## 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

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?

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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/ru00b2, which means g is inversely proportional to ru00b2
Or, gru00b2 = constant.

The Earth’s radius is approximately 6.4×10^6 m

So, 0.975 x (R’)u00b2 = 9.80 x (6.4×10^6)u00b2, where R’ is the distance from Earth’s centre.
R’ = 2.03×10^7 m

Distance above Earth’s surface = 2.03×10^7 – 6.4×10^6 = 1.39×10^7

The distance above the surface of the earth is [tex]1.36times 10^7[/tex].

Explanation:

Given that,

Acceleration due to the gravity g’= 0.995 m/su00b2 (at height )

Acceleration due to the gravity g= 9.80 m/su00b2 (at surface )

We know that,

The formula of gravity at height

[tex]g’=dfrac{g}{(1+dfrac{h}{R})^2}[/tex]

Where, h = height

We substitute the value into the formula

[tex]0.995=dfrac{9.8}{(1+dfrac{h}{6.378times10^{6}})^2}[/tex]

[tex]h=6.378times 10^6(sqrt{dfrac{9.8}{0.995}}-1)[/tex]

[tex]h=1.36times 10^7 m[/tex]

Hence, The distance above the surface of the earth is [tex]1.36times 10^7[/tex].