Question

 

Calculate that imaginary angular velocity of the earth for which effective acceleration due to gravity at the equator becomes zero. In this condition what will be the length (in hours) of the day?

(Re = 6400 km, g = 10 ms–2)

Solution

Correct option is

1.4 hours

 

Let g be the acceleration due to gravity if the earth were at rest. The acceleration due to gravity at the equator in presence of earth’s rotation is given by  

                             

Where ω is the angular velocity of the earth. For g’ = 0, we should have

                   

If in this situation the period of rotation of the earth about its axis be T, then  

                 

The earth would complete its rotation in 1.4 hours, instead of 24 hours

SIMILAR QUESTIONS

Q1

The distance of two planets the sun are 1013 and 1012 meter respectively. Find the ratio of time-periods and speeds of the two planets.

Q2

Two masses, 800 kg and 600 kg, are at a distance 0.25 m apart. Compute the magnitude of the intensity of the gravitational field at a point distant 0.20 m from the 800 kg mass and 0.15 m from the 600 kg mass (G = 6.66× 10 –11 N m2 kg–2).

Q3

Three particles, each of mass m, are situates at the vertices of an equilateral triangle of side length a. The only force acting on the particles are their mutual gravitational forces. It is desired that each particle move in a circle while maintaining the original mutual separation a. Find the initial velocity that should be given to each particle and also the time-period of the circular motion.

Q4

The weight of a person on the earth is 80 kg. What will be his weight on the moon? Mass of the moon = 7.34 × 1022kg, radius = 1.75 × 106 m and gravitational constant G = 6.67 × 10 –11 Nm2/kg2. What will be the mass of the person at the moon? If this person can jump 2 meter high on the earth, how much high can he jump at the moon? If he can walk 100 m in 1 minute on the earth, then how much will he walk in 1 minute on the moon?

Q5

What will be the acceleration due to gravity on the surface of the moon if its radius is 1/4th the radius of the earth and its mass is (1/80)th the mass of the earth. 

Q6

Determine the speed with which the earth would have to rotate on its axis so that a person on the equator would weigh 3/5 th as much as at present. Take the equatorial radius as 6400 km.

Q7

At what height above the earth’s surface the acceleration due to gravity will be 1/9 th of its value at the earth’s surface? Radius of earth is 6400 km.

Q8

The radius of the earth is approximately 6000 km. What will be your weight at 6000 km above the surface of the earth? At 12000 km above? At 18000 km above?

Q9

Calculate the gravitational field strength and the gravitational potential at the surface of the moon. The mass of the moon is  kg and its radius is 

Q10

The intensity of gravitational field at a point situated at earth’s surface is 2.5 N/kg. Calculate the gravitational potential at that point. Given: radius of earth,