The variation of acceleration due to gravity | vedclass

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

Question

with depth

$g_{1}=g\left(1-\frac{(d)}{R}\right)$

As depth $d$ goes on increasing $g_{1}$ goes on decreasing, it remains maximum at the surface o

Vedclass · Accepted Answer

Vedclass · Answer

with depth

$g_{1}=g\left(1-\frac{(d)}{R}ight)$

As depth $d$ goes on increasing $g_{1}$ goes on decreasing, it remains maximum at the surface of Earth. The above equation is in the form of straight line.

With height

$g_{2}=g\left(1-\frac{2 h}{R}ight)$

$=g-\frac{2 g h}{R}$

$g_{2} \propto \frac{1}{R}(	ext { Hyperbola })$

Acceleration due to gravity goes on decreasing as the $h$ above Earth surface increases.

Hence the correct answer is option $B$.

The variation of acceleration due to gravity $g$ with distance $d$ from centre of the earth is best represented by ($R =$ Earth's radius)

Similar Questions

A body weighs $700\,gm\,wt.$ on the surface of the earth. How much will it weigh on the surface of a planet whose mass is $\frac {1}{7}$ and radius half of that of the earth ....... $gm\, wt$

$Assertion$ : The escape speed does not depend on the direction in which the projectile is fired.
$Reason$ : Attaining the escape speed is easier if a projectile is fired in the direction the launch site is moving as the earth rotates about its axis.

Suppose the gravitational force varies inversely as the nth power of distance. Then the time period of a planet in circular orbit of radius $R$ around the sun will be proportional to

If the distance between the centres of Earth and Moon is $D$ and mass of Earth is $81\, times$ that of Moon. At what distance from the centre of Earth gravitational field will be zero?

A body of mass $m$ is situated at distance $4R_e$ above the Earth's surface, where $R_e$ is the radius of Earth how much minimum energy be given to the body so that it may escape