Find the accelerationof the moon with respect to the eath from the following data, Distance between the earth and the moon `=3.85xx10^5` km and the time taken by the moon to complete one revolution around the earth `=27.3days

Explanation

To find the acceleration of the moon with respect to the Earth, we can follow these steps:

### Step 1: Convert the time period from days to seconds
The time taken by the moon to complete one revolution around the Earth is given as 27.3 days. We need to convert this into seconds.

\[ \text{Time in seconds} = 27.3 \, \text{days} \times 24 \, \text{hours/day} \times 3600 \, \text{seconds/hour} \]

Calculating this:

\[ \text{Time in seconds} = 27.3 \times 24 \times 3600 = 2.36 \times 10^6 \, \text{seconds} \]

### Step 2: Calculate the velocity of the moon
The distance between the Earth and the moon is given as \(3.85 \times 10^5\) km. We need to convert this distance into meters:

\[ \text{Distance} = 3.85 \times 10^5 \, \text{km} = 3.85 \times 10^5 \times 1000 \, \text{m} = 3.85 \times 10^8 \, \text{m} \]

Now, we can calculate the velocity \(v\) of the moon using the formula:

\[ v = \frac{d}{t} \]

Where \(d\) is the distance and \(t\) is the time period we calculated earlier.

\[ v = \frac{3.85 \times 10^8 \, \text{m}}{2.36 \times 10^6 \, \text{s}} \approx 163.5 \, \text{m/s} \]

### Step 3: Calculate the centripetal acceleration
The centripetal acceleration \(a\) of the moon can be calculated using the formula:

\[ a = \frac{v^2}{r} \]

Where \(v\) is the velocity we found in the previous step and \(r\) is the distance from the Earth to the moon.

Substituting the values:

\[ a = \frac{(163.5 \, \text{m/s})^2}{3.85 \times 10^8 \, \text{m}} \]

Calculating this gives:

\[ a \approx \frac{26726.25}{3.85 \times 10^8} \approx 6.93 \times 10^{-5} \, \text{m/s}^2 \]

### Final Answer
The acceleration of the moon with respect to the Earth is approximately \(6.93 \times 10^{-5} \, \text{m/s}^2\).

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