Question

The ratio between the values of acceleration due to gravity at a height 1 km above and at a depth of 1 km below the earth surface is

Answer 1

Explanation:

as depth increases g increases as height increases g decreases

Answer 2

Given :

Height (h) = 1 m

Depth (d) = 1 m

To Find :

The ratio between the values of acceleration due to gravity at a height 1 km above and at a depth of 1 km below the earth surface.

Solution :

The acceleration due to gravity (g') at a height 'h' above the surface of the earth is -

      g' = g(1 - $\frac{2h}{R}$)

or,  g' = g(1 - $\frac{2 * 1}{6400 }$)                (as radius of earth (R) = 6400 Km)

or,  g' = g($\frac{6400 -2 }{6400}$)

or,  g' = 980 × $\frac{6398}{6400}$             (Taking g = 980 cm/sec²)

∴    g' =  979.69 cm/sec²

Therefore, the acceleration due to gravity at a height 1 km above the earth surface is 979.69 cm/sec² or 9.797 m/sec²

The acceleration due to gravity at a depth(d) of 1 m below the earth's surface (g$_d$) can be given as,

     g$_d$ = g(1 - $\frac{d}{R}$)

or,  g$_d$ = 980 (1 - $\frac{1}{6400}$)

or,  g$_d$ = 980 × $\frac{6399}{6400}$

∴    g$_d$ = 979.85 cm/sec²

Therefore, the acceleration due to gravity at a depth(d) of 1 m below the earth's surface (g$_d$) is 979.85 cm/sec² or 9.799 m/sec²