Escape Velocity

Two planets $P_1$ and $P_2$ with equal mass have radii $R_1$ and $R_2$, respectively, where $R_2 = \frac{R_1}{2}$. The escape speeds of $P_1$ and $P_2$ are $v_1$ and $v_2$, respectively. Then $\frac{v_2}{v_1}$ is:

The escape velocity from a spherical planet $A$ is $10\text{ km/s}$. The escape velocity from another planet $B$ whose density and radius are $10\%$ of those of planet $A$, is $\underline{\hspace{3em}}\text{ m/s}$.

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

A particle of mass $m$ is projected with a velocity $v = KV_e$ ($K < 1$) from the surface of the earth. The maximum height above the surface reached by the particle is ($V_e$ = escape velocity, $R$ is radius of earth)

A body is projected vertically upwards from earth's surface. If its kinetic energy of projection is equal to half of its minimum value required to escape from the gravitational influence then the height up to which it rises is ($R$ = radius of earth)

A body is projected vertically from earth's surface of radius R with velocity equal to half the escape velocity. The maximum height reached by the body is

A body of mass 'm' is raised to a height '$10R$' from the surface of the earth, where 'R' is the radius of the earth. Suppose it is released now. The speed with which it will strike the surface of the earth is nearly

[$g$ = acceleration due to gravity]

If a body of mass 1 kg falls on the earth from infinity, it attains velocity (v) and kinetic energy (k) on reaching the surface of earth. The values of v and k respectively are __________. (Take radius of earth to be 6400 km and g = 9.8 m/s2)

The masses and radii of the earth and moon are $M_1$, $R_1$ and $M_2$, $R_2$ and respectively, Their centres are at a distance 'd' apart. The minimum speed with which body of mass 'm' should be projected from a distance $2d/3$ from the centre of $M_1$ so as to escape to infinity is

The escape velocity of a satellite from the surface of earth does NOT depend on

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason $\mathbf{R}$

Assertion A: The kinetic energy needed to project a body of mass $m$ from earth surface to infinity is $\frac{1}{2}mgR$, where R is the radius of earth.

Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface.

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

Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is 11.2 km/s, the escape velocity in km/s from the planet will be:

A body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\text{rd}}$ of escape velocity. The maximum height reached by the body is ( $R =$ radius of earth)

An object is kept at rest at a distance of $3R$ above the earth's surface where $R$ is earth's radius. The minimum speed with which it must be projected so that it does not return to earth is: (Assume $M =$ mass of earth, $G =$ Universal gravitational constant)

The radius of the planet is double that of the earth, but their average densities are same. $V_p$ and $V_E$ are the escape velocities of planet and earth respectively. If $\frac{V_P}{V_E} = x$, the value of ' $x$ ' is

The escape velocity from earth surface is $11\text{ km/s}$. The escape velocity from a planet having twice the radius and same mean density as earth is

The escape velocity for earth is $v$. A planet having 9 times mass that of earth and radius, 16 times that of earth, has the escape velocity of:

To project a body of mass $m$ from earth's surface to infinity, the required kinetic energy is (assume, the radius of earth is $R_E, g =$ acceleration due to gravity on the surface of earth):

The mass of the moon is $\frac{1}{144}$ times the mass of a planet and its diameter is $\frac{1}{16}$ times the diameter of a planet. If the escape velocity on the planet is $v$, the escape velocity on the moon will be:

Escape velocity of a body from earth is $11.2\text{ km/s}$. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:

The radius and mean density of the planet are four times as that of the earth. The ratio of escape velocity at the earth to the escape velocity at a planet is

A body is projected in vertically upward direction from the surface of the earth of radius ' $R$ ' into space with velocity ' $nV_e$ ' $(n < 1)$. The maximum height from the surface of earth to which a body can reach is

A body starts from rest from a distance $R_0$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ( $R =$ radius of earth, $M =$ mass of earth)

Earth has mass ' $M_1$ ' radius ' $R_1$ ' and for moon mass ' $M_2$ ' and radius ' $R_2$ '. Distance between their centres is ' $r$ '. A body of mass ' $M$ ' is placed on the line joining them at a distance $\frac{r}{3}$ from the centre of the earth. To project a mass ' $M$ ' to escape to infinity, the minimum speed required is

Given below are two statements:

Statement I: For a planet, if the ratio of mass of the planet to its radius increases, the escape velocity from the planet also increases.

Statement II: Escape velocity is independent of the radius of the planet.

In the light of above statements, choose the most appropriate answer form the options given below

A planet having mass $9M_e$ and radius $4R_e$, where $M_e$ and $R_e$ are mass and radius of earth respectively, has escape velocity in $\text{km/s}$ given by:

(Given escape velocity on earth $V_e = 11.2 \times 10^{3}\text{ m/s}$)

The ratio of escape velocity of a planet to the escape velocity of earth will be:-

Given: Mass of the planet is 16 times mass of earth and radius of the planet is 4 times the radius of earth.

Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$.

Assertion A: Earth has atmosphere whereas moon doesn't have any atmosphere. Reason R: The escape velocity on moon is very small as compared to that on 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 $\frac{v_e}{3}\sqrt{x}\text{ m s}^{-1}$ will be the minimum velocity required by a rocket to pull out of gravitational force of $P$, where $v_e$ is escape velocity on earth. The value of $x$ is

The escape velocities of two planets $\text{A}$ and $\text{B}$ are in the ratio $1 : 2$. If the ratio of their radii respectively is $1 : 3$, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be:

A body is projected vertically upwards from earth's surface of radius '$R$' with velocity equal to ${\frac{1}{3}}^{\text{rd}}$ of escape velocity. The maximum height reached by the body is

A body is projected vertically from earth's surface with velocity equal to half the escape velocity. The maximum height reached by the satellite is ( $R$ = radius of earth)

A body (mass $m$) starts its motion from rest from a point distant $R_0\ (R_0 > R)$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ( $G =$ universal constant of gravitation, $M =$ mass of earth, $R$ = radius of earth)

A body of mass '$m$' kg starts falling from a distance 3R above earth's surface. When it reaches a distance '$R$' above the surface of the earth of radius '$R$' and Mass '$M$', then its kinetic energy is

A body is released from a height equal to the radius $(R)$ of the earth. The velocity of the body when it strikes the surface of the earth will be

(Given $g =$ acceleration due to gravity on the earth.)

An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be

A space ship of mass $2 \times 10^{4}\text{ kg}$ is launched into a circular orbit close to the earth surface. The additional velocity to be imparted to the space ship in the orbit to overcome the gravitational pull will be (if $g = 10\text{ m/s}^{2}$ and radius of earth $= 6400\text{ km}$ ):

The escape velocity of a body on the earth surface is $11.2\text{ km/s}$. If the same body is projected upward with velocity $22.4\text{ km/s}$, the velocity of this body at infinite distance from the centre of the earth will be:

The escape velocity of a body on a planet 'A' is 12 kms${}^{-1}$. The escape velocity of the body on another planet 'B', whose density is four times and radius is half of the planet 'A', is:

A body of mass $m$ is projected with velocity $\lambda v_e$ in vertically upward direction from the surface of the earth into space. It is given that $v_e$ is escape velocity and $\lambda < 1$. If air resistance is considered to be negligible, then the maximum height from the centre of earth, to which the body can go, will be: (R: radius of earth)

A body is projected vertically upwards from the surface of earth with a velocity equal to one third of escape velocity. The maximum height attained by the body will be:

(Take radius of earth $= 6400\text{ km}$ and $g = 10\text{ m s}^{-2}$)

The masses and radii of the moon and the earth are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. What should be the minimum speed with which a body of mass '$m$' should be projected from a point midway between their centres, so as to escape to infinity?

The radius of a planet is twice the radius of the earth. Both have almost equal average mass densities. If '$V_P$' and '$V_E$' are escape velocities of the planet and the earth respectively, then

The mass of a planet is six times that of the earth. The radius of the planet is twice that of the earth. If the escape velocity from the earth is '$V_e$', then the escape velocity from the planet is

The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is:

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

Assertion A: The escape velocities of planet A and B are same. But A and B are of unequal mass. Reason R: The product of their mass and radius must be same. M1R1 = M2R2

In the light of the above statements, choose the most appropriate answer from the options given below:

The radius in kilometer to which the present radius of earth (R = 6400 km) to be compressed so that the escape velocity is increased 10 times is ___________.

A particle of mass 'm' is projected with a velocity $v$ = kVe = (k < 1) from the surface of the earth. (Ve = escape velocity)

The maximum height above the surface reached by the particle is:

The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that $v_i$ $=$ $\sqrt{\frac{x}{y}} \times$ $v_e$. The value of x will be ____________.

A particle of mass '$m$' is kept at rest at a height $3R$ from the surface of earth, where '$R$' is radius of earth and '$M$' is the mass of earth. The minimum speed with which it should be projected, so that it does not return back is ( $g =$ acceleration due to gravity on the earth's surface)

A body is projected from earth's surface with thrice the escape velocity from the surface of the earth. What will be its velocity when it will escape the gravitational pull?

A body is projected vertically upwards from the surface of earth with a velocity sufficient enough to carry it to infinity. The time taken by it to reach height h is ___________ s.

The masses and radii of the earth and moon are (M1, R1) and (M2, R2) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses:

Suppose two planets (spherical in shape) in radii R and 2R, but mass M and 9M respectively have a centre to centre separation 8 R as shown in the figure. A satellite of mass 'm' is projected from the surface of the planet of mass 'M' directly towards the centre of the second planet. The minimum speed 'v' required for the satellite to reach the surface of the second planet is $\sqrt{\frac{a}{7}\frac{GM}{R}}$ then the value of 'a' is ____________.

[Given: The two planets are fixed in their position]

The escape velocity of a body from any planet, whose mass is six times the mass of earth and radius is twice the radius of earth will (v${}_e$ = escape velocity of a body from the earth's surface)

Planet A has mass M and radius R. Planet B has half the mass and half the radius of Planet A. If the escape velocities from the Planets A and B are vA and vB, respectively, then $\frac{v_A}{v_B} = n$ $4$. The value of n is:

A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:

A body is thrown from the surface of the earth velocity $v/\text{s}$. The maximum height above the earth's surface upto which it will reach is ($R =$ radius of earth, $g =$ acceleration due to gravity)

An asteroid is moving directly towards the centre of the earth. When at a distance of 10R (R is the radius of the earth) from the earths centre, it has a speed of 12 km/s. Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is 11.2 km/s)? Give your answer to the nearest integer in kilometer/s _____.

Earth has mass $M_1$ and radius $R_1$. Moon has mass $M_2$ and radius $R_2$. Distance between their centre is $r$. A body of mass $M$ is placed on the line joining them at a distance $\frac{r}{3}$ from centre of the earth. To project the mass $M$ to escape to infinity, the minimum speed required is

Two planets have masses M and 16 M and their radii are $a$ and 2$a$, respectively. The separation between the centres of the planets is 10$a$. A body of mass m is fired from the surface of the larger planet towards the smaller planet along the line joining their centres. For the body to be able to reach at the surface of smaller planet, the minimum firing speed needed is:

A rocket has to be launched from earth in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon :-

A body is projected vertically from the surface of the earth of radius ' $R$ ' with velocity equal to half of the escape velocity. The maximum height reached by the body is

A satellite is revolving in a circular orbit at a height h from the earth's surface, such that h < < R where R is the radius of the earth. Assuming that the effect of earth's atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is:

A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is -

Two stars of masses 3 $\times$ 1031 kg each, and at distance 2 $\times$ 1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is - (Take Gravitational constant; G = 6.67 $\times$ 10–11 Nm2 kg–2)

The ratio of escape velocity at earth (ve) to the escape velocity at a planet (vp) whose radius and mean density are twice as that of earth is

A satellite is revolving in a circular orbit at a height $'h'$ from the earth's surface (radius of earth $R;\ h << R$). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to: (Neglect the effect of atmosphere.)

A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = 5.98 $\times$ 1024 kg) have to be compressed to be a black hole?

The radius of a planet is twice the radius of earth. Both have almost equal average mass densities. VP and VE are escape velocities of the planet and the earth, respectively, then

A particle of mass 'm' is kept at rest at a height 3R from the surface of earth, where 'R' is radius of earth and 'M' is mass of earth. The minimum speed with which it should be projected, so that it does not return back, is

(g is acceleration due to gravity on the surface of earth)

If $v_e$ is escape velocity and $v_o$ is orbital velocity of a satellite for orbit close to the earth's surface, then these are related by

The mass of a spaceship is $1000\text{ kg}.$ It is to be launched from the earth's surface out into free space. The value of $g$ and $R$ (radius of earth) are $10\text{ m/s}^{2}$ and $6400\text{ km}$ respectively. The required energy for this work will be:

A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is

A planet in a distant solar system is $10$ times more massive than the earth and its radius is $10$ times smaller. Given that the escape velocity from the earth is $11\text{ km s}^{-1}$ $,$ the escape velocity from the surface of the planet would be

The earth is assumed to be a sphere of radius R. A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is fv, where v is its escape velocity from the surface of the Earth. The value of f is

The escape velocity for a body projected vertically upwards from the surface of earth is $11\text{ km/s}.$ If the body is projected at an angle of $45^\circ$ with the vertical, the escape velocity will be

The kinetic energy needed to project a body of mass $m$ from the earth surface (radius $R$) to infinity is

The escape velocity of a body depends upon mass as

With what velocity should a particle be projected so that its height becomes equal to radius of earth?

For a planet having mass equal to mass of the earth but radius is one fourth of radius of the earth. Then escape velocity for this planet will be

The escape velocity of a sphere of mass $m$ is given by ($G$ = universal gravitational constant, $M_e$ = mass of earth and $R_e$ = radius of earth)