Acceleration Due To Gravity

In older times, people used to think that the Earth was flat. Imagine that the Earth is indeed not a sphere of radius $R$, but an infinite plate of thickness $H$. What value of $H$ is needed to allow the same gravitational acceleration to be experienced as on the surface of the actual Earth? (Assume that the Earth's density is uniform and equal in the two models).

A (nonrotating) star collapses onto itself from an initial radius ${\text{R}}_{\text{i}}$ with its mass remaining unchanged. Which curve in figure best gives the gravitational acceleration ${\text{a}}_{\text{g}}$ on the surface of the star as a function of the radius of the star during the collapse?

If a tunnel is cut at any orientation through the earth, then a ball released from one end will reach the other end in how much time (neglect earth rotation)? 

A spherical uniform planet is rotating about its axis. The velocity of a point on its equator is $V$. Due to the rotation of the planet about its axis the acceleration due to gravity $g$ at equator is $\frac{1}{2}$ of $g$ at poles. The escape velocity of a particle on the planet in terms of $V$ from the pole of the planet is,

The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this planet if it is a second pendulum on earth?

If the radius of the earth be increased by a factor of 5, by what factor its density be changed to keep the value of g same?

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:

At what depth from the earth surface the acceleration due to gravity will be half the value at the surface. (Radius of earth is $6400 \mathrm{km}$)

The acceleration due to gravity at a height $1\mathrm{km}$ above the earth is the same as at a depth ‘$d$’ below the surface of earth. Then

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.

If the radius of the earth were to shrink by one percent, its mass remaining same, the value of $\mathrm{g}$ on the earths surface would

Weight of an object can be explained as                                                                         

At a height of $10 \mathrm{km}$ above the surface of earth, the value of acceleration due to gravity is the same as that of a particular depth below the surface of earth. Assuming uniform mass density of the earth, the depth is,

The gravitational field due to a mass distribution is, $E=\frac{K}{x^3}$ in the $x$-direction ($K$ is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance $x$ is,

A spiral galaxy can be approximated as an infinitesimally thin disc of a uniform surface mass density (mass per unit area) located at $z=0,$ Two stars $A$ and $B$ start from rest from heights $2z_0$ and $z_0$ ($z_0<<$ radial extent of the disc), respectively and fall towards the disc, cross over to the other side and execute periodic oscillations. The ratio of time periods of $A$ and $B$ is

Acceleration due to gravity at a height of $h$ is $25\%$ of that at the surface of the Earth. The value of $h$ in terms of radius of earth $R$ is
radius of the Earth. 

The mass and density of the moon, if acceleration due to gravity on its surface is $1.62 \mathrm{m} {\mathrm{s}}^{-2}$ and its radius is $1.74\times {10}^6 \mathrm{m}$, respectively. $($Take, $G=6.67\times {10}^{-11} \mathrm{N}-{\mathrm{m}}^2 {\mathrm{kg}}^{-2}$)

Two planets have the same average density but their radii are $R_1$ and $R_2.$ If acceleration due to gravity on these planets be $g_1$ and  $g_2$ respectively, then :-

Let $A$ and $B$ be the points respectively above and below the earth's surface each at a distance equal to half the radius of the earth. If the acceleration due to gravity at these points be $g_B$ and $g_A$ respectively, then $g_{\mathit{B}}:g_{\mathit{A}}$ is

The radii of two planets are respectively $R_1$ and $R_2$ and their densities are respectively ${\rho }_1$ and ${\rho }_1$. The ratio of the accelerations due to gravity at their surfaces is