Home
Class 11
PHYSICS
The height at which the acceleration due...

The height at which the acceleration due to gravity becomes `(g)/(9)` (where g =the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :

A

`2 R`

B

`R/(sqrt(3))`

C

`R/2`

D

`sqrt(2) R`

Text Solution

Verified by Experts

The correct Answer is:
4
Acceleration due to gravity at height `h` from earth surface.
`g'=g/((1+h/R)^(2))`
`g/9=g/((1+h/R)^(2))`
`impliesh=2R`
Promotional Banner

Similar Questions

Explore conceptually related problems

Let g be the acceleration due to gravity on the earth's surface.

The height at which the acceleration due to gravity becomes g//9 in terms of R the radius of the earth is

The acceleration due to gravity is _____ at the surface of the earth and ____ at the centre of the earth.

The ratio of acceleration due to gravity at a height 3R above earth 's surface to the acceleration due to gravity on the surface of the earth is (where R=radius of earth)

The ratio of acceleration due to gravity at a height 3 R above earth's surface to the acceleration due to gravity on the surface of earth is (R = radius of earth)

If g is the acceleration due to gravity on the surface of the earth , its value at a height equal to double the radius of the earth is

If g is acceleration due to gravity at the surface of the earth, then its value at a depth of 1/4 of the radius of the earth is

What will be the acceleration due it gravity at a depth in Earth. where g is acceleration due to gravity on the earth ?

If g_(e) is acceleration due to gravity on earth and g_(m) is acceleration due to gravity on moon, then