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
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
Show Hint
- $4800 \, km$
- $640 \, km$
- $2560\, km$
- $5260\, km$
The Correct Option is A
Approach Solution - 1
\(\frac {GM}{R^2}\)[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
Approach Solution -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.
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