Acceleration due to gravity at the surface of the earth is g. At what depth, the acceleration due to gravity becomes $\frac{g}{2}$ , if the radius of the earth is R?
$R$
$\frac{R}{2}$
$\frac{R}{4}$
$2 R$

\frac{R}{2}

\frac{R}{4}

R

2 R

Open in App

Solution

Verified by Toppr

Was this answer helpful?

Similar Questions

Acceleration due to gravity at the surface of the earth is g. At what depth, the acceleration due to gravity becomes

\frac{g}{2}

, if the radius of the earth is R?

View Solution

ntThe height from the surface of the Earth at which the acceleration due to gravity becomes g/9 , where g is the acceleration due to gravity on the surface of the Earth, in terms of radius of Earth R is?n nt(A) 2R (B) R/2 (C)R/2 (D) 2Rn

View Solution

The height at which the acceleration due to gravity becomes

\frac{g}{9}

(where g = the acceleration due to gravity on the surface of th earth) in terms of

R

, radius of the earth is

View Solution

At what angular speed the Earth should rotate, so that effective acceleration due to gravity at

45^{\circ}

latitude becomes half of effective acceleration due to gravity at pole? (Radius of earth is

R

and acceleration due to gravity without taking variation of Earth’s rotation at surface is

‘ g^{'}

)

View Solution

At what height above the surface of earth acceleration due to gravity becomes

1 m m / s^{2}

?
[Take

R

as radius of earth and acceleration due to gravity at the surface of earth

g = 10 m / s^{2}

]

View Solution

Acceleration due to gravity at the surface of the earth is g. At what depth, the acceleration due to gravity becomes g2, if the radius of the earth is R?RR2R42R

Acceleration due to gravity at the surface of the earth is g. At what depth, the acceleration due to gravity becomes $\frac{g}{2}$ , if the radius of the earth is R?
$R$
$\frac{R}{2}$
$\frac{R}{4}$
$2 R$