Chapter 4: Problem 33
Venus's circular velocity is \(35.03 \mathrm{km} / \mathrm{s}\), and its orbital radius is \(1.082 \times 10^{8} \mathrm{km} .\) Use this information to calculate the mass of the Sun.
Short Answer
Expert verified
The mass of the Sun is approximately \(1.9888 \times 10^{30} \, \text{kg}\).
Step by step solution
01
- Understand the relationship
The relationship between the circular orbital velocity of a planet, the radius of its orbit, and the mass of the Sun can be given by the formula for circular orbital velocity: \(v = \frac{\text{GM}}{r}\)where \(v\) is the orbital velocity, \(G\) is the gravitational constant, \(M\) is the mass of the Sun, and \(r\) is the orbital radius of the planet.
02
- Rearrange the formula to solve for mass
Rearrange the formula to solve for the mass of the Sun (\(M\)): \(M = \frac{v^2 \times r}{G}\)
03
- Substitute known values
Substitute the given values and the gravitational constant (\(G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)) into the formula: \(v = 35.03 \times 10^3 \, \text{m/s}\), \(r = 1.082 \times 10^{11} \, \text{m}\), \(G = 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)
04
- Calculate the mass of the Sun
Perform the calculation:\[M = \frac{(35.03 \times 10^3)^2 \times 1.082 \times 10^{11}}{6.67430 \times 10^{-11}}\]\[M = \frac{1.2271 \times 10^{9} \times 1.082 \times 10^{11}}{6.67430 \times 10^{-11}}\]\[M \rightarrow 1.9888 \times 10^{30} \, \text{kg}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Circular orbital velocity
Circular orbital velocity is the speed at which a planet or object orbits around a more massive body, such as the Sun.
This concept is crucial when calculating the mass of such a central body.
The formula for circular orbital velocity is given by:
\(v = \sqrt{\frac{GM}{r}}\) Here:
This concept is crucial when calculating the mass of such a central body.
The formula for circular orbital velocity is given by:
\(v = \sqrt{\frac{GM}{r}}\) Here:
- \(v\): Orbital velocity
- \(G\): Gravitational constant
- \(M\): Mass of the central body (like the Sun)
- \(r\): Orbital radius (distance from the center)
Gravitational constant
The gravitational constant, denoted as \(G\), is a key constant in physics that appears in Newton's law of gravitation.
Its value is:
\(G = 6.67430 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)
This constant helps us relate the gravitational force between two bodies to their masses and the distance between them.
When calculating the mass of the Sun using orbital data, \(G\) plays a major role in the formulas involved.
Its value is:
\(G = 6.67430 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)
This constant helps us relate the gravitational force between two bodies to their masses and the distance between them.
When calculating the mass of the Sun using orbital data, \(G\) plays a major role in the formulas involved.
Orbital radius
The orbital radius is the distance of a planet or satellite from the center of the object it orbits.
In our exercise with Venus, the orbital radius \(r\) is given as \(1.082 \times 10^{8} \, \text{km}\).
We need to convert this into meters for our calculations:
We use the formula:
\(r = 1.082 \times 10^{8} \times 10^3 \, \text{m}\)\rightarrow \( r = 1.082 \times 10^{11} \, \text{m} \)
Knowing this and the orbital velocity, we can find the mass of the Sun.
In our exercise with Venus, the orbital radius \(r\) is given as \(1.082 \times 10^{8} \, \text{km}\).
We need to convert this into meters for our calculations:
We use the formula:
\(r = 1.082 \times 10^{8} \times 10^3 \, \text{m}\)\rightarrow \( r = 1.082 \times 10^{11} \, \text{m} \)
Knowing this and the orbital velocity, we can find the mass of the Sun.
Mass of the Sun
The mass of the Sun is a fundamental astronomical value.
Using the orbital velocity of Venus and its orbital radius, we can determine the Sun's mass.
From the rearranged formula:
\(M = \frac{v^2 \times r}{G}\)
We substitute the known values:
\( M \approx 1.9888 \times 10^{30} \ \text{kg} \) This mass is crucial for understanding the dynamics of the solar system.
Using the orbital velocity of Venus and its orbital radius, we can determine the Sun's mass.
From the rearranged formula:
\(M = \frac{v^2 \times r}{G}\)
We substitute the known values:
- \(v = 35.03 \times 10^3 \ \text{m/s}\)
- \(r = 1.082 \times 10^{11} \ \text{m}\)
- \(G = 6.67430 \times 10^{-11} \ \text{m}^3 \text{kg}^{-1} \text{s}^{-2}\)
\( M \approx 1.9888 \times 10^{30} \ \text{kg} \) This mass is crucial for understanding the dynamics of the solar system.
Relational formulas in astronomy
Astronomical studies often rely on relational formulas to link different physical quantities.
These formulas help predict and verify various aspects of celestial mechanics.
For example:
These relationships are fundamental in astronomy, helping us understand and explore celestial phenomena.
These formulas help predict and verify various aspects of celestial mechanics.
For example:
- Orbital velocity formula: \(v = \sqrt{\frac{GM}{r}}\)
- Formula for orbital period: \(T = 2 \pi \sqrt{\frac{r^3}{GM}}\)
These relationships are fundamental in astronomy, helping us understand and explore celestial phenomena.