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 $g = 10\,m{s^{ - 2}})$
$2.5 \times {10^{ - 3}}\,rad/s$
$5.0 \times {10^{ - 1}}\,rad/s$
$10 \times {10^1}\,rad/s$
$7.8 \times {10^{ - 2}}\,rad/s$
Assuming the earth to be a sphere of uniform density, the acceleration due to gravity inside the earth at a distance of $r$ from the centre is proportional to
A person whose mass is $100\, {kg}$ travels from Earth to Mars in a spaceship. Neglect all other objects in sky and take acceleration due to gravity on the surface of the Earth and Mars as $10$ ${m} / {s}^{2}$ and $4 \,{m} / {s}^{2}$ respectively. Identify from the below figures, the curve that fits best for the weight of the passenger as a function of time.
The value of acceleration due to gravity at Earth’s surface is $9.8\, m\,s^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9\, m\,s^{-2}$, is close to: (Radius of earth $= 6.4\times10^6\, m$)
If the Earth has no rotational motion, the weight of a person on the equator is $W$. Determine the speed with which the earth would have to rotate about its axis so that the person at the equator will weight $\frac{3}{4}\,W$ . Radius of the Earth is $6400\, km$ and $g = 10\, m/s^2$
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?