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?

Respuesta :

Answer:

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

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

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

Where, h = height

r= radius of the earth

We substitute the value into the formula

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

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

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

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

The distance above the surface of the earth will be 1.36×10⁷ m.

What is the distance?

Distance is a numerical representation of the distance between two objects or locations.

The distance can refer to a physical length or an estimate based on other factors in physics or common use. |AB| is a symbol for the distance between two points A and B.

The given data in the problem is;

g' is the acceleration due to the gravity at height = 0.995 m/s²

g is the acceleration due to the gravity at the surface  = 9.80 m/s²

The formula for the gravity at height h is found as;

[tex]\rm g'=\frac{g}{(1+\frac{h}{R})^2 } \\\\\ 0.995=\frac{9.81}{(1+\frac{h}{6.378\times 10^6})^2 } \\\\ \rm h= 6.378 \times 10^6 (\sqrt{\frac{9.8}{0.995} } -1 \\\\ h= 1.36 \times 10^7 \ m[/tex]

Hence the distances above the surface of the earth will be 1.36×10⁷ m.

To learn more about the distance refer to the link;

https://brainly.com/question/26711747