The height at which the weight of a body beco | vedclass

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

Question

(D) The weight of a body at height $h$ is given by $W_h = m g_h$ and on the surface is $W = m g$.Given that $W_h = \frac{W}{9}$,we have $m g_h = \frac

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The weight of a body at height $h$ is given by $W_h = m g_h$ and on the surface is $W = m g$.Given that $W_h = \frac{W}{9}$,we have $m g_h = \frac{m g}{9}$,which implies $g_h = \frac{g}{9}$.The acceleration due to gravity at height $h$ is $g_h = \frac{GM}{(R+h)^2}$ and on the surface is $g = \frac{GM}{R^2}$.Substituting these into the equation $g_h = \frac{g}{9}$:$\frac{GM}{(R+h)^2} = \frac{1}{9} \cdot \frac{GM}{R^2}$$\frac{1}{(R+h)^2} = \frac{1}{9 R^2}$Taking the square root on both sides:$\frac{1}{R+h} = \frac{1}{3 R}$$R + h = 3 R$$h = 2 R$Therefore,the height is $2 R$.

The height at which the weight of a body becomes $\left(\frac{1}{9}\right)^{\text{th}}$ of its weight on the surface of the Earth is $(R = \text{radius of Earth})$: (in $R$)

Similar Questions

$T$ 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:

If the radius of the Earth becomes double while its mass remains unchanged,what will be the change in the weight of an object of mass $m$ on the surface of the Earth?

If the radius of the Earth were to shrink by half,its mass remaining the same,what would be the weight of an object on its surface?

Identify the incorrect statement. The acceleration due to gravity $g$ decreases if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module