Determine the speed with which the earth would have to roate on its axis , so that a person on the equator would weigh (3)/(5) th as nuch as at person. Take R = 6400 km.
Answers
TO FIND :
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 of its weight.
SOLUTION :
◆We know,
W' = W - m R ω² cos λ
◆Weight , W = mg
where, 'm' is mass
'g' is the gravity.
◆New weight on the equator
= W' = 3/5 mg --- (1)
Now, we know
◆mg' = mg - m R ω² cos λ
λ - distance from equator
ω - Angular velocity
R - Radius of earth - 6.4×10^6
◆Substituting,
3/5 mg = mg - m R ω² cos 0
(As λ = 0 , weight is at equator)
3/5 mg = mg - m R ω² ×1
3/5 mg = mg - m R ω²
m R ω² = mg - 3/5 mg
m R ω² = 2/5 mg
ω² = 2 g / 5 R
Substituting values,
ω² = (2 x 10) / (5 x 6.4x10⁶)
ω = √ (6.25x10⁻⁷) = 7.8 x 10⁻⁴ rad/sec.
ANSWER:
Speed with which the earth would have to rotate on its axis , so that a person on the equator would weigh (3)/(5) as much as at person is
ω = 7.8 x 10⁻⁴ rad/sec.