Question

The value of 'g' changes with the height ? Explain ?
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Answer 1

value of g is inversely proportional to

${r}^{2}$

.So. increasing the height means increasing the value of r (that is measured between centres of both objects). if value of r increases then g decreases and vice versa.

Answer 2

Yes, the value of 'g' changes with height.

The acceleration involved when an object falls towards the earth is due to the earth's gravitational force. This acceleration is known as the acceleration due to the gravitational pressure (or acceleration because of gravity). It is denoted using 'g' and its unit $ms^{-2}$.

'g' is affected by the change in the peak of the object in the following manner:

Let A be the point on the surface of the Earth and B be the point at an altitude 'h', let, M be the mass of the earth and R be the radius of the earth.

Consider the earth as a round body.

The acceleration because of gravity at point A on the surface of the Earth is

$y = \frac{CM}{R^{2} }$    ...(1)

Let, the body be placed at B at a height h from the surface of the Earth.

The acceleration due to gravity at B is,

$g^{1} = \frac{CM}{(R \ + \ h)^{2} }$   ...(2)

Dividing equation (1) by (2), we get

$\frac{g}{g^{1} } = \frac{\frac{CM}{R^{2} } }{[\frac{CM}{(R \ + \ h)^{2} }] }$

$\frac{g}{g^{1} } = \frac{CM*(R \ + \ h)^{2} }{CM*R^{2} }$

$\frac{g}{g^{1} } = \frac{(R \ + \ h)^{2} }{R^{2} }$

$\frac{g}{g^{1} } = (\frac{R}{R} \ + \ \frac{h}{R})^{2} = ( 1 \ + \ \frac{h}{R}) ^{2}$

$g^{1} = g(1 \ + \ \frac{h}{R}) ^{-2}$

Now, expanding using the binomial theorem, we get

$g^{1} = g(1 - \frac{2h}{R})$

The value of acceleration due to gravity 'g' decreases with an increase in height above the surface of the Earth.

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