Suppose that the rotation of the earth was increased such that the centripetal acceleration was equal to the gravitational acceleration at the equator (i) Find the speed of a person standing at the equator and (ii) How long would a day be ?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between centripetal acceleration and gravitational acceleration.
Centripetal acceleration \( a_c \) is given by the formula:
\[ a_c = \frac{V^2}{R} \]
where \( V \) is the linear speed of a person at the equator and \( R \) is the radius of the Earth.

At the equator, we want the centripetal acceleration to be equal to the gravitational acceleration \( g \):
\[ \frac{V^2}{R} = g \]

### Step 2: Rearrange the equation to solve for \( V \).
From the equation above, we can rearrange it to find \( V \):
\[ V^2 = g \cdot R \]
\[ V = \sqrt{g \cdot R} \]

### Step 3: Substitute the values for \( g \) and \( R \).
We know:
- The gravitational acceleration \( g \approx 9.8 \, \text{m/s}^2 \)
- The radius of the Earth \( R \approx 6400 \times 10^3 \, \text{m} \)

Now substituting these values into the equation:
\[ V = \sqrt{9.8 \, \text{m/s}^2 \cdot 6400 \times 10^3 \, \text{m}} \]

### Step 4: Calculate \( V \).
Calculating the above expression:
\[ V = \sqrt{9.8 \times 6400 \times 10^3} \approx \sqrt{62752000} \approx 7925 \, \text{m/s} \approx 7.9 \, \text{km/s} \]

### Step 5: Find the time period of rotation (length of a day).
The time period \( T \) can be calculated using the formula:
\[ T = \frac{2\pi R}{V} \]

Substituting the values:
\[ T = \frac{2\pi \cdot (6400 \times 10^3)}{7925} \]

### Step 6: Calculate \( T \).
Calculating the above expression:
\[ T \approx \frac{2 \cdot 3.14 \cdot 6400 \times 10^3}{7925} \approx \frac{40212320}{7925} \approx 5070 \, \text{seconds} \]

### Step 7: Convert seconds to hours.
To convert seconds into hours:
\[ \text{Hours} = \frac{5070}{3600} \approx 1.41 \, \text{hours} \]

### Final Answers:
(i) The speed of a person standing at the equator would be approximately \( 7.9 \, \text{km/s} \).

(ii) The length of a day would be approximately \( 5070 \, \text{seconds} \) or about \( 1.41 \, \text{hours} \).

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