Question

At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.995 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80 m/s2?

Answer 1

If we assume the Earth to be of uniform density and spherical, then Newton's inverse square law of g-field can be applied.

i.e. g = GM/r², which means g is inversely proportional to r²
Or, gr² = constant.

The Earth's radius is approximately 6.4x10^6 m

So, 0.975 x (R')² = 9.80 x (6.4x10^6)², where R' is the distance from Earth's centre.
R' = 2.03x10^7 m

Distance above Earth's surface = 2.03x10^7 - 6.4x10^6 = 1.39x10^7

Answer 2

Answer:

The distance above the surface of the earth is $1.36\times 10^7$.

Explanation:

Given that,

Acceleration due to the gravity g'= 0.995 m/s² (at height )

Acceleration due to the gravity g= 9.80 m/s² (at surface )

We know that,

The formula of gravity at height

$g'=\dfrac{g}{(1+\dfrac{h}{R})^2}$

Where, h = height

r= radius of the earth

We substitute the value into the formula

$0.995=\dfrac{9.8}{(1+\dfrac{h}{6.378\times10^{6}})^2}$

$h=6.378\times 10^6(\sqrt{\dfrac{9.8}{0.995}}-1)$

$h=1.36\times 10^7\ m$

Hence, The distance above the surface of the earth is $1.36\times 10^7$.