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Step-by-Step Solution
Step 1: Identify the Known Information
• The weight of the body at the North Pole is given as 49 N.
• Acceleration due to gravity (assumed constant at the pole) is $g = 9.8 \, \text{m/s}^2$.
• The radius of Earth is $R = 6400 \,\text{km} = 6.4 \times 10^6 \,\text{m}$.
• Let $m$ be the mass of the body.
Since weight $w_p = mg$ at the North Pole and $w_p = 49 \,\text{N}$:
$ m = \dfrac{w_p}{g} = \dfrac{49}{9.8} = 5 \,\text{kg}.$
Step 2: Write the Expression for Weight at the Equator
When the body is at the equator, it experiences a slightly reduced effective acceleration due to gravity because of Earth's rotation. This effective acceleration $g_e$ can be written as:
$ g_e = g - R \omega^2, $
where $\omega$ is the angular speed of Earth’s rotation. Consequently, the weight at the equator is:
$ w_e = m \, g_e = m \bigl(g - R \,\omega^2\bigr). $
Step 3: Calculate the Angular Speed of Earth
Earth completes one rotation in about 24 hours, i.e., $T = 24 \times 3600 = 86400 \,\text{s}$. Hence,
$ \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{86400} \approx 7.27 \times 10^{-5} \,\text{rad/s}. $
Step 4: Compute $R \omega^2$
First, evaluate $\omega^2$:
$ \omega^2 = (7.27 \times 10^{-5})^2 \approx 5.29 \times 10^{-9} \,\text{(rad/s)}^2. $
Then,
$ R \omega^2 = (6.4 \times 10^6 \,\text{m}) \times (5.29 \times 10^{-9}) \approx 0.034 \,\text{m/s}^2. $
Step 5: Determine the Effective Acceleration $g_e$ at the Equator
Subtract the centripetal term from $g$:
$ g_e = g - R\omega^2 = 9.8 \,\text{m/s}^2 - 0.034 \,\text{m/s}^2 = 9.766 \,\text{m/s}^2. $
Step 6: Calculate the Weight at the Equator
Using the mass $m = 5\,\text{kg}$,
$ w_e = m \times g_e = 5 \,\text{kg} \times 9.766 \,\text{m/s}^2 \approx 48.83\,\text{N}. $
Thus, the weight of the body at the equator is about $48.83\,\text{N}$.
