iCON Education - Gravitation AP-EAPCET Test

The acceleration due to gravity at a height of $(\sqrt{2}-1) \mathrm{R}$ from the surface of the Earth is

(Acceleration due to gravity on the surface of the Earth $=10 \mathrm{~ms}^{-2}$ and $R$ is radius of the Earth)

$2.5 \mathrm{~ms}^{-2}$

$7.5 \mathrm{~ms}^{-2}$

$5 \mathrm{~ms}^{-2}$

$10 \mathrm{~ms}^{-2}$

The escape velocity of a body from a planet of mass $M$ and radius $R$ is $14 \mathrm{~km} \mathrm{~s}^{-1}$. The escape velocity of the body from another planet having same mass and diameter 8 R (in $\mathrm{km} \mathrm{s}^{-1}$ ) is

The potential energy of a satellite of mass ' $m$ ' revolving around the Earth at a height of $R_e$ from the surface of the Earth is

( $R_e=$ Radius of Earth, $\mathrm{g}=$ acceleration due to gravity)

The time period of a simple pendulum on the surface of the Earth is $T$. If the pendulum is taken to a height equal to half of the radius of the Earth, then its time period is

If the escape velocity of a body from the surface of the Earth is $11.2 \mathrm{~km} \mathrm{~s}^{-1}$, then the orbital velocity of a satellite in an orbit which is at a height equal to the radius of the Earth is

$11.2 \mathrm{~km} \mathrm{~s}^{-1}$

$2.8 \mathrm{~km} \mathrm{~s}^{-1}$

$22.4 \mathrm{~km} \mathrm{~s}^{-1}$

$5.6 \mathrm{~km} \mathrm{~s}^{-1}$

An artificial satellite is revolving around a planet of radius $R$ in a circular orbit of radius ' $a$ '. If the time period of revolution of the satellite. $T \propto a^{3 / 2} g^x R^y$, then the values of $x$ and $y$ are respectively

[ $g=$ acceleration due to gravity]

$-\frac{1}{2}, \frac{1}{2}$

A mass of $6 \times 10^{24} \mathrm{~kg}$ is to be compressed in the form of a solid sphere such that the escape velocity from its surface is $3 \times 10^4 \mathrm{~ms}^{-1}$. The radius of the sphere is

(Universal gravitational constant $=6.66 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2}$ )

Two satellites $A$ and $B$ are revolving around the Earth in orbits of heights $1.25 R_E$ and $19.25 R_E$ from the surface of Earth respectively, where $R_E$ is the radius of the Earth. The ratio of the orbital speeds of the satellites $A$ and $B$ is

Two solid spheres each of radius ' $R$ ' made of same material are placed in contact with each other. If the gravitational force acting between them is $F$, then

If the angular velocity of a planet about its axis is halved, the distance of the stationary satellite of this planet from the centre of the planet becomes $2^n$ times the initial distance. Then, the value of ' $n$ ' is

An infinite number of objects each 1 kg mass are placed on the $X$-axis on both sides of $x=0$ at $\pm 1 \mathrm{~m}$, $\pm 2 \mathrm{~m}, \pm 4 \mathrm{~m}, \pm 8 \mathrm{~m} \ldots \ldots$ and so on. The magnitude of the resultant gravitational potential (in SI units) at $x=0$ is

( $G=$ Universal gravitational constant)

The time period of revolution of a satellite close to planet's surfaces is 80 min . The time period of another satellite, which is at a height of 3 times the radius of the planet from surface is

The gravitational potential energy of a body on the surface of the earth is $E$. If the body is taken from the surface of the earth to a height equal to $150 \%$ of the radius of the earth. Its gravitational potential energy is

A satellite moving round the earth in a circular orbit has kinetic energy $E$. Then, the minimum amount of energy to be added so that it escapes from the earth.

A particle is projected from the surface of the earth with a velocity equal to twice the escape velocity. When particle is very far from the earth. Its speed would be

The time period of revolution of a satellite $T$ around the carth depends on the radius of the circular orbit $R$. mass of the earth $M$ and universal gravitational constant $G$. The expression for $T$, using dimensional analysis is ( $K$ is constant of proportionality)

$K \sqrt{\frac{R^2}{G M}}$

$K \sqrt{\frac{R}{G M}}$

$K \sqrt{\frac{R^3}{G M}}$

$K \sqrt{\frac{R^3}{G M^2}}$

If the time period of revolution of a satellite is $T$, the its kinetic energy is proportional to

What is the height from the surface of earth, where acceleration due to gravity will be $1 / 4$ of that of the earth? $\left(R_E=6400 \mathrm{~km}\right)$

The acceleration due to gravity at a height of 6400 km from the surface of the earth is $2.5 \mathrm{~ms}^{-2}$. The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth= 6400 km )

$1.11 \mathrm{~ms}^{-2}$

$1.5 \mathrm{~ms}^{-2}$

$2.22 \mathrm{~ms}^{-2}$

$1.25 \mathrm{~ms}^{-2}$

Maximum height reached by a rocket fired with a speed equal to $50 \%$ of the escape speed from the surface of the earth is ( $R=$ Radius of the earth)

Two satellites of masses $m$ and 1.5 m are revolving around the earth with different speeds in two circular orbits of heights $R_E$ and $2 R_E$ respectively, where $R_F$ is the radius of the earth. The ratio of the minimum and maximum gravitational forces on the earth due to the two satellites is

Statement (A) Two artificial satellites revolving in the same circular orbit have same period of revolution.

Statement (B) The orbital velocity is inversely proportional to the square root of radius of the orbit.

Statement (C) The escape velocity of the body is independent of the altitude of the point of projection.

A uniform solid sphere of radius $R$ produces a gravitational acceleration of $a_0$ on its surface. The distance of the point from the centre of the sphere where the gravitational acceleration becomes $\frac{a_0}{4}$ is

A projectile is thrown straight upward from the earth's surface with an initial speed $v=\alpha v_e$ where $\alpha$ is a constant and $v_e$ is the escape speed. The projectile travels upto a height 800 km from earth's surface, before it comes to rest. The value of the constant $\alpha$ is (radius of the earth $=6400 \mathrm{~km}$)

If the Earth stops rotating in its orbit about the sun, there will be variation in the weight of our bodies at

At what depth below surface of the Earth, the acceleration due to gravity will be half of its value that at $1600 \mathrm{~km}$ above the surface of the Earth?

$4.8 \times 10^6 \mathrm{~m}$

$3.19 \times 10^6 \mathrm{~m}$

$1.59 \times 10^6 \mathrm{~m}$

$5.5 \times 10^6 \mathrm{~m}$

The gravitational potential energy is maximum at

the centre of the earth

twice the radius of the earth

A geostationary satellite is taken to a new orbit, such that its distance from centre of the earth is doubled. Then, find the time period of this satellite in the new orbit.

The distance through which one has to dig the Earth from its surface, so as to reach the point where the acceleration due to gravity is reduced by 40% of that at the surface of the Earth, is (radius of Earth is 6400 km)

Infinite number of masses each of 3kg are placed along a straight line at the distances of 1 m, 2m, 4m, 8m, ...... from a point O on the same line. If G is the universal gravitational constant, then the magnitude of gravitational field intensity at O is

A particle is kept on the surface of a uniform sphere of mass 1000 kg and radius 1 m. The work done per unit mass against the gravitational force between them is

[G = 6.67 $\times$ 10$^{-11}$ Nm$^2$ kg$^{-2}$]

$\mathrm{3.35\times10^{-10}~Jkg^{-1}}$

$\mathrm{-3.35\times10^{-10}~Jkg^{-1}}$

$\mathrm{6.67\times10^{-8}~Jkg^{-1}}$

$\mathrm{-3.35\times10^{-8}~Jkg^{-1}}$

The acceleration due to gravity at a height (1/20)th of the radius of Earth above the Earth's surface is 9 ms$^{-2}$. Its value at an equal depth below the surface of earth is