The value of the acceleration due to gravity is g 1 at a height h =( R /2)( R = radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth. The ratio (( d / R )) equals :

Question

The value of the acceleration due to gravity is g 1 at a height h =( R /2)( R = radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth. The ratio (( d / R )) equals : (A) (7/9) (B) (4/9) (C) (1/3) (D) (5/9)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d6525fdc23e0e6ee24404687172308d9-.png" /> g 2=( GM ( R - d )/ R 3) dots (2) g1=g2 ( GM /((3 R )2)2)=( GM ( R - d )/ R 3) ⇒ (4/9)=(( R - d )/ R ) 4 R =9 R -9 d 5R=9d ⇒ ( d / R )=(5/9)

Q. The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2} (R =$ radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $(\frac{d}{R})$ equals :

$\frac{7}{9}$

$\frac{4}{9}$

$\frac{1}{3}$

$\frac{5}{9}$

$g_{2} = \frac{GM ( R - d )}{R ^{3}} \dots (2)$
$g_{1} = g_{2}$
$\frac{GM}{( \frac{3 R}{2} ) ^{2}} = \frac{GM ( R - d )}{R ^{3}}$
$\Rightarrow \frac{4}{9} = \frac{( R - d )}{R}$
$4 R = 9 R - 9 d$
$5 R = 9 d$
$\Rightarrow \frac{d}{R} = \frac{5}{9}$

Q. The value of the acceleration due to gravity is g1​ at a height h=2R​(R= radius of the earth) from the surface of the earth. It is again equal to g1​ at a depth d below the surface of the earth. The ratio (Rd​) equals :

97​

94​

31​

95​