The acceleration due to gravity at a height of $(\sqrt{2}-1) R$ from the surface of the earth is (Acceleration due to gravity on the surface of the earth $= 10 \ m \ s^{-2}$ and $R$ is radius of the earth). (in $m \ s^{-2}$)

  • A
    $2.5$
  • B
    $7.5$
  • C
    $5$
  • D
    $10$

Explore More

Similar Questions

If the earth rotates faster than its present speed,the weight of an object will

If a planet has a mass and radius both half that of the Earth,the acceleration due to gravity at its surface would be ......... $m/s^2$ ($g$ on Earth $= 9.8\, m/s^2$)

As we go from the equator to the poles,the value of $g$

The mass of a planet is $\frac{1}{10}$ that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is: (in $m \ s^{-2}$)

The variation of acceleration due to gravity $(g)$ with distance $(r)$ from the center of the earth is correctly represented by ... (Given $R =$ radius of earth)

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