Question

At a certain depth "$d$" below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $3 R$ above earth surface Where $R$ is Radius of earth (Take $R =6400\, km$ ) The depth $d$ is equal to

• A. $4800 \, km$
• B. $640 \, km$
• C. $2560\, km$
• D. $5260\, km$

Hint:

The acceleration due to gravity at a certain depth below the Earth's surface decreases, and at a height above the surface, it decreases as well, following the inverse square law.

Answer 1

The Correct Option is A

\(\frac {GM}{R^2}\)​[1−\(\frac dR\)​] = \(\frac {4 \times GM}{(4R)^2}\)
1−\(\frac dR\)​ = \(\frac 14\)
⇒ \(\frac dR\)​ = \(\frac 34\) 
​⇒ d=\(\frac 34\)​R 
⇒ d= \(\frac 34\) x 6400
⇒  d=4800 km

So, the correct option is (A): 4800 km

Answer 2

Using the formula for gravitational force, we have:

\[ \frac{GM}{R^2} \left( 1 - \frac{d}{R} \right) = \frac{4 \times GM}{(4R)^2} \]

Simplifying:

\[ 1 - \frac{d}{R} = \frac{1}{16} \]

\[ \frac{d}{R} = 1 - \frac{1}{16} = \frac{15}{16} \]

\[ d = \frac{15}{16} \times R = \frac{15}{16} \times 6400 \, \text{km} = 4800 \, \text{km} \]

Thus, the depth is 4800 km.