Variation of g with Depth: Formula & Meaning
- Ratio to surface g
- Result
- Depth below surface
- 10 km
- You entered
This calculator works out the acceleration due to gravity (g) at a given depth below the Earth’s surface. You enter the depth, and it returns the reduced value of g, based on the standard formula that assumes a uniform Earth density. The result is shown in meters per second squared (m/s²).
Formula
The standard formula for the variation of g with depth is:
g(d) = g₀ × (1 − d / R)
Where:
- g(d) = acceleration due to gravity at depth d (m/s²)
- g₀ = acceleration due to gravity at the surface, approximately 9.80665 m/s²
- d = depth below the surface (m or km, as long as units match R)
- R = radius of the Earth, approximately 6,371 km (6.371 × 10⁶ m)
This formula assumes the Earth has a constant density throughout. In reality, the Earth’s core is much denser than the crust, so the actual decrease is slower than this linear formula predicts. The formula works well for depths up to a few hundred kilometers, and is a common approximation in introductory physics.
Worked Example
Using the default depth of 10 km:
- Start with surface gravity: g₀ = 9.80665 m/s²
- Divide depth by Earth’s radius: 10 / 6371 = 0.001569
- Subtract from 1: 1 − 0.001569 = 0.998431
- Multiply by surface gravity: 9.80665 × 0.998431 = 9.791 m/s²
The calculator shows 9.791 m/s² (rounded to 3 decimal places), which matches the manual calculation. The ratio output is 0.9984, meaning gravity at 10 km depth is 99.84% of surface gravity.
Why Does Gravity Decrease with Depth?
Gravity arises from the mass of the Earth pulling on an object. When you are on the surface, the entire Earth’s mass pulls you downward. When you go underground, the mass above you pulls upward, partially canceling the pull from the mass below. The net effect is that the effective gravitational force decreases.
Mathematically, only the mass within the sphere of radius (R − d) contributes to the net gravitational pull. Assuming uniform density, that mass is proportional to (R − d)³, and the distance to the center is (R − d). Since gravity is proportional to mass divided by distance squared, the (R − d)³ cancels with (R − d)², leaving a linear dependence: g ∝ (R − d).
How to Use the Calculator
Enter the depth in kilometers (or any unit, as long as you adjust the radius accordingly). The calculator uses Earth’s mean radius of 6,371 km. For depths in meters, divide by 1000 first or use 6,371,000 as the radius in a manual calculation.
The output is the acceleration due to gravity at that depth. The ratio output tells you what fraction of surface gravity remains. For example, at 100 km depth, the ratio is 1 − 100/6371 ≈ 0.9843, so gravity is about 98.43% of surface value.
Common Mistakes and Misconceptions
- Using the wrong radius: Make sure depth and radius are in the same units. If depth is in km, use R = 6371 km. If depth is in m, use R = 6,371,000 m.
- Assuming the formula works for all depths: The linear formula fails near the center of the Earth, where real gravity actually decreases to zero, but not linearly because density increases dramatically in the core.
- Confusing depth with altitude: The formula for altitude (above surface) is different: g = g₀ × (R / (R + h))². Depth and altitude are not interchangeable.
- Forgetting that g is a vector: This formula gives magnitude only. Direction always points toward the center of the Earth.
Limitations of the Formula
The uniform-density assumption is a significant simplification. The Earth’s core has a density of about 13,000 kg/m³, while the crust is about 2,700 kg/m³. This means the real decrease of g with depth is slower than the linear formula predicts. For example, at the core-mantle boundary (about 2,900 km deep), the linear formula gives g ≈ 9.81 × (1 − 2900/6371) ≈ 5.34 m/s², but actual measurements and seismic models show g is about 10.7 m/s² at that depth—actually higher than surface gravity due to the dense core.
For most practical purposes—mining operations, deep drilling, or physics homework—the linear formula is adequate for depths up to a few hundred kilometers. For planetary interiors, more complex models are needed.
Frequently Asked Questions
Does gravity increase or decrease as you go underground?
Under the uniform-density assumption, gravity decreases linearly with depth. However, in the real Earth, gravity initially increases slightly as you go down because the denser mantle and core exert a stronger pull. The maximum occurs at the core-mantle boundary, then it drops to zero at the center.
What is the value of g at the center of the Earth?
At the exact center, gravitational forces from all directions cancel out, so g = 0 m/s². This is true regardless of density distribution, because the mass is symmetrically distributed around you.
How does this formula compare to the altitude formula?
For altitude (h above surface), the formula is g = g₀ × (R / (R + h))², which decreases quadratically. For depth, the formula is linear. At the same distance from the surface, gravity decreases faster with altitude than with depth because in the altitude case, you are moving away from the entire Earth’s mass, while in the depth case, part of the mass remains above you.
Why is the formula linear for depth but quadratic for altitude?
For depth, the mass inside the sphere shrinks as (R−d)³ while distance squared shrinks as (R−d)², leaving a linear factor. For altitude, mass stays constant but distance squared grows as (R+h)², giving an inverse-square law.
Can I use this formula for other planets?
Yes, if you know the planet’s radius and surface gravity. Replace g₀ with the planet’s surface gravity and R with the planet’s radius. For example, on Mars (R ≈ 3,390 km, g₀ ≈ 3.71 m/s²), the formula would give g at depth d as 3.71 × (1 − d/3390).