(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

Difficult

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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Gravitation

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Escape Velocity and Escape Energy

Does the escape speed of a body from the Earth depend on:
$(a)$ the mass of the body,
$(b)$ the location from where it is projected,
$(c)$ the direction of projection,
$(d)$ the height of the location from where the body is launched?

Medium

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If an object is projected vertically upwards with a speed equal to half the escape speed of Earth,then the maximum height attained by it is ($R$ is the radius of Earth).

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$A$ 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 $= 5.98 \times 10^{24} \, kg$) have to be compressed to be a black hole?

MediumAIPMT 2014

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$A$ body of mass $m$ is situated at a distance equal to $2R$ ($R$ is the radius of the Earth) from the Earth's surface. The minimum energy required to be given to the body so that it may escape out of the Earth's gravitational field is:

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$A$ body is projected vertically upwards from the surface of the earth with a velocity sufficient to carry it to infinity. The time taken by it to reach a height of three times the radius of the earth is (acceleration due to gravity $g = 9.8 \ m/s^2$ and radius of the earth $R = 6400 \ km$). (in $min$)

MediumAP EAMCET 2018

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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:

Similar Questions

Does the escape speed of a body from the Earth depend on:$(a)$ the mass of the body,$(b)$ the location from where it is projected,$(c)$ the direction of projection,$(d)$ the height of the location from where the body is launched?

If an object is projected vertically upwards with a speed equal to half the escape speed of Earth,then the maximum height attained by it is ($R$ is the radius of Earth).

$A$ 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 $= 5.98 \times 10^{24} \, kg$) have to be compressed to be a black hole?

$A$ body of mass $m$ is situated at a distance equal to $2R$ ($R$ is the radius of the Earth) from the Earth's surface. The minimum energy required to be given to the body so that it may escape out of the Earth's gravitational field is:

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Does the escape speed of a body from the Earth depend on:
$(a)$ the mass of the body,
$(b)$ the location from where it is projected,
$(c)$ the direction of projection,
$(d)$ the height of the location from where the body is launched?