Gravitation practice · Medium

If $\rho$ is the density of the planet, the time period of nearby satellite is given by

Newton’s universal law of gravitation applies to

The height above the earth’s surface at which the acceleration due to gravity becomes 9% of its value at the surface is ($R$ is radius of earth)

Two bodies of mass 10 kg and 5 kg moving in concentric orbits of radii R and r such that their periods are the same. Then the ratio between their centripetal acceleration is

The weight of an object in the coal mine, sea level, at the top of the mountain are ${W_1},\;{W_2}\,and\,{W_3}$ respectively, then

For a satellite orbiting in an orbit, close to the surface of Earth, to escape, what is the percentage increase in the kinetic energy required?

Two small bodies of masses $10 \mathrm{~kg}$ and $20 \mathrm{~kg}$ are kept a distance $1.0 \mathrm{~m}$ apart and released. Assuming that only mutual gravitational forces are acting, find the speeds of the particles when the separation decreases to $0.5 \mathrm{~m}$.

If $v_o$ be the orbital velocity of a satellite in a circular orbit close to the earth's surface and $v_e$ is the escape velocity from the earth, then relation between the two is

Assertion: The acceleration of a particle near the earth surface differs slightly from the gravitational acceleration $a_g = GM/R^2.$
Reason: The earth is not a uniform sphere and because the earth rotates.

Two satellites of same mass are launched in the same orbit of radius $r$ around the earth so as to rotate opposite to each other. If they collide inelastically and stick together as wreckage, the total energy of the system just after collision is

Gravitational acceleration on the surface of a planet is $\frac{{\sqrt 6 }}{{11}}\mathrm{\;}g,\;$, where $g$ is the gravitational acceleration on the surface of earth. The average mass density of the planet is $\frac{2}{3}$ times
that of the earth. If the escape speed on the surface of the earth is taken on be $11\mathrm{\;km}{\mathrm{s}^{ - 1}},$
the escape speed on the surface of the planet in $\mathrm{km}{\mathrm{s}^{ - 1}}\mathrm{\;}$ will be

A body of mass $m$ is projected from earth surface with speed ${v = \,2\sqrt {gR} }$ (where $R$ is the radius of earth). Find its speed at very far (infinite) from earth’s surface:-

The gravitational force between two point masses ${m_1}\,and\,{m_2}$ at separation $r$ is given by $F = k\frac{{{m_1}{m_2}}}{{{r^2}}}$. The constant $k$

For a uniform ring of mass $M$ and radius $R$ at its centre,

Assertion: Plane of space satellite is always equatorial plane.

Reason: On the equator value of $g$ is minimum.

Assertion: Plane of geostationary satellites always passes through equator.

Reason: Geostationary satellites always lies above Moscow.

If the earth stops rotating about its own aixs, $g$ remains unchanged at

The gravitation field in a region is given by $\overrightarrow g = 5N/kg\hat i\, + \,12N/kg\hat j.$ The change in the gravitational potential energy of a particle of mass 1 kg when it is taken from the origin to a point (7 m, -3 m) is:

Gravitational potential is expressed as $V = m{x^2}$ field at $x = 1m$ is (in S.I)

A boy can jump to a height $h$ on ground level. What should be the radius of a sphere of density $d$ such that on jumping on it, he escapes out of the gravitational field of the sphere

For elliptical orbits, in the equation $T^2=\left(\frac{4 \pi^2}{{GM}_s}\right) R^3, R$ refers to

The effect of rotation of the earth on the value of acceleration due to gravity is

A planet is revolving round the sun. Its distance from the sun at Apogee is $r_A$ and that at Perigee is $r_{{r}}$. The mass of planet and sun is $m$ and $M$ respectively, $v_A$ and $V_P$ is the velocity of planet at Apogee and Perigee respectivley and ${T}$ is the time period of revolution of planet round the sun.
(a) $T^2=\frac{\pi^2}{2 G m}\left(r_A+r_p\right)^2$
(b) $T^2=\frac{\pi^2}{2 G M}\left(r_A+r_p\right)^3$
(c) $v_A r_A=v_p r_p$
(d) $v_A < v_p, r_A > r_p$

When a satellite moves around the earth in a certain orbit, the quantity which remains constant is:

An object of mass $X$ is split into two portions $y$ and $(X-Y)$, which are then placed at some distance. Then what will be the ratio of $Y / X$ maximizes the gravitational force between the two portions?

A planet has twice the density of earth but the acceleration due to gravity on its surface is exactly the same as on the surface of the earth. Its radius in terms of radius of earth $R$ will be

The angular momentum of the earth revolving round the sun is proportional to $r^n$, where $r$ is the distance between the centres of earth and the sun. The value of $n$ is

Assertion: If rotation of earth about its own axis is suddenly stops then acceleration due to gravity will increase at all places on the earth.
Reason: At height $\mathrm{h}$ from the surface of earth, acceleration due to gravity is $\mathrm{g}_{\mathrm{h}}=\mathrm{g}\left(1-\frac{2 \mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)$.

A body moves in a circular orbit of radius $r$ under the action of a central force. Potential due to the central force is given by $V(r)$ – $kr$ ($k$ is a positive constant). Period of revolution of the body is proportional to:

The escape velocity of a sphere of mass $m$ from earth having mass $M$ and radius $R$ is given by