Escape Velocity

Two planets A and B have the same material density. If the radius of A is twice that of B, then the ratio of the escape velocity  $\frac{{\text{V}}_{\text{A}}}{{\text{V}}_{\text{B}}}\text{is}$

Given below are two statements:
Statement I : For a planet, if the ratio of mass of the planet to its radius increase, the escape velocity from the planet also increase.

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

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

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.

A planet having mass $9 {\mathrm{M}}_{\mathrm{e}}$ and radius $4{\mathrm{R}}_{\mathrm{e}},$ where ${\mathrm{M}}_{\mathrm{e}}$ and ${\mathrm{R}}_{\mathrm{e}}$ are mass and radius of earth respectively, has escape velocity in $\mathrm{km} {\mathrm{s}}^{-1}$ given by: (Given escape velocity on earth $V_{\mathrm{e}}=11.2\times {10}^3 \mathrm{m} {\mathrm{s}}^{-1}$)

A space ship of mass $2\times {10}^4 \mathrm{kg}$ 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 $g=10 \mathrm{m} {\mathrm{s}}^{-2}$ and radius of earth $=6400 \mathrm{km}$ ):

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason 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 a planet has mass equal to $16$ times the mass of earth, and radius equal to $4$ times that of earth. The ratio of escape speed of planet to that of earth is

Assertion $\left(A\right)$: Earth has atmosphere and moon doesn't.

Reason $\left(R\right)$: Escape speed on moon is less than that of Earth.

If earth has a mass nine times and radius twice to that of a planet$P$. Then $\frac{v_e}{3}\sqrt{x} \mathrm{m} {\mathrm{s}}^{-1}$ be the minimum velocity required by a rocket to pull out of gravitational force of$P$, the value of $x$ is

A man can jump up to a height of $h_0=1 \mathrm{m}$ on the surface of the earth. What should be the radius of a spherical planet so that the man makes a jump on its surface and escapes out of its gravity? Assume that the man jumps with same speed as on earth and the density of planet is same as that of earth. Take escape speed on the surface of the earth to be $11.2 \mathrm{km} {\mathrm{s}}^{-1}$ and radius of earth to be $6400 \mathrm{km}$.

If an object is projected from surface of Earth with a speed $k$ times of escape speed at surface of earth, then interstellar speed of object will be ($g=$ acceleration due to gravity at surface of earth, $R=$ radius of earth)

A body is thrown with a velocity equal to $k\left(k>1\right)$ times the escape velocity $\left(v_e\right)$. The interstellar speed of the body is

The $\mathrm{KE}$ required to project a body of mass '$m$' from a height equal to the radius $\left(R\right)$ of earth's surface to infinity is

A super dense particle of mass $\frac{M}{2}$ is projected from the surface of the planet of mass $M$ and radius $R$. Escape velocity for this particle of mass $\frac{M}{2}$ from surface of planet is

A spaceship is launched into a circular orbit close to the earth's surface. The additional velocity now to be imparted to the spaceship in the orbit to overcome the gravitational pull (Radius of the earth $=6400 \mathrm{km}$ and $\mathrm{g}=9.8 \mathrm{m} {\sec }^{-2}$) is given as $n\times {10}^3 \mathrm{m} {\sec }^{-1}$.Then find $n$.

(Take $\sqrt{2}=1.4$ and $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

Find the escape velocity from a planet having nine times the radius and one-third of density as the earth if the escape velocity from the earth is $11 \mathrm{km} {\mathrm{s}}^{-1}$.

A missile is launched with a velocity less than the escape velocity. The sum of kinetic energy and potential energy is :

Gravitational acceleration on the surface of a planet is $\frac{\sqrt{6}}{11}\mathit{ }g,$ where $g$ is the gravitational acceleration on the surface of the earth. The average mass density of the planet is $1.5$ times that of the earth. If the escape speed on the surface of the earth is taken to be $11 \mathrm{km} {\mathrm{s}}^{-1}$ the escape speed on the surface of the planet in $\mathrm{km} {\mathrm{s}}^{-1}$ will be

Two bodies each of mass $M$ are kept fixed with a separation $2L$. A particle of mass $m$ is projected from the mid-point of the line joining their centres perpendicular to the line. The gravitational constant is $G$. The correct statement $\left(s\right)$ is (are)

The escape velocity of a spherical planet of radius $R$ and density $\rho$ is $V$. If the radius of this planet is changed to $3R$ and density is changed to $3\rho$, then the escape velocity of the planet will be changed to