Question - Calculating Gravitational Force Between Celestial Bodies
Solution:
To solve for the gravitational force between the Sun and Proxima Centauri, we can use Newton's law of gravitation given by \( F = \frac{G \cdot m_1 \cdot m_2}{r^2} \) where \( F \) is the gravitational force, \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the two bodies, and \( r \) is the distance between the centers of the two masses.
\( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) (Gravitational constant)
\( m_1 = 2.00 \times 10^{30} \, \text{kg} \) (Mass of the Sun)
\( m_2 = 2.40 \times 10^{29} \, \text{kg} \) (Mass of Proxima Centauri)
\( r = 4.00 \times 10^{13} \, \text{km} \) (Distance between Sun and Proxima Centauri)
We need to convert \( r \) from kilometers to meters:
\( r = 4.00 \times 10^{13} \times 10^3 \, \text{m} \) (since \( 1 \, \text{km} = 10^3 \, \text{m} \))
Now we substitute the values into the equation:
\( F = \frac{(6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \cdot (2.00 \times 10^{30} \, \text{kg}) \cdot (2.40 \times 10^{29} \, \text{kg})}{(4.00 \times 10^{16} \, \text{m})^2} \)
\( F = \frac{(6.674 \times 10^{-11}) \cdot (2.00 \times 10^{30}) \cdot (2.40 \times 10^{29})}{16.00 \times 10^{32}} \)
\( F = \frac{(6.674 \times 2.40) \cdot 10^{48}}{16.00 \times 10^{32}} \)
\( F = \frac{16.0184 \times 10^{48}}{16.00 \times 10^{32}} \)
\( F = 1.00115 \times 10^{16} \, \text{N} \)
Note: The final answer has been rounded to five significant figures.
Tutor Solution:
The question deals with the gravitational force between two celestial bodies, the Sun and Proxima Centauri. In part (i), it is asking for an explanation of how Newton's law of gravitation applies, given that we have the diameters of two stars.
Newton's law of gravitation states that every point mass attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the separation between their centers. This law can be applied to objects that are spherically symmetric, treating them as point masses located at their centers. Since the diameters of the Sun and Proxima Centauri are given, we can assume that both stars are approximately spherical in shape and their masses can be considered to be concentrated at their centers. This is why Newton's law of gravitation can be used to calculate the gravitational force between the two stars.
In part (ii), the question requests the calculation of that gravitational force \( F \) between the Sun and Proxima Centauri. We can use the formula derived from Newton's law of gravitation:
\[ F = \frac{G M_1 M_2}{r^2} \]Where \( F \) is the gravitational force, \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \) N m²/kg²), \( M_1 \) and \( M_2 \) are the masses of the two bodies (the Sun and Proxima Centauri, respectively), and \( r \) is the distance between the centers of the two bodies.
Plugging in the given values:
\[ M_{\text{Sun}} = 2.0 \times 10^{30} \text{ kg} \] \[ M_{\text{Proxima Centauri}} = 4.2 \times 10^{29} \text{ kg} \] \[ r = 4.0 \times 10^{13} \text{ km} = 4.0 \times 10^{16} \text{ m (since } 1 \text{ km} = 10^{3} \text{ m)} \]The calculation would be:
\[ F = \frac{6.67 \times 10^{-11} \times (2.0 \times 10^{30}) \times (4.2 \times 10^{29})}{(4.0 \times 10^{16})^2} \]Carrying out the computation, you will arrive at a force \( F \) which has a large value, as indicated in the image; however, the specific result of the force calculation isn't completely visible in the image. The calculations shown lead to an intermediate step of \( (8.0 \times 10^{60}) / (1.6 \times 10^{34}) \), and afterward, the result is indicated as very large but not explicitly stated.
In the final part of the question, it asks why this force has a negligible effect on the motion of the Sun. The gravitational force, while large, is negligible for the Sun's motion because the Sun's mass is so much greater than the force acting upon it. Additionally, there are many other forces and factors influencing the Sun's motion within the Milky Way galaxy, including its momentum, the gravitational pull from other celestial bodies, and its position within the gravitational "well" of the galaxy. Therefore, the force from Proxima Centauri on the Sun is not significant enough to have considerable influence, and the Sun's path is largely unaffected by it.
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com