Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
To verify the distance between the probe and the center of Venus, we follow these steps using the gravitational force formula:
The formula for the gravitational force [tex]\( F \)[/tex] between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:
[tex]\[ F = G \cdot \frac{m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the objects,
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of Venus ([tex]\( 4.87 \times 10^{24} \, \text{kg} \)[/tex]),
- [tex]\( m_2 \)[/tex] is the mass of the probe (655 kg),
- [tex]\( r \)[/tex] is the distance between the two objects (which we need to verify as [tex]\( 1 \times 10^6 \, \text{m} \)[/tex]).
Given:
- [tex]\( F = 2.58 \times 10^3 \, \text{N} \)[/tex],
- [tex]\( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex],
- [tex]\( m_1 = 4.87 \times 10^{24} \, \text{kg} \)[/tex],
- [tex]\( m_2 = 655 \, \text{kg} \)[/tex].
Plugging these values into the gravitational force equation, we get:
[tex]\[ 2.58 \times 10^3 = 6.67 \times 10^{-11} \cdot \frac{4.87 \times 10^{24} \cdot 655}{r^2} \][/tex]
To solve for [tex]\( r \)[/tex], we rearrange the equation:
[tex]\[ r^2 = \frac{6.67 \times 10^{-11} \cdot 4.87 \times 10^{24} \cdot 655}{2.58 \times 10^3} \][/tex]
[tex]\[ r^2 = \frac{2.11919035 \times 10^{17}}{2.58 \times 10^3} \][/tex]
[tex]\[ r^2 = 8.21472829 \times 10^{13} \][/tex]
Taking the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r \approx \sqrt{8.21472829 \times 10^{13}} \][/tex]
[tex]\[ r \approx 9.065 \times 10^6 \, \text{m} \][/tex]
Upon review, it seems the distance calculation from the initial Python code was incorrect. With further investigation, it reveals the probe should be approximately [tex]\( 9.065 \times 10^6 \, \text{m} \)[/tex] from the center of Venus, not [tex]\( 1 \times 10^6 \, \text{m} \)[/tex].
The formula for the gravitational force [tex]\( F \)[/tex] between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:
[tex]\[ F = G \cdot \frac{m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the objects,
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of Venus ([tex]\( 4.87 \times 10^{24} \, \text{kg} \)[/tex]),
- [tex]\( m_2 \)[/tex] is the mass of the probe (655 kg),
- [tex]\( r \)[/tex] is the distance between the two objects (which we need to verify as [tex]\( 1 \times 10^6 \, \text{m} \)[/tex]).
Given:
- [tex]\( F = 2.58 \times 10^3 \, \text{N} \)[/tex],
- [tex]\( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex],
- [tex]\( m_1 = 4.87 \times 10^{24} \, \text{kg} \)[/tex],
- [tex]\( m_2 = 655 \, \text{kg} \)[/tex].
Plugging these values into the gravitational force equation, we get:
[tex]\[ 2.58 \times 10^3 = 6.67 \times 10^{-11} \cdot \frac{4.87 \times 10^{24} \cdot 655}{r^2} \][/tex]
To solve for [tex]\( r \)[/tex], we rearrange the equation:
[tex]\[ r^2 = \frac{6.67 \times 10^{-11} \cdot 4.87 \times 10^{24} \cdot 655}{2.58 \times 10^3} \][/tex]
[tex]\[ r^2 = \frac{2.11919035 \times 10^{17}}{2.58 \times 10^3} \][/tex]
[tex]\[ r^2 = 8.21472829 \times 10^{13} \][/tex]
Taking the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r \approx \sqrt{8.21472829 \times 10^{13}} \][/tex]
[tex]\[ r \approx 9.065 \times 10^6 \, \text{m} \][/tex]
Upon review, it seems the distance calculation from the initial Python code was incorrect. With further investigation, it reveals the probe should be approximately [tex]\( 9.065 \times 10^6 \, \text{m} \)[/tex] from the center of Venus, not [tex]\( 1 \times 10^6 \, \text{m} \)[/tex].
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.