Question

variation of g with height from the Earth surface

Answer 1

Hey Mate !!

If the object is placed at a distance h above the surface of the earth, the force of gravitation on it due to earth is

$F= \frac{GMm}{(R + {h}^{2}) }$

where M is the mass of the earth and R is its Radius. Thus,

$g = \frac{F}{m} = \frac{GM}{(R + {h}^{2}) }$

We see that the value of g decreases as one goes up. we can write,

$g = \frac{GM}{ {R}^{2}( {1 + \frac{h}{ {R}^{2}}) }^{2} } = \frac{ g_0 }{ ({1 + \frac{R}{h} )}^{2} }$

Where g0 = GM/R² is the value of g at the surface of the earth. If h<<R.

$g = g_0 {(1 + \frac{h}{R}) }^{ - 2} ≈ g_0(1 - \frac{2h}{R})$

If one goes a distance h inside the earth such as in mines, the value of g again decreases. The Force by the earth is, by equation.

$F = \frac{GMm}{{R}^{3} } (R - h)$

Or,

$g = \frac{F}{m} = \frac{GM}{ {R}^{2} } ( \frac{R - h}{R} )$

$= g_0 (1 - \frac{h}{R} )$

The value of g us maximum at the surface of the earth and decreases with the increase in height as well as with depth similar .

___________________ThankYou !!