The escape velocity of a body from the earth' | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The escape velocity from the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.At a height $h = R$ from the surface,the distance from

Vedclass · Accepted Answer

$\frac{v_e}{\sqrt{2}}$

Vedclass · Answer

(A) The escape velocity from the surface of the earth is given by $v_e = \sqrt{\frac{2GM}{R}}$.At a height $h = R$ from the surface,the distance from the center of the earth is $r = R + h = R + R = 2R$.The escape velocity $v'$ at this height is given by $v' = \sqrt{\frac{2GM}{r}} = \sqrt{\frac{2GM}{2R}}$.Simplifying this,we get $v' = \sqrt{\frac{1}{2} \cdot \frac{2GM}{R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}}$.Substituting $v_e$ into the equation,we get $v' = \frac{v_e}{\sqrt{2}}$.

The escape velocity of a body from the earth's surface is $v_e$. The escape velocity of the same body from a height equal to $R$ from the earth's surface will be

Similar Questions

The escape velocity for a body projected vertically upwards from the surface of earth is $11.2 \ km/s$. If the body is projected at an angle of $45^o$ with the vertical,the escape velocity will be ......... $km/s$.

The escape velocity from the moon's surface is less than that on the earth's surface,because:

The ratio of the radius of the Earth to that of the Moon is $10$. The ratio of acceleration due to gravity on the Earth and on the Moon is $6$. The ratio of the escape velocity from the Earth's surface to that from the Moon is

Is it necessary to provide an initial speed of $11.2 \, km/s$ to a rocket launched from Earth?

$A$ particle released at a large distance from a planet reaches the planet only under gravitational attraction and passes through a smooth tunnel through its centre. If $v_e$ is the escape velocity of a body at the planet,then the particle's speed at the centre of the planet is

Vedclass Test Series

Exam Paper Generator

Online Exam Module