Variation of g

The height in terms of radius of the earth ($R$), at which the acceleration due to gravity becomes $\dfrac g9$, where $g$ is acceleration due to gravity on earth's surface, is ________.

When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in g is approximately $\alpha$ %. The value of $\alpha$ is ________.
(Take radius of the earth = 6400 km.)

Two satellites P and Q are moving in different circular orbits around the Earth (radius $R$). The heights of P and Q from the Earth surface are $h_P$ and $h_Q$, respectively, where $h_P = R/3$. The accelerations of P and Q due to Earth's gravity are $g_P$ and $g_Q$, respectively. If $g_P/g_Q = 36/25$, what is the value of $h_Q$?

The approximate height from the surface of earth at which the weight of the body becomes $\dfrac13$ of its weight on the surface of earth is :

[Radius of earth $R=6400$ km and $\sqrt3=1.732$]

The height of any point P above the surface of earth is equal to diameter of earth. The value of acceleration due to gravity at point P will be : (Given g = acceleration due to gravity at the surface of earth).

The variation of acceleration due to gravity (g) with distance (r) from the center of the earth is correctly represented by :

(Given R = radius of earth)

The elongation of a wire on the surface of the earth is $10^{-4}$ m. The same wire of same dimensions is elongated by $6\times10^{-5}$ m on another planet. The acceleration due to gravity on the planet will be ______ ms$^{-2}$. (Take acceleration due to gravity on the surface of earth = 10 ms$^{-2}$)

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: If we move from poles to equator, the direction of acceleration due to gravity of earth always points towards the center of earth without any variation in its magnitude.

Reason R: At equator, the direction of acceleration due to the gravity is towards the center of earth.

In the light of above statements, choose the correct answer from the options given below

Given below are two statements :

Statement I : The law of gravitation holds good for any pair of bodies in the universe.

Statement II : The weight of any person becomes zero when the person is at the centre of the earth.

In the light of the above statements, choose the correct answer from the options given below.

The radii of two planets A and B are in the ratio $2:3$. Their densities are $3\rho$ and $5\rho$ respectively. The ratio of their acceleration due to gravity is :

The length of a seconds pendulum at a height $h=2R$ from earth surface will be :

(Given R = Radius of earth and acceleration due to gravity at the surface of earth, $g=\pi^2$ ms$^{-2}$)

An object is taken to a height above the surface of earth at a distance $\dfrac54$R from the centre of the earth. Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be :

The percentage decrease in the weight of a rocket, when taken to a height of 32 km above the surface of earth will, be :

(Radius of earth = 6400 km)

If the radius of earth shrinks by 2% while its mass remains same. The acceleration due to gravity on the earth's surface will approximately :

Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height $h$ above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of $h$ is -

(consider radius of earth $R_E$ = 6400 km and g on earth 10 m/s$^2$)

If the acceleration due to gravity experienced by a point mass at a height h above the surface of earth is same as that of the acceleration due to gravity at a depth αh (h << R$_e$) from the earth surface. The value of α will be _______.

(use R$_e$ = 6400 km)

The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius) :

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 weigh $\dfrac34$W. Radius of the Earth is 6400 km and $g=10$ m/s$^2$.

A bullet is fired vertically upwards with velocity $v$ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is $v_{esc}=v\sqrt N$, then the value of $N$ is (ignore energy loss due to atmosphere)

A planet of radius $R=\frac1{10}\times$(radius of Earth) has the same mass density as Earth. Scientists dig a well of depth $\frac R5$ on it and lower a wire of the same length and of linear mass density $10^{-3}\ \mathrm{kg\,m^{-1}}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth $=6\times10^6$ m and the acceleration due to gravity on Earth is $10\ \mathrm{m\,s^{-2}}$)