Given that `T^(2) = kR^(3)`, express the constant `k` of the above relation in days and kilometres. Given, `k= 10^(-13)s^(2) m^(-3)`. The Moon is at a distance of `3.84 xx 10^(5) km` from the earth. Obtain its time period of revolution in days.

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Convert the constant \( k \) from \( s^2 \, m^{-3} \) to \( day^2 \, km^{-3} \)

Given:
\[ k = 10^{-13} \, s^2 \, m^{-3} \]

We need to convert seconds to days and meters to kilometers.
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