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

  • A
    $\frac{v_e}{\sqrt{2}}$
  • B
    $\frac{v_e}{2}$
  • C
    $\frac{v_e}{2\sqrt{2}}$
  • D
    $\frac{v_e}{4}$

Explore More

Similar Questions

$A$ body is projected vertically from the Earth's surface of radius $R$ with a velocity equal to half the escape velocity. The maximum height reached by the body is

$A$ 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

Define escape energy. Write its unit and dimensional formula.

The escape velocity of a body from the surface of the earth is $11.2 \,km/s$. The escape velocity of a body from a planet having the same mean density as the earth but twice the radius of earth is: (in $\,km/s$)

The masses and radii of the Earth and the Moon are $M_1, R_1$ and $M_2, R_2$ respectively. Their centres are at a distance $d$ apart. The minimum speed with which a particle of mass $m$ should be projected from a point midway between the two centres so as to escape to infinity is

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo