Calculate the acceleration due to gravity on Mercury due to Sun.
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![**Problem: Calculate the acceleration due to gravity on Mercury due to the Sun.**
In this exercise, we will determine the acceleration that the Sun induces on Mercury due to gravitational attraction. This requires applying the universal law of gravitation. The method involves using known values such as the gravitational constant, the mass of the Sun, and the average distance between the Sun and Mercury.
For students, it is important to understand the formula used in this scenario:
\[ F = \frac{G \cdot M_s \cdot M_m}{r^2} \]
Where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\)),
- \( M_s \) is the mass of the Sun (\(1.989 \times 10^{30} \, \text{kg}\)),
- \( M_m \) is the mass of Mercury (\(3.301 \times 10^{23} \, \text{kg}\)),
- \( r \) is the average distance between the Sun and Mercury (\(5.791 \times 10^{10} \, \text{m}\)).
Ultimately the acceleration \( a \) due to the gravity \( F \) is given by Newton's second law:
\[ a = F/M_m. \]
For instructional purposes, encourage students to solve the equation step by step to obtain the value of the acceleration due to gravity experienced by Mercury because of the Sun. This process illustrates key concepts in both gravity and orbital mechanics, enhancing the understanding of celestial dynamics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffc44715e-e997-4081-aac6-58f75fd336c7%2F89bbb390-43dd-4f67-8113-1aaabbf52d64%2Fr55vhzy_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem: Calculate the acceleration due to gravity on Mercury due to the Sun.**
In this exercise, we will determine the acceleration that the Sun induces on Mercury due to gravitational attraction. This requires applying the universal law of gravitation. The method involves using known values such as the gravitational constant, the mass of the Sun, and the average distance between the Sun and Mercury.
For students, it is important to understand the formula used in this scenario:
\[ F = \frac{G \cdot M_s \cdot M_m}{r^2} \]
Where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant (\(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\)),
- \( M_s \) is the mass of the Sun (\(1.989 \times 10^{30} \, \text{kg}\)),
- \( M_m \) is the mass of Mercury (\(3.301 \times 10^{23} \, \text{kg}\)),
- \( r \) is the average distance between the Sun and Mercury (\(5.791 \times 10^{10} \, \text{m}\)).
Ultimately the acceleration \( a \) due to the gravity \( F \) is given by Newton's second law:
\[ a = F/M_m. \]
For instructional purposes, encourage students to solve the equation step by step to obtain the value of the acceleration due to gravity experienced by Mercury because of the Sun. This process illustrates key concepts in both gravity and orbital mechanics, enhancing the understanding of celestial dynamics.
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