how many times faster than its present speed the earth should rotate so that the apparent weight of an object at equator becomes zero? Given radius of earth =6.37*106m. what would be the duration of the day in that case?
how many times faster than its present speed the earth should rotate so that the apparent weight of an object at equator becomes zero? Given radius of earth =6.37*106m. what would be the duration of the day in that case?
Let the angular velocity of earth be ω, and the time period be T. The radius of earth is R.
Now, T = 24 h = 86400 s
R = 6370000 m
ω = 2π/T = 7.27 × 10-5 rad/s
The relation between angular velocity and acceleration due to gravity is given by,
g/ = g - ω2R [g/ is the apparent acceleration due to gravity]
Now, the angular velocity of earth is changed in such a way that the g/ becomes zero.
Let the new angular velocity of earth be ω/.
So, 0 = g – (ω/)2R
=> ω/ = (g/R)1/2
=> ω/ =1.24 × 10-3 rad/s
And,
ω// ω = (1.24 × 10-3)/(7.27 × 10-5) = 17.06
Thus, the angular velocity of earth’s rotation must increase by 17.06 times to nullify gravity at the equator.
The duration of a day would be, T = 2π/ω/ = 5064.5 s = 84 min
Let the angular velocity of earth be ω, and the time period be T. The radius of earth is R.
Now, T = 24 h = 86400 s
R = 6370000 m
ω = 2π/T = 7.27 × 10-5 rad/s
The relation between angular velocity and acceleration due to gravity is given by,
g/ = g - ω2R [g/ is the apparent acceleration due to gravity]
Now, the angular velocity of earth is changed in such a way that the g/ becomes zero.
Let the new angular velocity of earth be ω/.
So, 0 = g – (ω/)2R
=> ω/ = (g/R)1/2
=> ω/ =1.24 × 10-3 rad/s
And,
ω// ω = (1.24 × 10-3)/(7.27 × 10-5) = 17.06
Thus, the angular velocity of earth’s rotation must increase by 17.06 times to nullify gravity at the equator.
The duration of a day would be, T = 2π/ω/ = 5064.5 s = 84 min
