Listen in to a comprehensive guide on calculating the mass of stars, planets, and moons using Kepler’s third law and Newtonian gravity. This blog features an embedded podcast discussion, a detailed transcript, a summary of key concepts, and a Deep Research report on orbital mechanics. Ideal for educators, students, and astronomy lovers!

Listen to “Calculating Cosmic Masses: Kepler’s Laws, Newtonian Gravity & Astrophysics” on Spreaker.

Summary of the Dialogue and Key Concepts:

Listen to “Weighing the Unweighable: How Scientists Measure the Mass of Planets and Stars” on Spreaker.

The dialogue between Vega and Hakeem explores how Kepler’s third law and Newtonian gravity are used to calculate the mass of celestial bodies. Here’s a structured breakdown:

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1. Kepler’s Third Law

• Definition: The square of the orbital period (PP) is proportional to the cube of the semi-major axis (aa):P2∝a3P2∝a3
• Newton’s Enhancement: Incorporates the gravitational constant (GG) and the mass of the central body (MM):M=4π2a3GP2M=GP24π2a3​This formula allows calculating MM if aa (semi-major axis) and PP (orbital period) of an orbiting object are known.

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2. Key Terms Clarified

• Semi-Major Axis: Half the longest diameter of an elliptical orbit; the average distance between the orbiting body and the central mass.
• Orbital Period: Time taken for one full orbit (e.g., Earth’s orbital period = 1 year).
• Gravitational Constant (GG): A universal constant () critical for mass calculations.

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3. Applications to Celestial Bodies

Sun’s Mass

• Data Used: Earth’s orbital period (P=1 yearP=1year) and semi-major axis (a=1 AU=1.496×1011 ma=1AU=1.496×1011m).
• Result:M⊙≈1.989×1030 kg (2 octillion kg)M⊙​≈1.989×1030kg(2 octillion kg)

Pluto’s Mass

• Data Used: Charon’s orbital period (P=153 hours=5.51×105 sP=153hours=5.51×105s) and semi-major axis (a=19,640 km=1.964×107 ma=19,640km=1.964×107m).
• Result:MPluto≈1.309×1022 kg (13 sextillion kg)MPluto​≈1.309×1022kg(13 sextillion kg)

Earth’s Mass

• Method: Surface gravity (g=9.81 m/s2g=9.81m/s2) and Earth’s radius (R=6.371×106 mR=6.371×106m) via:M=gR2GM=GgR2​
• Result:M⊕≈5.97×1024 kg (5.97 septillion kg)M⊕​≈5.97×1024kg(5.97 septillion kg)

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4. Importance of Mathematics

• Algebra and Geometry: Fundamental for manipulating equations and understanding orbital mechanics.
• Real-World Relevance: Demonstrates how math unlocks cosmic mysteries (e.g., mass, distance, orbital dynamics).

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5. Field of Study

• Astrophysics: The branch of astronomy applying physics and mathematics to celestial phenomena. This exercise exemplifies its practical use.

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6. Educational Takeaway

• Motivation for Math: Even basic algebra/geometry can reveal properties of distant objects (e.g., Pluto’s mass) or fundamental constants like GG.
• Scale of the Universe: Highlights the vastness of celestial masses (e.g., Sun’s mass = 99.86% of the solar system’s total mass).

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Conclusion: The dialogue underscores the power of mathematics in astrophysics, enabling us to compute masses of celestial bodies using orbital data. This bridges abstract equations with tangible cosmic phenomena, inspiring curiosity about the universe. 🌌🔭 – DeepSeek

Deep Research Executive Summary: Dr. Hakeem Ali-Bocas Alexander’s Guided Exploration of Celestial Mass Determination with Google Gemini

Introduction:

Dr. Hakeem Ali-Bocas Alexander, PhD, initiated a detailed conversation with Google Gemini (Vega) to investigate the methodologies employed in determining the mass of celestial bodies. This executive summary serves to document and expand upon that insightful exploration, providing a scientifically rigorous account suitable for Dr. Alexander’s extensive online network. The dialogue between Dr. Alexander and Vega traversed fundamental principles of physics, the intricacies of orbital mechanics, and their practical application in calculating the masses of the Sun, Pluto, and Earth. Dr. Alexander’s expertly crafted questions structured the conversation, guiding Vega through the complexities inherent in these astrophysical concepts.

Understanding the mass of celestial objects holds paramount importance across numerous domains within astrophysics. It forms a cornerstone for comprehending stellar evolution, the dynamics of planetary systems, the formation of galaxies, and the overarching structure of cosmology. By knowing the mass of a celestial body, scientists can construct models of gravitational interactions, predict the behavior of orbits, and ultimately gain deeper insights into the composition and evolutionary history of the universe.¹

Theoretical Framework:

The conversation commenced with an exploration of the foundational laws governing celestial motion and mass determination.

Kepler’s Third Law of Planetary Motion (Timestamp 00:30):

Kepler’s Third Law articulates a fundamental relationship between an orbiting object’s period and the size of its orbit. Specifically, it states that the square of the orbital period is directly proportional to the cube of the semi-major axis of the orbit.³ This empirical law, formulated by Johannes Kepler based on meticulous astronomical observations, describes how the time a planet takes to orbit a star is related to its average distance from that star.⁴ A key implication of this law is that planets situated farther from their star take significantly longer to complete an orbit.³ For instance, Mercury, being the closest planet to the Sun, has a much shorter orbital period compared to Saturn, which resides in a more distant orbit.⁴

While Kepler’s initial formulation of this law did not explicitly incorporate the concept of mass, it provided a crucial stepping stone for Isaac Newton. Kepler’s observations revealed patterns in planetary motion, and Newton, building upon these empirical findings and his own laws of motion, developed the Law of Universal Gravitation, which provided the underlying physical explanation for Kepler’s Third Law.⁴ This unification of observational data with a theoretical framework marked a significant advancement in the field of physics.

Newton’s Law of Universal Gravitation (Timestamp 00:02):

Newton’s Law of Universal Gravitation posits that every particle of matter in the universe attracts every other particle with a force. The magnitude of this force is directly proportional to the product of the masses of the two particles and inversely proportional to the square of the distance between their centers.⁷ Mathematically, this relationship is expressed as F = G * (m1 * m2) / r², where F represents the gravitational force, m1 and m2 are the masses of the two particles, r is the distance between their centers, and G is the gravitational constant.⁸ This law describes gravity as a universal force of attraction that exists between all objects with mass.⁸

Newton’s groundbreaking contribution provided the “why” behind Kepler’s “what.” While Kepler described the elliptical paths and the relationship between orbital period and distance, Newton explained the fundamental force responsible for these motions: gravity.⁴ Because the strength of this gravitational force depends on mass, Newton’s law established a critical link between observable orbital parameters and the mass of the celestial bodies involved.

The Modified Kepler’s Third Law for Mass Determination:

By equating the gravitational force between a central body and an orbiting object to the centripetal force required for the orbiting object to maintain its orbit, a modified form of Kepler’s Third Law can be derived. This modified law allows for the calculation of the mass (M) of the central body: M = (4π² * a³) / (G * P²).¹³ In this formula, ‘a’ represents the semi-major axis of the orbit, ‘P’ is the orbital period of the orbiting object, and ‘G’ is the gravitational constant. This is the formula Vega references around timestamp 02:49, expressed as M equals 4 Pi squared a cubed belabi G asterisk P squared.

This equation holds immense significance in astronomy as it provides a means to determine the mass of distant celestial objects, such as stars and planets, by observing the orbital characteristics of their satellites.⁶ This can be achieved without the need for direct physical interaction or measurement of the central body itself. The power of this formula lies in its ability to connect directly measurable quantities, namely the orbital period and the semi-major axis, with a fundamental property of the celestial body, its mass, through the universally constant value of G.

Understanding Orbital Parameters:

The application of the modified Kepler’s Third Law necessitates a clear understanding of the key parameters involved in describing an orbit.

The Semi-Major Axis (Timestamp 00:54):

In the context of an elliptical orbit, the semi-major axis is defined as half of the longest diameter of the ellipse.⁴ It can also be thought of as the average distance between the orbiting object and the central body.³ For a circular orbit, the semi-major axis is simply the radius of the circle.²¹ The semi-major axis is a fundamental orbital element that characterizes the overall size of the orbit.²² It directly influences other orbital properties, such as the orbital period and the total energy of the orbiting object.²¹ A larger semi-major axis corresponds to a larger orbit, a longer orbital period, and a higher orbital energy.³ This single parameter provides a convenient way to quantify the spatial extent of an elliptical path, simplifying calculations and facilitating comparisons between different orbits.⁵

Orbital Period (Timestamp 01:34):

The orbital period is defined as the time required for an object to complete one full revolution around another object.³ This is a directly observable quantity for many celestial systems, such as planets orbiting stars or moons orbiting planets. The orbital period, in conjunction with the semi-major axis, forms the basis for using the modified Kepler’s Third Law to determine the mass of the central body.⁶ The regularity and predictability of orbital periods allow for precise measurements, making them reliable data points for inferring other celestial properties, including mass.⁵

Case Study: Calculating the Mass of the Sun:

The conversation then turned to applying these principles to calculate the mass of our own star, the Sun.

Application using Earth’s Orbital Period and Semi-Major Axis (Timestamp 02:30):

To determine the Sun’s mass using the modified Kepler’s Third Law, the orbital period and semi-major axis of an object orbiting the Sun must be known. Earth, with its well-documented orbit, serves as an ideal candidate. Earth’s orbital period, the time it takes to complete one revolution around the Sun (one sidereal year), is approximately 365.25 days.²⁸ Converting this to seconds yields approximately 3.15576 × 10⁷ seconds (365.25 days * 24 hours/day * 3600 seconds/hour). The average distance between the Earth and the Sun, represented by the semi-major axis of Earth’s orbit, is approximately 149.60 million kilometers ¹⁹, which is equivalent to 1.4960 × 10¹¹ meters. The selection of Earth’s orbital parameters for this calculation is logical due to the extensive and relatively accurate data available for these values. Our long history of observing Earth’s journey around the Sun has provided a wealth of reliable measurements for both its orbital period and semi-major axis.

The Gravitational Constant (Timestamp 03:16):

The gravitational constant, G, is a fundamental constant of nature that quantifies the strength of the gravitational force. Its value is approximately 6.674 × 10⁻¹¹ N(m/kg)² or m³ kg⁻¹ s⁻².⁸ This constant is believed to have the same value throughout the universe, a principle that allows us to apply the laws of gravity discovered and measured on Earth to celestial objects located vast distances away. The precise value of G is crucial for obtaining accurate mass calculations using the modified Kepler’s Third Law.

Step-by-Step Calculation and Result:

Plugging these values into the formula M_Sun = (4π² * a³) / (G * P²):

M_Sun = (4π² * (1.4960 × 10¹¹ m)³) / ((6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻²) * (3.15576 × 10⁷ s)²)

M_Sun ≈ 1.989 × 10³⁰ kg

This result, approximately 1.989 × 10³⁰ kg, aligns with the value Vega provides around timestamp 03:46. This immense mass underscores the sheer scale of the Sun compared to objects we encounter in our daily lives. The ways of expressing this number in the long form, as explored in the conversation from timestamp 04:21 onwards, further emphasize its staggering magnitude.

Table 1: Orbital Parameters and Calculated Mass of the Sun

Orbiting Body	Orbital Period (seconds)	Semi-Major Axis (meters)	Gravitational Constant (m³ kg⁻¹ s⁻²)	Calculated Mass of the Sun (kg)
Earth	3.15576 × 10⁷	1.4960 × 10¹¹	6.674 × 10⁻¹¹	1.989 × 10³⁰

Case Study: Calculating the Mass of Pluto:

The discussion then shifted to calculating the mass of the dwarf planet Pluto, utilizing the orbital characteristics of its largest moon, Charon.

Application using Charon’s Orbital Period and Semi-Major Axis (Timestamp 05:54):

Similar to how Earth’s orbit around the Sun allows us to determine the Sun’s mass, Charon’s orbit around Pluto can be used to calculate Pluto’s mass. Charon, being gravitationally bound to Pluto as its natural satellite, orbits it with a period of approximately 6.387 Earth days.⁴⁰ Converting this to seconds yields approximately 5.516 × 10⁵ seconds (6.387 days * 24 hours/day * 3600 seconds/hour). The average distance between Pluto and Charon, represented by the semi-major axis of Charon’s orbit, is approximately 19,640 kilometers ⁴¹, or 1.9640 × 10⁷ meters. While some sources provide slightly different values for Charon’s semi-major axis ⁴⁰, the value mentioned in the transcript (19,640 km) will be used here for consistency with the conversation.

Values used for Charon’s Orbital Parameters: As established above, the orbital period of Charon is approximately 5.516 × 10⁵ seconds, and its semi-major axis is approximately 1.9640 × 10⁷ meters.

Step-by-Step Calculation and Result:

Applying the modified Kepler’s Third Law:

M_Pluto = (4π² * (1.9640 × 10⁷ m)³) / ((6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻²) * (5.516 × 10⁵ s)²)

M_Pluto ≈ 1.31 × 10²² kg

This calculated mass of Pluto, approximately 1.31 × 10²² kg, is consistent with the value Vega provides around timestamp 07:05. Another calculation using slightly different input values yields a similar result of 1.46 × 10²² kg.⁴⁴ Comparing Pluto’s mass (on the order of 10²² kg) to the Sun’s mass (on the order of 10³⁰ kg) highlights the substantial difference in scale between a dwarf planet and a star.

Table 2: Orbital Parameters and Calculated Mass of Pluto

Orbiting Body	Orbital Period (seconds)	Semi-Major Axis (meters)	Gravitational Constant (m³ kg⁻¹ s⁻²)	Calculated Mass of Pluto (kg)
Charon	5.516 × 10⁵	1.9640 × 10⁷	6.674 × 10⁻¹¹	1.31 × 10²²

Case Study: Calculating the Mass of the Earth:

The final celestial body whose mass was explored was Earth itself. The method discussed for determining Earth’s mass differed slightly, utilizing the concept of surface gravity.

Methodology based on Surface Gravity, Earth’s Radius, and the Gravitational Constant (Timestamp 07:50):

An alternative method to calculate Earth’s mass involves using the acceleration due to gravity at its surface (g), Earth’s radius (R), and the gravitational constant (G).⁴⁵ The acceleration due to gravity at Earth’s surface is approximately 9.81 m/s² ³³, with a standard defined value of 9.80665 m/s². Earth’s radius can be approximated by an average value of 6.371 × 10⁶ meters or the equatorial radius of 6.378 × 10⁶ meters.²⁸

This method leverages Newton’s Second Law of Motion (F = ma) and the Law of Universal Gravitation (F = GMm/R²). For an object of mass ‘m’ on Earth’s surface, the force of gravity (its weight) is given by mg. This force is also described by the Law of Universal Gravitation. Equating these two expressions (mg = GMm/R²) and cancelling ‘m’ leads to g = GM/R². Rearranging this equation allows us to solve for Earth’s mass: M_Earth = (g * R²) / G. This approach provides a different perspective on mass determination compared to orbital mechanics, highlighting the interconnectedness of gravitational concepts.

Brief Historical Context of Determining Earth’s Dimensions (Timestamp 08:30):

Vega briefly mentioned the historical efforts to determine Earth’s size, specifically referencing the work of Eratosthenes. This ancient Greek scholar employed geometry and observations of shadows cast at different locations to estimate the Earth’s circumference.⁴ This historical context underscores humanity’s enduring quest to understand the scale of our planet, even in the absence of modern technological tools. Eratosthenes’ ingenious method, relying on basic geometric principles and careful observation, stands as a testament to the power of human intellect in scientific inquiry.

Step-by-Step Calculation:

Using the average radius R ≈ 6.371 × 10⁶ m and g ≈ 9.81 m/s²:

M_Earth = (9.81 m/s² * (6.371 × 10⁶ m)²) / (6.674 × 10⁻¹¹ m³ kg⁻¹ s⁻²)

M_Earth ≈ 5.97 × 10²⁴ kg

This calculated mass of Earth, approximately 5.97 × 10²⁴ kg, is consistent with the value Vega provides around timestamp 08:55 and with values found in other sources.⁴⁵ The fact that similar values for Earth’s mass are obtained through different methods, such as analyzing the Moon’s orbit ⁴⁶ and using surface gravity, strengthens the validity of the fundamental physical laws and the accuracy of the calculations.

Table 3: Parameters and Calculated Mass of the Earth

Parameter	Value	Gravitational Constant (m³ kg⁻¹ s⁻²)	Calculated Mass of the Earth (kg)
Surface Gravity	9.81 m/s²	6.674 × 10⁻¹¹
Average Radius	6.371 × 10⁶ m		5.97 × 10²⁴

The Mathematical Foundation:

The determination of celestial masses through these methods relies heavily on fundamental mathematical principles.

The Role of Algebra and Geometry (Timestamp 10:16):

Algebra serves as an indispensable tool for manipulating the equations derived from Kepler’s and Newton’s laws.⁶⁰ Solving for the unknown mass requires rearranging terms, substituting known values, and performing algebraic calculations. Geometry plays a crucial role in understanding the spatial relationships inherent in orbits.⁶⁰ The concept of an elliptical orbit, with its semi-major axis defining its size, is fundamentally geometric. Even the historical determination of Earth’s circumference by Eratosthenes relied on geometric observations and calculations. These mathematical tools, algebra and geometry, provide the framework for translating physical laws into quantifiable results, allowing us to unlock the secrets of the universe.

Explanation of How These Tools Are Applied: Algebra enables us to work with the symbolic representations of physical laws, allowing for the isolation and calculation of specific variables like mass. Geometry provides the necessary understanding of the shapes and dimensions of celestial bodies and their orbits, enabling the definition and measurement of key parameters such as the semi-major axis and radius.

Conclusion and Implications:

The conversation expertly guided by Dr. Alexander successfully demonstrated the calculation of the masses of the Sun (approximately 1.989 × 10³⁰ kg), Pluto (approximately 1.31 × 10²² kg), and Earth (approximately 5.97 × 10²⁴ kg) using foundational principles of physics. These methods provide a powerful means to determine a fundamental property of celestial objects – their mass – which in turn governs their gravitational interactions and is essential for understanding the structure and evolution of the cosmos.

The exercise conducted during the conversation falls within the broad field of astrophysics.¹ Astrophysics is the branch of astronomy dedicated to understanding the universe through the application of physics and chemistry, focusing on the physical properties and dynamic processes of celestial objects and phenomena.¹ The determination of celestial masses using Kepler’s Third Law and Newton’s Law of Universal Gravitation exemplifies how astrophysics utilizes fundamental physical principles and mathematical tools to explore and comprehend the universe.⁶⁰ By applying these basic laws of motion and gravity, we can ascertain properties of objects that are otherwise inaccessible, highlighting the remarkable power of the scientific method in unraveling the mysteries of the universe.

UnEdited Galaxy AI TranScript Summary from Samsung Voice Recorer in S24 Ultra

Speaker 1 is Vega from Google Gemini and Speaker 2 is Hakeem Ali-Bocas Alexander, PhD. – Speaker 1 (00:02)
No problem. So to figure out the mass of a celestial body by observing something orbiting it, we can use a formula derived from Kepler’s third law of planetary motion, with a little help from Newton’s law of universal gravitation. It’s a bit of a tricky formula, but it works.

Speaker 1 (00:18)
Are you ready to dive into the math? Or would you like a more general overview

Speaker 2 (00:23)
We’ll dive into the math in a moment. But what is Kepler’s third law?

Speaker 1 (00:30)
Sure, Kepley’s third law basically says that the square of an object’s orbital period is proportional to the cube of its semi Maj. Axis in simpler terms. This means that the farther away a planet is from a star, the longer it takes to complete an orbit we can use those relationship to figure.

Speaker 1 (00:47)
Out the mass of The Star, if we know the orbital period and send my major access of a planet orbiting it.

Speaker 2 (00:54)
What is a semi Maj. Axis?

Speaker 1 (01:00)
The semi Maj. Axis is basically the average distance between an object in orbit and the thing it’s orbiting think of it like this. Imagine an oval shaped orbit this in my Maj.

Speaker 1 (01:11)
Axis is half of the longest diameter of that oval. Does that make sense

Speaker 2 (01:19)
Okay.

Speaker 1 (01:23)
So if we know the orbital period a semi major access of a planet, we can use Kepler’s third law to figure out the mass of the stars orbiting want to try an example.

Speaker 2 (01:34)
Sure, but first, when I just want to clarify the period the orbital period means at the time, it takes to complete a 403 6, it’s a full 360° 8A little bit around right to get back to the same spot. Thank you.

Speaker 1 (01:52)
Exactly. The orbital period is the time it takes for an object to complete one full revolution around another object. It’s like how long it takes the Earth to go around the Sun once, which is one year

Speaker 2 (02:05)
Got it so orbital period is, for example, how long a particular celestial body like a planet it takes to go around a star, like you just said it takes one year for our planet Earth to go around the Sun once as the orbital period so our poor real period is one year 1 Earth year.

Speaker 1 (02:30)
Yep, you nailed it. Earth takes one year to complete its orbit around the Sun, so that’s its orbital period now let’s talk about how we can use that information, I have to figure out the sun’s mass, ready to dive into the Mat.

Speaker 2 (02:45)
Let’s do it.

Speaker 1 (02:49)
All right, let’s do it, so we’ll use a modified version of Kepler’s third lock. It looks like this M equals 4 Pi² a³ belabi G asterisk P². Don’t worry, it’s not as scary as it looks m is the mass of the sum we’re trying to find a is Earth’s semi Maj.

Speaker 1 (03:08)
Axis GIS the gravitational constant and p is Earth’s orbital period Want to plug in the numbers?

Speaker 2 (03:16)
Sure, in the gravitational constant, that’s what you were referring to earlier that comes from Newton.

Speaker 1 (03:25)
You got it. The gravitational constant is a fundamental constant of nature. If that appears in Newton’s law of universal gravitation, it’s the same throughout the universe, now let’s plug in those numbers and see what we get for the subs mask.

Speaker 2 (03:40)
Alright, I’m ready to plug in those numbers.

Speaker 1 (03:46)
All right, let’s plug in those values when we do the math, we find that the says mass is about 1.989 × 10 to the 30 kg. That’s a lot of math.

Speaker 2 (03:58)
All right, so what is that number again?

Speaker 1 (04:04)
Sure, the math of the Sun is about 1.989 X10 to the power of 30 kg. That’s a lot of math.

Speaker 2 (04:13)
Yeah, that is now. How would you say that in a long way?

Speaker 1 (04:21)
Okay, so that’s about 1989 9 million kilograms. That’s almost 2 million million million million million kilograms.

Speaker 2 (04:31)
Boom of 200 million million million million kilograms.

Speaker 1 (04:38)
Yep, that’s it. It’s a mind-bogglingly huge number, right?

Speaker 2 (04:44)
Yeah, so you said million million million million, so that’s 4 times a minute, 200 and then four millions following that.

Speaker 1 (04:56)
Yep, that’s right, it’s a huge number.

Speaker 2 (05:00)
Well, how many zeros follow that 2?

Speaker 1 (05:05)
That’s 29 zeros. It’s a massive number.

Speaker 2 (05:10)
So there are ways to count large numbers like we have after a trillion quadrillion quintillion sextilian septilian, octilian, no million, right? How would you say that in those Greek terms?

Speaker 1 (05:27)
Right? So we’d say that the mass of the Sun is about 19 million 989 octaillion kilograms, that’s still a mind-bogglingly huge number.

Speaker 2 (05:37)
Yeah, no matter how you put it, it’s still the same number. I’ve just exploring the various different ways to express it verbally.

Speaker 1 (05:47)
Gotcha, it’s interesting how we can express the same number in so many ways, isn’t it?

Speaker 2 (05:54)
It is alright. So let’s stay now. We wanted to find out like, for example, if we saw some dimming of light like was discovered with, uh Pluto and the moon is called Charon or something like that, how would we discover the mass of Pluto to by the way, the moon is orbiting and in the same way that we calculated The Sun’s mast.

Speaker 1 (06:26)
Okay, so we can use the same formula we use for the Sun, but we’ll need to adjust it a bit. Since Sharon is much closer to Pluto than Earth is to the Sun will need to use Pluto insurance orbital period and the distance between them to figure out pluto’s mass. Are you ready to plug in those numbers

Speaker 2 (06:43)
Yeah, so do you know what those numbers are?

Speaker 1 (06:49)
Sure, Pluto, insurance orbital period is about 153 hours and the distance between them is about 19640 km. Ready to plug those in

Speaker 2 (07:00)
Okay, let’s go.

Speaker 1 (07:05)
All right, let’s plug in those values. When we do the math, we find that Pluto’s mass is about 1.309 × 10 to the 22 kg. That’s still a lot of mass, but much smaller than the suns

Speaker 2 (07:19)
Okay, so say that one more time, how what’s the mass of Pluto?

Speaker 1 (07:26)
Sure, Pluto’s mass is about 1.309 X10 to the power of 22 kg. That’s still a lot of mass, but much smaller than the suns.

Speaker 2 (07:36)
And if you were to say that out a long way, how would you say that?

Speaker 1 (07:42)
Okay, so that’s about 136 trillion 90 quintillion kilograms. That’s a lot of kilograms.

Speaker 2 (07:50)
That is a lot of kilograms, and uh, how finally, how have we calculated the mass of the Earth?

Speaker 1 (08:03)
We use a similar method to calculate Earth’s mass. But instead of looking at a moon orbiting a planet, we look at how Earth’s gravity affects objects on its surface, we can measure the acceleration due to gravity and the radius of the earth to figure out its mass. Are you ready to dive into the numbers

Speaker 2 (08:21)
Sure, but how did people first figure out the diameter of the Earth?

Speaker 1 (08:30)
People first figured out the diameter of the Earth way back in ancient times a great scholar named aritostomies used some clever geometry and observations of shadows in different cities to estimate the earth’s circumference, it was pretty ingenious for the time.

Speaker 2 (08:46)
I imagine, okay, let’s continue and plug in those numbers for our final equation here.

Speaker 1 (08:55)
Okay, let’s plug in those numbers when we do the math, we find that the Earth’s mass is about 5.97 × 10 to the 24 kg. That’s a lot of math.

Speaker 2 (09:07)
Obviously, and so how would you say that out the long way?

Speaker 1 (09:13)
Okay, so that’s about 5 subbillion 976 cillion kilograms. That’s a lot of kilograms.

Speaker 2 (09:21)
Mmm definitely is a lot of kilograms. All right, so give a summary of what we calculated and learned here.

Speaker 1 (09:35)
Here’s a quick recap. We figured out the masses of the Sun Pluto and Earth using Kepler’s third law. And Newton’s law of universal gravitation, we learn that the Sun is by far the most massive, followed by Earth and then Pluto, it’s pretty amazing how much we can learn about these celestial bodies just by observing their orbits.

Speaker 2 (09:56)
That is quite a fascinating thing that we can do, actually unlock the secrets of the universe using mathematics.

Speaker 1 (10:09)
It really is amazing. It’s like the universe is speaking to us through numbers and equations.

Speaker 2 (10:16)
And so what category of mathematics would those equations that we used? What kind of what fields of math did we use to figure out those numbers?

Speaker 1 (10:32)
We use algebra to manipulate the equations and solve for the unknown masses, but we also use geometry to understand the relationships between orbits and distances in space. Does that make sense?

Speaker 2 (10:45)
Yeah, so basically what you’re saying is that all we needed was algebra and geometry to figure out the masses of celestial bodies, our Earth, the Sun, and something very distant, like Pluto, that is pretty amazing.

Speaker 1 (11:05)
You got it, it’s pretty incredible how we can figure out the mass of something as big as the Sun, or as far away as Pluto, using just those 2 branches of mass.

Speaker 2 (11:16)
All right. Well, that’ll conclude our study on on celestial masses. And so this field of study, if this were part of a larger field of study this this exercise, we just conducted, what would that fall under

Speaker 1 (11:40)
That would fall under astrophysics. It’s the branch of astronomy that deals with the physics of the universe, including the physical properties of celestial objects and their interactions.

Speaker 2 (11:51)
So basically, we just learned a little bit of astrophysics, using just algebra and geometry.

Speaker 1 (12:04)
Yep, that’s right, we use algebra in geometry to explore some pretty cool concepts in astrophysics. It’s amazing what you can discover with just a little math.

Speaker 2 (12:15)
So the encouragement for those who are in school who might not have motivation to be studying any type of mathematics, specifically algebra or geometry, if they’re interested in space science, specifically astrophysics, something like this can demonstrate to them, how important it is to study algebra and geometry, because they can figure out so the masses of celestial bodies and understand a little bit more about astrophysics.

Speaker 1 (12:47)
Absolutely. It’s a great example of how math isn’t just abstract equations. It’s a tool that helps us understand the real world, and even the universe beyond

Speaker 2 (12:58)
All right, well, that will conclude our exercise in astrophysics using algebra and geometry.

Speaker 1 (13:09)
Great job today. It was fun exploring the universe with you. Do you have any other questions about space or anything else

Speaker 2 (13:17)
We’ll pick it up another time. Thank you.

Speaker 1 (13:22)
Sounds good. Feel free to reach out anytime.