In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be ( $g = 10 m s^{- 2}$ and radius of earth is 6400 kms)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B

$g^{'} = g - ω^{2} R c o s^{2} λ$
For weightlessness at equator $λ = 0 a n d g^{'} = 0$
$∴ 0 = g - ω^{2} R \Rightarrow w = \sqrt{\frac{g}{R}} = \frac{1}{800} \frac{r a d}{s}$

Suggest Corrections

Similar questions

Q. In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be

(g = 10 m s^{- 2} a n d r a d i u s o f e a r t h i s 6400 k m)

Q. The angular velocity of rotation of the earth in order to make the effective acceleration due to gravity equal to zero at equator should be

g = 10 m / s^{2}, R = 6400 k m

Q. The speed of earth 's rotation about its axis is

ω

.Its speed is increased to

x

times to make the effective acceleration due to gravity equal to zero at the equator, then

x

is around (

g = 10 m s^{- 2}

R = 6400

km)

Q. Find the imaginary angular velocity of the earth for which the effective acceleration due to gravity at the equator shall be zero. In this condition, find the length (in hours) of a day also.

(T a k e, g = 10 m s^{- 2}

for the acceleration due to gravity if the earth were at rest and radius of earth equal to 6400 Km.)

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on $60^{\circ}$ latitude becomes zero is (Radius of earth = 6400 At the poles g = 10 $m s^{- 2}$ )

Join BYJU'S Learning Program

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ms−2 and radius of earth is 6400 kms)

The correct option is B g′=g−ω2Rcos2λ For weightlessness at equator λ=0 and g′=0 ∴0=g−ω2R⇒w=√gR=1800rads

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be ( $g = 10 m s^{- 2}$ and radius of earth is 6400 kms)

The correct option is B

$g^{'} = g - ω^{2} R c o s^{2} λ$
For weightlessness at equator $λ = 0 a n d g^{'} = 0$
$∴ 0 = g - ω^{2} R \Rightarrow w = \sqrt{\frac{g}{R}} = \frac{1}{800} \frac{r a d}{s}$