Question

at what depth from the surface of the earth is the value of acceleration due to gravity (one fourth) (one third) (half of the value at the surface of the surface?​

Answer 1

Answer:

a) 4800 km

b) 4266.67 km

c) 3200 km

Explanation:

As $g_{d} = g(1-\frac{d}{R} )$

Considering R = 6400 km,

a) $\frac{g}{4} = g(1-\frac{d}{R} )$

⇒ $\frac{1}{4} = (1-\frac{d}{R} )$

⇒ $\frac{d}{R} = \frac{3}{4}$

⇒ $d =\frac{3}{4} * 6400$ = 4800 km

∴ d = 4800 km deep from the surface of the Earth, the value of g will be 1/4rth of the value of g on surface.

b)  $\frac{g}{3} = g(1-\frac{d}{R} )$

⇒ $\frac{1}{3} = (1-\frac{d}{R} )$

⇒ $\frac{d}{R} = \frac{2}{3}$

⇒ $d =\frac{2}{3} * 6400$ = 4266.667 km

∴ d = 4266$\frac{2}{3}$ km deep from the surface of the Earth, the value of g will be 1/3rd of the value of g on surface.

c)  $\frac{g}{2} = g(1-\frac{d}{R} )$

⇒ $\frac{1}{2} = (1-\frac{d}{R} )$

⇒ $\frac{d}{R} = \frac{1}{2}$

⇒ $d =\frac{1}{2} * 6400$ = 3200 km

∴ d = 3200 km deep from the surface of the Earth, the value of g will be half of the value of g on surface.

Thanks !