CameraIcon

SearchIcon

MyQuestionIcon

1

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

Introduction

Calculate the...

Question

Calculate the linear velocity of a point at equator on the earth surface due to its rotation (radius = 6400 km)

Open in App

Solution

Time period ,T=24hour=246060=86400s
Radius of earth,r=6400Km=6400000m
linear velocity,v=ωr (ω-angukar velocity)
=(2π/T)r
=(2π/86400)*6400000
=465.185m/s (π=3.14)

Suggest Corrections

thumbs-up

17

Similar questions

Q. What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes

\frac{3}{5}

of the present weight at the equator. Equatorial radius of the earth is

6400 k m

.

Q. Calculate the acceleration due to gravity at a point

64 k m

32 k m

below the surface of earth.
Given Radius of Earth

= 6400 k m

. Acceleration due to gravity at the surface of earth

= 9.8 m s^{- 2}

Q. Plot a graph showing the variation of acceleration due to gravity with the center of the earth.
The weight of a body on the surface of earth is

250 N

. Calculate its weight at distance equal to half of the radius of earth below the surface of earth. (Radius of earth

= 6400 k m

Q. At a point above the surface of earth, the gravitational potential is

- 5.12 \times 10^{7} J / k g

and the acceleration due to gravity is

6.4 m / s^{2}

. Assuming the mean radius of the earth to be

6400 k m

, calculate the height of this point above the earth's surface.

g

on the surface of the Earth is

9.8 m s^{- 2}

, its value at a height of

6400 k m

is: (Radius of the Earth

= 6400 k m

)

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Relative

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon