If radius of earth is 6347, km , then what will be difference between acceleration of free fall and

English

Gujarati

Hindi

7.Gravitation

medium

If radius of the earth is $6347\, km ,$ then what will be difference between acceleration of free fall and acceleration due to gravity near the earth's surface ?

A

$0.0340$

B

$0.3400$

C

$0.00334$

D

$0.24$

(AIIMS-2019)

Solution

The expression acceleration due to gravity near the earth surface is given by,

$g =\frac{ GM }{ R ^{2}}$

$=9.8$

The expression acceleration of free fall is given by,

$g _{\text {froe } fall }=\frac{ GM }{ R ^{2}}-\omega^{2} R$

$=9.8-\omega^{2} R$

Solve the difference as follows.

$g – g _{\text {free fill }}=\omega^{2} R$

$=\left(\frac{2 \pi}{ T }\right)^{2} R$

$=\frac{4 \pi^{2}}{(24 \times 60 \times 60)^{2}} \times 6347 \times 10^{3}$

$=0.03401$

Standard 11

Physics

Share

Similar Questions

The ratio of inertial mass and gravitational mass has been found to be $1$ for all material bodies. Were this ratio different for different bodies, the two bodies having same gravitational mass but different inertial mass would have?

easy

The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2 \,R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth)

medium

Weight of $1 \,kg$ becomes $1/6$ on moon. If radius of moon is $1.768 \times {10^6}\,m$, then the mass of moon will be

medium

If the change in the value of $‘g’$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it, then (both $x$ and $h$ being much smaller than the radius of the earth)

medium

(AIEEE-2005)

The ratio of the radius of the earth to that of the moon is $10$. The ratio of acceleration due to gravity on the earth and on the moon is $6$. The ratio of the escape velocity from the earth's surface to that from the moon is

easy