What woluld be the length of a day. If angular speed of earth is increased such that bodies typing on the equator by off?

Answer

To solve the problem of determining the length of a day if the angular speed of the Earth is increased such that bodies at the equator feel as if they are about to fly off, we can follow these steps:

### Step 1: Understand the condition for flying off
At the equator, for a body to feel weightless (or to "fly off"), the effective gravitational force must be zero. The effective gravitational force \( g' \) at the equator can be expressed as:
\[ g' = g - r \omega^2 \]
where:
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( r \) is the radius of the Earth (approximately \( 6.371 \times 10^6 \, \text{m} \)),
- \( \omega \) is the angular speed of the Earth.

### Step 2: Set the effective gravitational force to zero
To find the angular speed \( \omega \) at which the effective gravitational force becomes zero, we set:
\[ 0 = g - r \omega^2 \]
Rearranging gives:
\[ r \omega^2 = g \]
Thus,
\[ \omega^2 = \frac{g}{r} \]

### Step 3: Calculate the angular speed \( \omega \)
Substituting the known values:
- \( g = 9.81 \, \text{m/s}^2 \)
- \( r = 6.371 \times 10^6 \, \text{m} \)

We calculate \( \omega \):
\[ \omega = \sqrt{\frac{9.81}{6.371 \times 10^6}} \approx \sqrt{1.539 \times 10^{-6}} \approx 0.0012374 \, \text{rad/s} \]

### Step 4: Calculate the time period \( T \)
The time period \( T \) (length of the day) is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2 \times 3.14}{0.0012374} \approx 5075.16 \, \text{seconds} \]

### Step 5: Convert seconds to hours
To convert seconds into hours, we divide by the number of seconds in an hour (3600 seconds):
\[ \text{Time in hours} = \frac{5075.16}{3600} \approx 1.3 \, \text{hours} \]

### Final Answer
Thus, the length of the day, if the angular speed of the Earth is increased such that bodies at the equator feel as if they are about to fly off, would be approximately **1.3 hours**.

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