(A) The gravitational potential energy $U$ of a mass $m$ placed at the midpoint between $M_1$ and $M_2$ is the sum of the potentials due to both masse

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$\sqrt{\frac{4G(M_1 + M_2)}{d}}$

(A) The gravitational potential energy $U$ of a mass $m$ placed at the midpoint between $M_1$ and $M_2$ is the sum of the potentials due to both masses:$U = -\frac{G M_1 m}{d/2} - \frac{G M_2 m}{d/2} = -\frac{2 G m}{d} (M_1 + M_2)$To escape the system,the total energy must be at least zero. By the law of conservation of energy,the kinetic energy provided must equal the magnitude of the potential energy:$\frac{1}{2} m v_e^2 = |U| = \frac{2 G m}{d} (M_1 + M_2)$Solving for the escape velocity $v_e$:$v_e^2 = \frac{4 G (M_1 + M_2)}{d}$$v_e = \sqrt{\frac{4 G (M_1 + M_2)}{d}}$

Class 11 Physics Gravitation Escape Velocity and Escape Energy

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Masses and radii of Earth and Moon are $M_1, M_2$ and $R_1, R_2$ respectively. The distance between their centers is $d$. The minimum velocity given to a mass $m$ from the midpoint of the line joining their centers so that it escapes the gravitational field of both is:

A
$\sqrt{\frac{4G(M_1 + M_2)}{d}}$
B
$\sqrt{\frac{4G}{d} \frac{M_1 M_2}{(M_1 + M_2)}}$
C
$\sqrt{\frac{2G}{d} \left(\frac{M_1 + M_2}{M_1 M_2}\right)}$
D
$\sqrt{\frac{2G}{d} (M_1 + M_2)}$

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Escape Velocity and Escape Energy

$A$ body is projected vertically upwards from the Earth's surface with velocity $2 V_e$,where $V_e$ is the escape velocity from the Earth's surface. The velocity when the body escapes the gravitational pull is

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The escape velocity for a planet is $v_e$. $A$ tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet,its speed will be

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The ratio of the radii of two planets is $r$ and the ratio of accelerations due to gravity on the planets is $x$. Then the ratio of the escape velocities from the planets is

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$A$ satellite with kinetic energy $E_k$ is revolving around the Earth in a circular orbit. How much more kinetic energy should be given to it so that it may just escape into outer space?

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$A$ spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the spaceship $= 1000 \; kg$,mass of the Sun $= 2 \times 10^{30} \; kg$,mass of Mars $= 6.4 \times 10^{23} \; kg$,radius of Mars $= 3395 \; km$,radius of the orbit of Mars $= 2.28 \times 10^{8} \; km$,$G = 6.67 \times 10^{-11} \; N m^{2} kg^{-2}$.

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Masses and radii of Earth and Moon are $M_1, M_2$ and $R_1, R_2$ respectively. The distance between their centers is $d$. The minimum velocity given to a mass $m$ from the midpoint of the line joining their centers so that it escapes the gravitational field of both is:

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$A$ body is projected vertically upwards from the Earth's surface with velocity $2 V_e$,where $V_e$ is the escape velocity from the Earth's surface. The velocity when the body escapes the gravitational pull is

The escape velocity for a planet is $v_e$. $A$ tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet,its speed will be

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