At what height above the earth's surface does | vedclass

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

Question

(A) Given: $g_{h} = 1\% 	ext{ of } g_{s} = 0.01 g_{s}$.The formula for acceleration due to gravity at height $h$ is $g_{h} = g_{s} \left( \frac{R}{R+

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) Given: $g_{h} = 1\% 	ext{ of } g_{s} = 0.01 g_{s}$.The formula for acceleration due to gravity at height $h$ is $g_{h} = g_{s} \left( \frac{R}{R+h} ight)^{2}$.Substituting the given value: $0.01 g_{s} = g_{s} \left( \frac{R}{R+h} ight)^{2}$.Dividing both sides by $g_{s}$: $0.01 = \left( \frac{R}{R+h} ight)^{2}$.Taking the square root on both sides: $0.1 = \frac{R}{R+h}$.Rearranging the equation: $0.1(R + h) = R$.$0.1R + 0.1h = R$.$0.1h = 0.9R$.$h = \frac{0.9}{0.1} R = 9R$.Thus,at a height of $9R$,the acceleration due to gravity falls to $1\%$ of its value at the surface.

At what height above the earth's surface does the acceleration due to gravity fall to $1\%$ of its value at the earth's surface (in $,R$)?

Similar Questions

If the acceleration due to gravity at the Earth is $g$,the mass of the Earth is $80$ times that of the Moon,and the radius of the Earth is $4$ times that of the Moon,then the value of the acceleration due to gravity at the surface of the Moon will be:

What is the change in mass of a body,when taken $64 \ km$ below the surface of the earth? $[$Take radius of the earth as $6400 \ km]$

Since the Earth is not a perfect sphere,what is the effect on the acceleration due to gravity $(g)$?

What is the acceleration due to gravity at a distance of $2R$ from the surface of the Earth,where $R$ is the radius of the Earth?

Considering Earth to be a sphere of radius $R$ having uniform density $\rho$,the value of acceleration due to gravity $g$ in terms of $R$,$\rho$,and $G$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module