To calculate the gravitational force exerted by Ceres on your dog, we can use Newton's law of universal gravitation:
\[ F = G \frac{m_1 m_2}{r^2} \]
where:
- \( F \) is the force between the two objects,
- \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} , \text{N m}^2/\text{kg}^2 \),
- \( m_1 \) is the mass of Ceres,
- \( m_2 \) is the mass of your dog,
- \( r \) is the distance between the centers of the two objects.
Given:
- Mass of Ceres, \( m_1 = 8.7 \times 10^{20} , \text{kg} \)
- Mass of your dog, \( m_2 = 30 , \text{kg} \)
- Distance, \( r = 14,000 , \text{km} = 14,000,000 , \text{m} \)
Now we can substitute these values into the equation:
\[ F = G \frac{m_1 m_2}{r^2} \]
Substituting in the values:
\[ F = 6.674 \times 10^{-11} \frac{(8.7 \times 10^{20})(30)}{(14,000,000)^2} \]
Calculating the denominator:
\[ (14,000,000)^2 = 1.96 \times 10^{14} , \text{m}^2 \]
Now substituting back into the equation:
\[ F = 6.674 \times 10^{-11} \frac{(8.7 \times 10^{20})(30)}{1.96 \times 10^{14}} \]
Calculating the numerator:
\[ (8.7 \times 10^{20})(30) = 2.61 \times 10^{22} \]
So we have:
\[ F = 6.674 \times 10^{-11} \frac{2.61 \times 10^{22}}{1.96 \times 10^{14}} \]
Now, calculating the fraction:
\[ \frac{2.61 \times 10^{22}}{1.96 \times 10^{14}} \approx 1.33 \times 10^{8} \]
Now substituting back to find \( F \):
\[ F = 6.674 \times 10^{-11} \cdot 1.33 \times 10^{8} \]
Calculating the final result:
\[ F \approx 8.88 \times 10^{-3} , \text{N} \]
Thus, the force exerted by Ceres on your dog when it passes within 14,000 km is approximately:
\[ \boxed{0.00888 , \text{N}} \]
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