Acceleration Due to Gravity of the Earth

Let $\omega$ be the angular velocity of the earth's rotation about its axis. Assume that the acceleration due to gravity on the earth's surface has the same value at the equator and the poles. An object weighed at the equator gives the same reading as a reading taken at a depth $\mathrm{d}$ below earth's surface at a pole $\left(d\ll R\right)$. The value of $d$ is:

Gravitational force on point mass in side solid sphere.

(i) When a point mass $m$ is considered at distance $r$ from the centre of solid sphere inside it then gravitational force on point mass is proportional to.

An object of mass $100 \mathrm{kg}$ is sent inside the earth radially for a distance of $300 \mathrm{km}$, What is the gravitational force of attraction between the object and the earth? (Take radius of the earth $R_e = 6371 \mathrm{km}$, Mass of the earth $M_e = 5.97\times {10}^{24} \mathrm{kg}$.)

The force of attraction due to a solid sphere of uniform density, on a point mass situated inside it is zero.

Derive an expression for the gravitational force of a point mass $m$ kept inside a solid sphere of mass $M$ and radius $R$ at a distance of $r$ from the center of the sphere.

The ratio of the SI unit to CGS unit of acceleration is

The use of energy from sunlight by plants doing _____ is the basis of life on earth.

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 ?

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

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

The value of acceleration due to gravity at the surface of the earth is  $9.8 \mathrm{m}{\mathrm{s}}^{-2}$ and its mean radius is about $6.4\times {10}^6 \mathrm{m}$. Assuming that we could get more soil from somewhere, estimate how thick (in $\mathrm{k}\mathrm{m}$) would an added uniform outer layer on the earth have to have the value of acceleration due to gravity $10 \mathrm{m}{\mathrm{s}}^{-2}$ exactly? (Given the density of the earth's soil, $\rho =\mathrm{5,520} \mathrm{k}\mathrm{g}{\mathrm{m}}^{-3}$.)

The radius of the earth is about $\mathrm{6,400} \mathrm{k}\mathrm{m}$ and that of mars is about $\mathrm{3,200} \mathrm{k}\mathrm{m}$. The mass of the earth is about $10$ times the mass of mars. An object weighs $80 \mathrm{N}$ on the surface of earth. What will be its weight (in $\mathrm{N}$) on the surface of mars?

Imagine a new planet having the same density as that of the earth but it has a radius $3$ times that of the earth. If the acceleration due to gravity on the surface of the earth is $g$ and that on the new planet is $g\text{'}$, then what is the value of $g\text{'}/g$?

The density of a newly discovered planet is twice that of the earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is $R$ and the radius of the planet is $R\text{'}$. Then, what is the value of $R/R\text{'}$?

Choose the correct statements from the following options.

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)

The density of two planets are in ratio $1:2$ and their radii are in the ratio $3:2$ . The acceleration due to gravity on the surface of planets are in the ratio

Change in acceleration due to gravity is same upto a height h from each other the earth surface and below depth $\text{x}$ , then relation between x and h is (h and $\text{x}<<{\text{R}}_{\text{e}}$ )