Acceleration due to Gravity

The mass of body on surface of the earth is $100 \mathrm{kg}$. What will be its (i) mass and (ii) weight at an altitude of $1000 \mathrm{km}$?

$(R=6400 \mathrm{km}, g=9.8 \mathrm{m}/{\mathrm{s}}^2)$

Assuming the earth to be a homogeneous sphere, determine the density of the earth from following data:

$g=9.8 \mathrm{m}/{\mathrm{s}}^2, G= 6.673 \mathrm{x} {10}^{-11} {\mathrm{Nm}}^2/{\mathrm{kg}}^2, R=6400 \mathrm{km}.$

A body weighs $3.5 \mathrm{kgwt}$ on the surface of the earth. What will be its weight on the surface of a planet whose mass is $\frac{1}{7}\mathrm{th}$ of the mass of the earth and radius half of that of the earth ?

Discuss the variation of $\mathrm{g}$ with latitude.

Discuss the variation of $\mathrm{g}$ with depth.

Discuss the variation of $\mathrm{g}$ with altitude.

The acceleration due to gravity on the earth's surface is 

The value of acceleration due to gravity is maximum at ____.

Weight of body at the centre of the earth is _____.

A body falling freely under gravity passes through two points $30\mathrm{ }\mathrm{m}$ apart in $1 \mathrm{s}$. From what distance above the upper point it began to fall? $(\text{Take}\mathrm{ }g=9.8\mathrm{ }\mathrm{m}\mathrm{ }{\mathrm{s}}^{-2})$

As astronaut accidentally gets separated out of his small spaceship accelerating in interstellar space at a constant rate of $100\mathrm{ }\mathrm{m}\mathrm{ }{\mathrm{s}}^{-2}$. What is the acceleration of the astronaut the instant he is outside the spaceship? (Assume that there are no nearby stars to exert gravitational force on him)

What should be the angular speed of earth, so, that body lying on equator may appear weightlessness?

$[g=10 \mathrm{m} {\mathrm{s}}^{-2},\mathrm{ }R=6400\mathrm{ }\mathrm{k}\mathrm{m}]$

Calculate angular velocity of the earth so that acceleration due to gravity at ${60}^o$ latitude becomes zero. (radius of the earth $=6400\mathrm{ }km$ , gravitational acceleration at poles $=10 \mathrm{m} {\mathrm{s}}^{-2}$ , $\mathit{cos}{60}^o=0.5)$

Assertion: In a free fall, weight of a body becomes effectively zero.

Reason: Acceleration due to gravity acting on a body having free fall is zero.

When you move from equator to pole, the value of acceleration due to gravity $\left(g\right)$,

The acceleration due to gravity '$g$' and mean density of the earth $\rho$ are related by which of the following relations?

(Where, $G$ is the gravitational constant and $R$ is radius of earth.)

What should be the angular speed of earth, so that, body lying on equator may appear weightlessness?

$\left[g=10 \mathrm{m} {\mathrm{s}}^{-2},\mathrm{ }R=6400\mathrm{ }\mathrm{k}\mathrm{m}\right]$

The depth $d$ at which the value of acceleration due to gravity becomes $\frac{1}{n}$ times the value at the surface, is:

$[R=\mathrm{R}\mathrm{a}\mathrm{d}\mathrm{i}\mathrm{u}\mathrm{s}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{e}\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{h}]$

The weight of the body at the centre of the earth is -

If the speed of rotation of earth about its axis increases, then the weight of the body at the equator will,