where  is the universal gravitational constant $G=6.67x10^{-11}m^3/kgs^2$, $M$ is the mass of the earth: $M=5.97x10^{24}kg$, and $r$ is the radius from the center of earth.

Clearing for the distance r:

$r=\sqrt{\frac{GM}{g} }$

substituting known values

$r=\sqrt{\frac{(6.67x10^{-11}m^3/kgs^2)(5.97x10^{24}kg)}{8.80m/s^2} }$

$r=\sqrt{\frac{3.982^{14}m^3/s^2}{8.80m/s^2} }$

$r=\sqrt{4.525x10^{13}m^2 }$

$r=6.7268x10^6m$

and since the radius of earth from the core to the surface is about $6.371x10^6m$, the distance above the surface where g is reduce to $8.80m/s^2$  is:

$r=6.7268x10^6m-6.371x10^6m=355,800m$ wich in kilometers is: $355.8km$