The variation of acceleration due to gravity | vedclass

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

Question

(A) The acceleration due to gravity $(g)$ at a distance $(r)$ from the center of the earth is given by:$1$. Inside the earth $(r \leq R)$: $g = \frac{

Vedclass · Accepted Answer

Vedclass · Answer

(A) The acceleration due to gravity $(g)$ at a distance $(r)$ from the center of the earth is given by:$1$. Inside the earth $(r \leq R)$: $g = \frac{GMr}{R^3}$. Here,$g$ is directly proportional to $r$ $(g \propto r)$,which represents a straight line passing through the origin.$2$. Outside the earth $(r \geq R)$: $g = \frac{GM}{r^2}$. Here,$g$ is inversely proportional to the square of the distance $(g \propto 1/r^2)$,which represents a parabolic curve.Thus,the graph shows a linear increase up to $r = R$ and a non-linear decrease for $r > R$. This corresponds to the graph shown in option $A$.

The variation of acceleration due to gravity $(g)$ with distance $(r)$ from the center of the earth is correctly represented by ... (Given $R =$ radius of earth)

Similar Questions

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^o$ latitude becomes zero is (Radius of earth $= 6400 \, km$. At the poles $g = 10 \, m/s^2$)

The value of acceleration due to gravity at a depth of $ 1600 \,km $ is equal to: (Radius of Earth $ = 6400 \,km $) (in $\,ms^{-2}$)

Suppose that the angular velocity of rotation of the earth is increased. Then,as a consequence:

At a height of $10 \,km$ above the surface of the earth,the value of acceleration due to gravity is the same as that at a particular depth below the surface of the earth. Assuming uniform mass density for the earth,the depth is ............. $km$.

If $R$ is the radius of the earth and $g$ is the acceleration due to gravity on the earth's surface,the mean density of the earth is

Vedclass Test Series

Exam Paper Generator

Online Exam Module