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Ostrovityanka [42]
2 years ago
12

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

Physics
2 answers:
Sladkaya [172]2 years ago
6 0
Gravity changes as the altitude change.<span> 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
</span>

<span>Hope my answer would be a great help for you.    
If </span>you have more questions feel free to ask here at Brainly.

<span> </span>


OlgaM077 [116]2 years ago
4 0

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

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power by kinetic energy = 1/2(mv²)

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u

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El peso del signo = 397,97 N.

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