Use Surface-Volume to approximate gravity for planets and protons

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SUMMARY

This discussion presents a formula for approximating surface gravity (g) using the gravitational constant (G), mass (M), and surface area (A) of celestial bodies. The derived equation is g ≈ (4π G M) / A, which effectively correlates surface gravity with the geometric properties of spheres. Validation against known values for Earth, Moon, Sun, Jupiter, and protons demonstrates its accuracy. The discussion also explores the reverse approximation of G, confirming its consistency across various celestial bodies.

PREREQUISITES
  • Understanding of gravitational constant (G) ≈ 6.674 × 10⁻¹¹ m³/kg·s²
  • Knowledge of surface area calculations, particularly for spheres (A = 4πr²)
  • Familiarity with dimensional analysis in physics
  • Basic concepts of mass (M) and acceleration due to gravity (g)
NEXT STEPS
  • Research the implications of specific surface area (α = A / V) in gravitational calculations
  • Explore the application of the derived formula for non-spherical bodies
  • Investigate the historical context and existing formulations of gravitational equations
  • Learn about the limitations of approximations in gravitational physics
USEFUL FOR

Physicists, astrophysicists, and students studying gravitational theories, as well as anyone interested in the mathematical relationships between mass, surface area, and gravitational forces.

Owe Kristiansen
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TL;DR
Approximate surface gravity using the gravitational constant surface area and mass only

g ≈ (4π G M) / A
Title: Gravity Proportional to Mass over Surface Area — A Dimensional Approximation

I derived an approximation for surface gravity using the gravitational constant and surface area:
g ≈ (4π G M) / A

🔍 What Each Symbol Means:

• g: Surface gravity — the acceleration due to gravity at the surface (in m/s²)
• G: Gravitational constant ≈ 6.674 × 10⁻¹¹ m³/kg·s²
• M: Mass of the object (in kilograms)
• A: Surface area of the object (in square meters)
• 4π: A geometric factor that comes from the surface area of a sphere


It matches known values well for spheres.

Main Idea

Starting from the units of the gravitational constant:
[G] = m³ / (kg·s²)

I looked for a combination of measurable quantities:

• g: surface gravity (m/s²)
• A: surface area (m²)
• M: mass (kg)

The combination (g × A) / M gives:
(m/s² × m²) / kg = m³ / (kg·s²)

To match the geometry of a sphere, I introduced the factor 4π, as my calculations was 4pi off, leading to:
G ≈ (g × A) / (4π M)
→ g ≈ (4π G M) / A

This can also be written using:

• α = A / V: specific surface area (1/m)
• v = V / M: specific volume (m³/kg)


Then:
G ≈ (g × α × v) / (4π)

Validation Table (Using Surface Area)

Body | Mass (kg) | Surface Area (m²) | g (calc) | g (known)
---------|----------------|--------------------|----------|----------
Earth | 5.972e24 | 5.10e14 | 9.80 | 9.81
Moon | 7.35e22 | 3.79e13 | 1.62 | 1.62
Sun | 1.989e30 | 6.09e18 | 274 | 274
Jupiter | 1.898e27 | 6.15e16 | 25.9 | 24.8
Proton | 1.6726e-27 | 8.88e-30 | 1.58e-7 | —

---

Reverse Approximation of G

Using:
G ≈ (g × A) / (4π M)

Body | g (m/s²) | Surface Area (m²) | Mass (kg) | G (approx)
---------|----------|--------------------|----------------|-------------
Earth | 9.81 | 5.10e14 | 5.972e24 | 6.67e-11
Moon | 1.62 | 3.79e13 | 7.35e22 | 6.67e-11
Sun | 274 | 6.09e18 | 1.989e30 | 6.67e-11
Jupiter | 24.8 | 6.15e16 | 1.898e27 | 6.68e-11
Proton | 1.58e-7 | 8.88e-30 | 1.6726e-27 | 6.66e-11

---

Feedback Welcome

• Is this formulation already known or used?
• Could the specific volume form be useful for non-spherical bodies?
• Is the reverse approximation of G meaningful or just numerically coincidental?

Note: This article was created with the help of Microsoft Copilot. Some values may contain approximations or errors. Feedback and corrections are welcome.
 
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Your formula: g ≈ (4π G M) / A
Surface gravity formula: g = GM/r²
Surface area of sphere formula: A=4πr²

Isolating r² of SA formula: r² = A/4π
Therefore: g = GM/A/4π
Simplifying by multiplying by reciprocal of denominator: g = 4πGM/A

The formulas are identical. :wink:

Edit: Note that real bodies are not spherical and are not point masses, so this formula will always be an approximation for the real world.
 
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Thanks so much — I really appreciated your comment. I was aware that surface area and radius are connected through A = 4πr², but I hadn’t explicitly walked through how that substitution ties the formula directly back to Newton’s law. The way you laid it out made me feel like I hadn’t missed the mark — just approached it from a different angle. It was reassuring and really well explained. Thanks again for taking the time. 😊
 
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surface area and radius are connected through A = 4πr²
it took me about 2 weeks to figure this out

thank you, and good effort
 
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Owe Kristiansen said:
TL;DR Summary: Approximate surface gravity using the gravitational constant surface area and mass only

g ≈ (4π G M) / A

This article was created with the help of Microsoft Copilot
You were lucky this time. Modern LLM’s are not trained to do physics and they hallucinate a lot. They are not considered acceptable sources here at this time.
 
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Dale said:
You were lucky this time. Modern LLM’s are not trained to do physics and they hallucinate a lot. They are not considered acceptable sources here at this time.
Yes, I was, I did not use it for much though.
 
leoherry said:
surface area and radius are connected through A = 4πr²
it took me about 2 weeks to figure this out

thank you, and good effort
Thanks, ratios and units is what make me understand physics more than before.
 

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