Escape velocity on a planet is $v_e$. If the radius of the planet remains the same and the mass becomes $4$ times,the escape velocity becomes:

  • A
    $4\,v_e$
  • B
    $2\,v_e$
  • C
    $v_e$
  • D
    $\frac{1}{2}v_e$

Explore More

Similar Questions

$A$ spaceship of mass $2 \times 10^4 \, kg$ is launched into a circular orbit close to the Earth's surface. The additional velocity to be imparted to the spaceship in the orbit to overcome the gravitational pull will be $......$ (given $g = 10 \, m/s^2$ and radius of Earth $R = 6400 \, km$).

Given the mass of the moon is $1/81$ of the mass of the earth and its radius is $1/4$ of the earth's radius. If the escape velocity on the earth's surface is $11.2 \, km/s$,the value of the escape velocity on the surface of the moon is ......... $km/s$.

The ratio of escape velocity at earth $(v_e)$ to the escape velocity at a planet $(v_p)$ whose radius and mean density are twice as that of earth is:

$A$ meteor of mass $m$ having a speed $v$ at infinity reaches the surface of the earth with a speed of (where $v_e$ is the escape speed from the earth's surface).

When the total energy of a satellite-planet system is positive,the satellite will .......

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