The height of the point vertically above the | vedclass

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

Question

(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula $g' = g \left( \frac{R}{R + h} \right)^2$,where

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula $g' = g \left( \frac{R}{R + h} ight)^2$,where $g$ is the acceleration due to gravity at the surface and $R$ is the radius of the earth.Given that $g' = 1\% 	ext{ of } g$,we have $g' = \frac{1}{100} g$.Substituting this into the formula: $\frac{1}{100} g = g \left( \frac{R}{R + h} ight)^2$.Canceling $g$ from both sides: $\frac{1}{100} = \left( \frac{R}{R + h} ight)^2$.Taking the square root of both sides: $\frac{1}{10} = \frac{R}{R + h}$.Cross-multiplying gives: $R + h = 10R$.Therefore,$h = 10R - R = 9R$.

Column-$I$	Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$	$(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$	$(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$	$(c)$ At the equator

The height of the point vertically above the earth's surface,at which the acceleration due to gravity becomes $1\%$ of its value at the surface is (Radius of the earth $= R$). (in $,R$)

Similar Questions

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth

$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles

$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

What should be the angular speed of the Earth about its axis so that the acceleration due to gravity at the equator becomes zero?

If it is assumed that the spinning motion of the Earth increases,then the weight of a body on the equator

Imagine a new planet having the same density as that of Earth but it is $3$ times bigger than the Earth in size. If the acceleration due to gravity on the surface of Earth is $g$ and that on the surface of the new planet is $g^{\prime}$,then

At what height from the surface of the earth will the value of acceleration due to gravity fall to half of its value on the surface of the earth?

Vedclass Test Series

Exam Paper Generator

Online Exam Module