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(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$.Applying the law of conservation of energy between the Earth's surface and infi

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$\sqrt{3} V_e$

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(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$.Applying the law of conservation of energy between the Earth's surface and infinity:Total energy at surface = Total energy at infinity$-\frac{GMm}{R} + \frac{1}{2}m(2V_e)^2 = 0 + \frac{1}{2}mV^2$Here,$V$ is the final velocity at infinity.Substituting $V_e^2 = \frac{2GM}{R}$,we get:$-\frac{GM}{R} + \frac{1}{2}(4 \cdot \frac{2GM}{R}) = \frac{1}{2}V^2$$-\frac{GM}{R} + \frac{4GM}{R} = \frac{1}{2}V^2$$\frac{3GM}{R} = \frac{1}{2}V^2$$V^2 = \frac{6GM}{R} = 3 \left( \frac{2GM}{R} \right) = 3V_e^2$$V = \sqrt{3}V_e$

$A$ 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

Similar Questions

The escape velocity from the surface of the Earth is $V_e$. The escape velocity from the surface of a planet whose mass and radius are $3$ times those of the Earth will be:

"The value of the escape velocity $v_e$ for a stationary object on the surface of a planet is directly proportional to the mass and radius of the planet." Is this statement correct? If not,correct it.

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).

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?

The acceleration due to gravity on a planet is the same as that on Earth,and its radius is four times that of Earth. What will be the value of the escape velocity on that planet if it is $v_e$ on Earth?

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