What is the acceleration due to gravity at distance of 2 earth radii above earth's surface?

Answer 1

The force of gravity between two masses is inversely proportional to the square of the distance

between them.

So when you move from (1 earth radius away from the center) out to (10 of them), the force between

you and the earth becomes

1/102 = 1/100 = 1 percent

of what it was originally.

Notice that gravity doesn't "become weaker" out there. It still follows exactly the same physical laws.

And in precise accordance with those laws, the mutual gravitational force between you and the earth

becomes 99% less than it is at the earth's surface.

Answer 2

The distance from the earth's center of mass at 2 radii is 3 times the value

at the surface.

Since the acceleration (and force) of gravity vary as 1/R2 , they decrease to

1/9th of the surface value at 2 radii off the surface.

[ 9.8 meters (32.2 feet) per second2 ] / 9 = 1.09 meter (3.57 ft) per second2

(This is all ignoring any effects of being closer to the moon, which we haven't

even bothered to look at and decide whether or not it's significant. In any case,

the calc above refers exclusively to the acceleration of gravity due to the earth's

presence.)

Answer 3

The acceleration of gravity is inversely proportional to the square of the
distance from the mass.

At 3 Earth radii above the surface, your distance from the Earth's center is
3 times the distance on the surface, and the acceleration of gravity is

1/(3^2) = 1/9

of its value on the surface.

The surface value is 9.807 m/s^2 (rounded). So at 3 Earth radii up
(about 12,000 miles), it's

(9.807/9) = 1.09 m/s^2 (rounded) .

Answer 4

The acceleration of gravity is inversely proportional to the square of the distance from the mass.

At 3 Earth radii above the surface, your distance from the Earth's center is 4 times the distance on the surface (the surface is one Earth radius from the center, so if you are 3 radii above the surface you are a total of 3+1=4 radii from the center).

Thus, the acceleration of gravity is1/(4^2) = 1/16 of its value on the surface.

The surface value is 9.807 m/s^2 (rounded). So at 3 Earth radii up(about 12,000 miles), it's(9.807/16) = 0.61 m/s^2 (rounded) .

Answer 5

The acceleration due to gravity at a distance of 2 Earth radii above Earth's surface is approximately one-fourth (1/4) of the acceleration due to gravity at the Earth's surface. This is because the gravitational force decreases with the square of the distance.

Answer 6

Earth doesn't pull us down, but space pushes us into the earth.

Answer 7

a = 9.82 / ((2 / 1)^2) = 9.82 / 4 = 2.455 ((m/s)/s)

Answer 8

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