Define acceleration due to gravity. Give the magnitude of $g$ on the surface of earth.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) Acceleration due to gravity is defined as the uniform acceleration produced in a body falling freely under the sole influence of the earth's gravitational force. It is denoted by the symbol $g$.
The magnitude of acceleration due to gravity on the surface of the earth is approximately $9.8 \ m/s^2$.

Explore More

Similar Questions

Match Column-$I$ with Column-$II$.
Column-$I$ Column-$II$
$(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth
$(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles
$(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator

$A$ body weighs $W$ newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be:

If the density of the earth is doubled keeping the radius constant,find the new acceleration due to gravity (in $m/s^2$)? $(g = 9.8 \ m/s^2)$

If the Earth is assumed to be a sphere of radius $R$,and $g_{30}$ is the value of acceleration due to gravity at a latitude of $30^\circ$ and $g$ is the value at the equator,the value of $g - g_{30}$ is:

An iron ball and a wooden ball of the same radius are released from the same height in vacuum. They take the same time to reach the ground. The reason for this is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo