Question:
The angular speed of the Earth’s rotation is $7.3 \times 10^{-5}\text{ rad}\cdot\text{s}^{-1}$. Take the radius of the Earth at the equator to be $6400\text{ km}$. Then the ratio ($a_c / g$) of the magnitude of the centripetal acceleration $a_c$ at a point on the equator to $g$, is of the order
The angular speed of the Earth’s rotation is $7.3 \times 10^{-5}\text{ rad}\cdot\text{s}^{-1}$. Take the radius of the Earth at the equator to be $6400\text{ km}$. Then the ratio ($a_c / g$) of the magnitude of the centripetal acceleration $a_c$ at a point on the equator to $g$, is of the order
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An order of magnitude estimate can be done quickly by rounding numbers:
$\omega \approx 7 \times 10^{-5}$, so $\omega^2 \approx 5 \times 10^{-9}$.
With $R \approx 6 \times 10^6$, we get $a_c \approx 3 \times 10^{-2} = 0.03$.
Dividing by $g \approx 10$ yields $0.003 = 3 \times 10^{-3}$, confirming the order of magnitude.
$\omega \approx 7 \times 10^{-5}$, so $\omega^2 \approx 5 \times 10^{-9}$.
With $R \approx 6 \times 10^6$, we get $a_c \approx 3 \times 10^{-2} = 0.03$.
Dividing by $g \approx 10$ yields $0.003 = 3 \times 10^{-3}$, confirming the order of magnitude.
Updated On: Jun 16, 2026
- $10^{-3}$
- $10^{-5}$
- $10^{0}$
- $10^{-1}$
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Solution and Explanation
Step 1: Understanding the Question:
The question asks us to compute the ratio of the centripetal acceleration of a point on the Earth's equator to the standard acceleration due to gravity, and then determine its order of magnitude.
Centripetal acceleration arises from the Earth's rotation about its polar axis.
Step 2: Key Formula or Approach:
The formula for centripetal acceleration $a_c$ of a point rotating at a distance $R$ with angular speed $\omega$ is:
\[ a_c = \omega^2 R \]
We will substitute the given values into this formula and divide by the standard acceleration due to gravity $g \approx 9.8\text{ m/s}^2$.
Step 3: Detailed Explanation:
• The given angular speed of rotation is $\omega = 7.3 \times 10^{-5}\text{ rad/s}$.
• The equatorial radius of the Earth is $R = 6400\text{ km} = 6.4 \times 10^6\text{ m}$.
• Let us calculate the centripetal acceleration $a_c$ at the equator:
\[ a_c = (7.3 \times 10^{-5}\text{ rad/s})^2 \times (6.4 \times 10^6\text{ m}) \]
\[ a_c = (53.29 \times 10^{-10}) \times (6.4 \times 10^6) \]
\[ a_c = 341.056 \times 10^{-4} \approx 0.034\text{ m/s}^2 \]
• Now, we find the ratio of this centripetal acceleration to the acceleration due to gravity ($g \approx 9.8\text{ m/s}^2$):
\[ \frac{a_c}{g} = \frac{0.034\text{ m/s}^2}{9.8\text{ m/s}^2} \approx 0.00347 \]
• Expressing this ratio in scientific notation:
\[ \frac{a_c}{g} = 3.47 \times 10^{-3} \]
• The order of magnitude of this value is $10^{-3}$.
Step 4: Final Answer:
The ratio of the centripetal acceleration to $g$ is of the order of $10^{-3}$.
The question asks us to compute the ratio of the centripetal acceleration of a point on the Earth's equator to the standard acceleration due to gravity, and then determine its order of magnitude.
Centripetal acceleration arises from the Earth's rotation about its polar axis.
Step 2: Key Formula or Approach:
The formula for centripetal acceleration $a_c$ of a point rotating at a distance $R$ with angular speed $\omega$ is:
\[ a_c = \omega^2 R \]
We will substitute the given values into this formula and divide by the standard acceleration due to gravity $g \approx 9.8\text{ m/s}^2$.
Step 3: Detailed Explanation:
• The given angular speed of rotation is $\omega = 7.3 \times 10^{-5}\text{ rad/s}$.
• The equatorial radius of the Earth is $R = 6400\text{ km} = 6.4 \times 10^6\text{ m}$.
• Let us calculate the centripetal acceleration $a_c$ at the equator:
\[ a_c = (7.3 \times 10^{-5}\text{ rad/s})^2 \times (6.4 \times 10^6\text{ m}) \]
\[ a_c = (53.29 \times 10^{-10}) \times (6.4 \times 10^6) \]
\[ a_c = 341.056 \times 10^{-4} \approx 0.034\text{ m/s}^2 \]
• Now, we find the ratio of this centripetal acceleration to the acceleration due to gravity ($g \approx 9.8\text{ m/s}^2$):
\[ \frac{a_c}{g} = \frac{0.034\text{ m/s}^2}{9.8\text{ m/s}^2} \approx 0.00347 \]
• Expressing this ratio in scientific notation:
\[ \frac{a_c}{g} = 3.47 \times 10^{-3} \]
• The order of magnitude of this value is $10^{-3}$.
Step 4: Final Answer:
The ratio of the centripetal acceleration to $g$ is of the order of $10^{-3}$.
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