CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Variation of g (Acceleration Due to Gravity)

If g is the a...

Question

If g is the acceleration due to gravity on the surface of the earth, it's value at a height equal to triple the radius of earth is (assuming the earth to be a perfect sphere)

Open in App

Solution

Value of acceleration due to gravity at height comparable with radius of earth is given by

g' = g(R/(R+h))^2

In problem h = 3R

g' = g(R/4R)^2 = g/16

Suggest Corrections

thumbs-up

2

Similar questions

Q. What is the value of acceleration due to gravity at height equal to half the radius of earth from surface of earth. [take

g = 10 m / s^{2}

at earth surface]

Q. The value of acceleration due to gravity at certain height

h

above the surface of the earth is

\frac{g}{4}

g

is the value of acceleration due to gravity at the surface of the earth. The height

h

R

is the radius of earth)

Q. The height from earth's surface at which acceleration due to gravity becomes

\frac{g}{4}

g

is acceleration due to gravity on the surface of earth and

R

is radius of earth).

Q. The value of

g

(acceleration due to gravity) at earth's surface is

10 m s^{- 2}

. It's value at the centre of the earth which is assumed to be a sphere of radius

R

metre and uniform mass density is,

If earth suddenly stops rotating about its axis, then the value of acceleration due to gravity will become same at all the places on the surface of the earth.

[Assume earth as a perfect sphere]

Reason: The value of acceleration due to gravity is independent of rotation of the earth

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Variation of g (acceleration due to gravity) Tackle

PHYSICS

Watch in App

explore_more_icon

Explore more

Variation of g (Acceleration Due to Gravity)

Standard X Physics

Join BYJU'S Learning Program

CrossIcon