Home
Class 11
PHYSICS
The radius of the earth is 6400 km. What...

The radius of the earth is 6400 km. What will be the value of the centrifugal acceleration at the equatorial region due to the earth's diurnal motion?

Text Solution

Verified by Experts

If the radius of the earth is R and its angular velocity is `omega`, then at equatorial region,
centrifugal acceleration
= `omega^(2) R = ((2pi)^(2)/(T)) xx R = ((2pi)/(24 xx 60 xx 60))^(2) xx 6400 xx 10^(3)`
= 0.0338 m`cdot s^(-2)`.
Promotional Banner

Topper's Solved these Questions

  • CIRCULAR MOTION

    CHHAYA PUBLICATION|Exercise SECTION RELATED QUESTIONS|15 Videos
  • CIRCULAR MOTION

    CHHAYA PUBLICATION|Exercise HIGHER ORDER THINKING SKILL (HOTS) QUESTIONS|18 Videos
  • CALORIMETRY

    CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE|5 Videos
  • DOPPLER EFFECT IN SOUND

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|2 Videos

Similar Questions

Explore conceptually related problems

The radius of the earth is 6400 km. Its capacitance in microfarad is…………….

Radius of the earth is 6400 km. Determine its capacitance in muF .

If the radius of the earth shrinks by1.5% then the value of acceleration due to gravity changes by

If the radius of the earth is R , the height above the surface of the earth where the acceleration due to gravity will be 1% of its value on the earth's surface is

What is the value of the acceleration due to gravity on the surface of the earth ?

What is the value of the acceleration due to gravity at the centre of the earth ?

The average density of the earth is 5500kg*m^(-3) the gravitational constant is 6.7xx10^(-11)N*m^2*kg^(-2) and the radius of the earth is 6400 km. Using the given values, find the magnitude of the acceleration due to gravity on the surface of the earth.

Acceleration due to gravity decreases as we go below the surface of the earth. What will be the nature of the graph between the change of acceleration due to gravity and the depth below the surface of the earth ?

If the mass of the earth is 6xx10^(27) g and its radius 6400 km, what will be the acceleration due to gravity at a height 64 km above the earth's surface ? Given that G=6.67xx10^(-8) CGS unit.

If the earth is assumed to be a sphere of radius R, the height above the surface of the earth where the value of the acceleration due to gravity will be half its value on the earth's surface is