At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.765 m/s^2if the acceleration due to gravity at the surface has magnitude 9.80 m/s^2?answer in meters

1.6441×10^7m

Explanation:

The distance above the surface of the earth can be calculated by finding the difference between the distance between the masses of the object and the radius of the earth which is expressed below

r= RE + h

Where

h=distance above the surface of the earth

r= distance between the masses

RE= radius of the earth= 6.3781×106 m

acceleration due to gravity can be expressed as

(g)= GmE/r^2.

Where r= distance between the masses of object

mE= mass of the earth= 5.972 × 10^24 kg

G =gravitational constant= 6.67×10^-11 m3 s-2 kg-1),

g= acceleration due to the earth's gravity= 0.765

Making r subject of the formula

r=√GmE/g

r=√[(6.67×10^-11) ×5.972 × 10^24 )]/0.765

=√5.207×10^15

=2.282×10^7m

Substitute the values in below expresion

h= r - RE

=(2.282×10^7)-(6.3781×106 m)

=1.6441×10^7m

Hence, distance above the surface of the earth is 1.6441×10^7m

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.

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