The acceleration due to gravity at an altitud | vedclass

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

Question

(C) The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R}{R + h} \right)^2$. Given that $g' = \frac{g}{2}$,we have $\frac{

Vedclass · Accepted Answer

$1600$

Vedclass · Answer

(C) The acceleration due to gravity at height $h$ is given by $g' = g \left( \frac{R}{R + h} \right)^2$. Given that $g' = \frac{g}{2}$,we have $\frac{1}{2} = \left( \frac{R}{R + h} \right)^2$. Taking the square root on both sides,we get $\frac{1}{\sqrt{2}} = \frac{R}{R + h}$. This implies $R + h = R\sqrt{2}$,so $h = R(\sqrt{2} - 1)$. Given $R = 4000 \ mile$ and $\sqrt{2} \approx 1.414$,we have $h = 4000(1.414 - 1) = 4000(0.414) = 1656 \ mile$. Rounding to the nearest provided option,$h \approx 1600 \ mile$.

The acceleration due to gravity at an altitude $h$ above the Earth's surface is half of its value on the surface of the Earth. If the radius of the Earth $R = 4000 \ mile$,the altitude $h$ is approximately ......... $mile$.

Similar Questions

By approximately how many meters is the radius of the Earth at the equator greater than the radius of the Earth at the poles (in $,000 \ m$)?

In order to make the effective acceleration due to gravity equal to zero at the equator,the angular velocity of rotation of the earth about its axis should be $(g = 10\,m/s^2$ and radius of earth is $6400\,km)$.

Assuming the density of the Earth is constant, which graph correctly represents the variation of acceleration due to gravity $(g)$ with the distance $(r)$ from the center of the Earth (radius of the Earth $= R$)?

If the earth rotates faster than its present speed,the weight of an object will

The average density of the Earth:

Vedclass Test Series

Exam Paper Generator

Online Exam Module