If the radius of the Earth is R,then the heig | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g' = g \left( \frac{R}{R + h} \right)^2$.Given

Vedclass · Accepted Answer

$R$

Vedclass · Answer

(C) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula: $g' = g \left( \frac{R}{R + h} \right)^2$.Given that the value of $g$ becomes one-fourth,we have $g' = \frac{g}{4}$.Substituting this into the formula: $\frac{g}{4} = g \left( \frac{R}{R + h} \right)^2$.Canceling $g$ from both sides: $\frac{1}{4} = \left( \frac{R}{R + h} \right)^2$.Taking the square root of both sides: $\frac{1}{2} = \frac{R}{R + h}$.Cross-multiplying gives: $R + h = 2R$.Solving for $h$: $h = 2R - R = R$.

If the radius of the Earth is $R$,then the height $h$ at which the value of $g$ becomes one-fourth is:

Similar Questions

What should be the angular velocity of earth due to rotation about its own axis so that the weight at the equator becomes $(3/5)^{th}$ of its initial value? ($g$ = acceleration due to gravity, $R$ = radius of earth)

What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

What is the ratio of the acceleration due to gravity on two planets with radii $R_1$ and $R_2$ and densities $\rho_1$ and $\rho_2$?

The magnitudes of the gravitational field at distances $r_1$ and $r_2$ from the center of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then:

$T$ is the time period of a simple pendulum on the Earth's surface. Its time period becomes $xT$ when taken to a height $R$ (equal to the Earth's radius) above the Earth's surface. Then,the value of $x$ will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module