Gravitation

If $${g_E}$$ and $${g_M}$$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio $${{electro\,\,ch\arg e\,\,on\,\,the\,\,moon} \over {electronic\,\,ch\arg e\,\,on\,\,the\,\,earth}}\,\,to\,be$$

The height at which the acceleration due to gravity becomes $${g \over 9}$$ (where $$g=$$ the acceleration due to gravity on the surface of the earth) in terms of $$R,$$ the radius of the earth, is:

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 $${3 \over 4}$$ W. Radius of the Earth is 6400 km and g=10 m/s².

Suppose that the angular velocity of rotation of earth is increased. Then, as a consequence :

The value of acceleration due to gravity at Earth's surface is 9.8 ms^–2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms^–2, is close to : (Radius of earth = 6.4 × 10⁶ m)

A box weight 196 N on a spring balance at the north pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take g = 10 ms^–2 at the north pole and the radius of the earth = 6400 km) :

The height ‘h’ at which the weight of a body will be the same as that at the same depth ‘h’ from the surface of the earth is (Radius of the earth is R and effect of the rotation of the earth is neglected)

The mass density of a planet of radius R varies with the distance r from its centre as $$\rho$$(r) = $${\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$. Then the gravitational field is maximum at :

On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by $${{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:

The value of the acceleration due to gravity is g₁ at a height h = $${R \over 2}$$ (R = radius of the earth) from the surface of the earth. It is again equal to g₁ at a depth d below the surface of the earth. The ratio $$\left( {{d \over R}} \right)$$ equals :

The acceleration due to gravity on the earth’s surface at the poles is g and angular velocity of the earth about the axis passing through the pole is $$\omega$$. An object is weighed at the equator and at a height h above the poles by using a spring balance. If the weights are found to be same, then h is (h << R, where R is the radius of the earth)

A body weights 49N on a spring balance at the north pole. What will be its weight recorded on the same weighing machine, if it is shifted to the equator? [Use $$g = {{GM} \over {{R^2}}}$$ = 9.8 ms^$$-$$2 and radius of earth, R = 6400 km.]

If the angular velocity of earth's spin is increased such that the bodies at the equator start floating, the duration of the day would be approximately : [Take g = 10 ms^$$-$$2, the radius of earth, R = 6400 $$\times$$ 10³ m, Take $$\pi$$ = 3.14]

Consider a planet in some solar system which has a mass double the mass of earth and density equal to the average density of earth. If the weight of an object on earth is W, the weight of the same object on that planet will be :

If R_E be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : r < R_E)

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.

If earth has a mass nine times and radius twice to that of a planet P. Then $$\frac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}$$ will be the minimum velocity required by a rocket to pull out of gravitational force of $$\mathrm{P}$$, where $$v_{e}$$ is escape velocity on earth. The value of $$x$$ is

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:

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).