Let g be the acceleration due to gravity at t | vedclass

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

Question

(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$. Since $G$ and $M$ are constant,$g \propto R^{-2}$. Taking the derivative,$\frac{

Vedclass · Accepted Answer

$g$ increases by $4 \%$ and $K$ increases by $4 \%$

Vedclass · Answer

(C) The acceleration due to gravity is given by $g = \frac{GM}{R^2}$. Since $G$ and $M$ are constant,$g \propto R^{-2}$. Taking the derivative,$\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$. Given $\frac{\Delta R}{R} = -2 \%$,we get $\frac{\Delta g}{g} = -2(-2 \%) = +4 \%$. Thus,$g$ increases by $4 \%$.The rotational kinetic energy is $K = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$,where $L$ is the angular momentum. Assuming $L$ remains constant,$K \propto \frac{1}{I}$. Since $I = \frac{2}{5} MR^2$,we have $I \propto R^2$. Therefore,$K \propto R^{-2}$. Taking the derivative,$\frac{\Delta K}{K} = -2 \frac{\Delta R}{R}$. Given $\frac{\Delta R}{R} = -2 \%$,we get $\frac{\Delta K}{K} = -2(-2 \%) = +4 \%$. Thus,$K$ increases by $4 \%$. Therefore,both $g$ and $K$ increase by $4 \%$.

Let $g$ be the acceleration due to gravity at the Earth's surface and $K$ be the rotational kinetic energy of the Earth. Suppose the Earth's radius decreases by $2 \%$ keeping its mass the same,then:

Similar Questions

As we move above the surface of the Earth,the value of acceleration due to gravity $g$ . . . . . . .

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{2}$ below the surface of the Earth,where $R$ is the radius of the Earth,what is the new frequency of oscillations at this depth?

Two equal masses $m$ and $m$ are hung from a balance whose scale pans differ in vertical height by $h$. The error in weighing in terms of density of the earth $\rho$ is:

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n-1}$ times the value at the earth's surface is ($R=$ radius of the earth).

The radius of the Moon is $\frac{1}{4}$ times that of the Earth,and its mass is $\frac{1}{80}$ times that of the Earth. If the acceleration due to gravity on Earth is $g$,what is the acceleration due to gravity on the Moon?

Vedclass Test Series

Exam Paper Generator

Online Exam Module