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Calculate the height of the point vertically above the Earth's surface at the which the value of acceleration due to gravity is reduced to one - fourth of its value at the surface of the earth given that radius of the earth =6400 km?
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Calculate the height of the point vertically above the Earth's surface...
2400km. apply formulae g'=g(1-2h/R) where r is radius of Earth and g'=g/4
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Calculate the height of the point vertically above the Earth's surface...
Given:
Radius of the Earth (r) = 6400 km

To find:
Height (h) at which acceleration due to gravity is reduced to one-fourth of its value at the surface of the Earth.

Formula:
The formula to calculate the acceleration due to gravity at a certain height above the Earth's surface is given by:
g' = (g * (r / (r + h)))^2

Where:
g' = acceleration due to gravity at height h
g = acceleration due to gravity at the surface of the Earth
r = radius of the Earth
h = height above the Earth's surface

Solution:
To find the height at which the acceleration due to gravity is reduced to one-fourth, we need to substitute the given values into the formula.

Let's assume the acceleration due to gravity at the surface of the Earth is g.

So, g' = (g * (r / (r + h)))^2

Given that g' = (1/4)g, we can substitute this value into the equation:

(1/4)g = (g * (r / (r + h)))^2

Taking the square root of both sides, we get:

√((1/4)g) = g * (r / (r + h))

Simplifying further:

√(1/4) = r / (r + h)

1/2 = r / (r + h)

Cross-multiplying:

r = (r + h) / 2

2r = r + h

h = 2r - r

h = r

Therefore, the height at which the acceleration due to gravity is reduced to one-fourth of its value at the surface of the Earth is equal to the radius of the Earth, which is 6400 km.
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