Gravitational Acceleration Calculator: Find g

This calculator works out the gravitational acceleration (g) at a given height above a planet’s surface, using the standard formula from Newton’s law of universal gravitation. You enter the planet’s mass, its radius, and your altitude above the surface, and you get back the acceleration in m/s². It also shows the equivalent value in g-force units, so you can see how much weaker or stronger gravity is compared to Earth’s surface.

Formula

The calculator uses the equation for gravitational acceleration at a distance r from the center of a spherical mass:

g = G × M / r²

• g – gravitational acceleration in meters per second squared (m/s²)
• G – gravitational constant, 6.674 × 10â�»Â¹Â¹ N·m²/kg²
• M – mass of the planet (or moon, star, etc.) in kilograms (kg)
• r – distance from the center of the planet to the point where you measure g, in meters (m). Here, r = (planet radius + height) converted to meters.

The formula comes directly from Newton’s law of universal gravitation. It assumes the planet is a perfect sphere with uniform density, which is not exactly true for real planets, but it is accurate enough for most calculations.

Worked Example

Using the default values (Earth):

1. Mass M = 5.972 × 10²� kg
2. Radius R = 6371 km, height h = 0 km
3. Convert radius to meters: 6371 km × 1000 = 6,371,000 m
4. r = 6,371,000 m + 0 m = 6,371,000 m
5. r² = (6,371,000)² = 4.0589 × 10¹³ m²
6. g = (6.674 × 10�¹¹ × 5.972 × 10²�) / 4.0589 × 10¹³
7. Numerator = 3.9858 × 10¹�
8. g = 3.9858 × 10¹� / 4.0589 × 10¹³ = 9.820 m/s²

The calculator shows 9.820 m/s², and the g-force is 9.820 / 9.80665 = 1.001 g. The small difference from the often-quoted 9.81 m/s² is because Earth is not a perfect sphere and has local density variations.

When to Use This Calculator

Use this tool when you need the theoretical value of gravitational acceleration at a specific altitude. Common situations include:

• Physics homework problems involving satellite orbits or falling objects.
• Estimating the gravity on other planets or moons for science projects.
• Checking how much less you would weigh on top of a mountain or in an airplane.
• Calculating the gravitational pull at the altitude of the International Space Station (about 400 km).

For Earth at sea level, the result is close to 9.82 m/s², but the actual measured value varies from about 9.78 at the equator to 9.83 at the poles due to Earth’s rotation and shape. The calculator gives the ideal spherical value.

How to Read the Result

The main output is in m/s². A value of 9.82 means that, ignoring air resistance, a falling object gains 9.82 m/s of speed every second. The second output converts this to g-force, where 1 g is exactly 9.80665 m/s² (the standard gravity defined by international agreement).

If you enter a height of 400 km (ISS altitude), you get about 8.68 m/s², or 0.885 g. Astronauts feel weightless not because gravity is absent, but because they are in continuous free fall.

Common Mistakes and Limitations

• Unit confusion: The formula requires meters, not kilometers. This calculator converts automatically, but if you use the formula manually, remember to multiply km by 1000.
• Using Earth’s radius instead of the distance from the center: The r in the formula is the distance from the planet’s center, not the altitude above the surface. Always add the planet’s radius to your height.
• Ignoring rotation: The formula gives the gravitational component. On a rotating planet, the effective weight is slightly less at the equator due to centrifugal force. This difference is about 0.3% for Earth.
• Non-spherical planets: Real planets are oblate (flattened at poles). The formula assumes a perfect sphere, so results for real planets are approximations.
• Local geology: Dense rock or underground mass concentrations can change local g by a few hundredths of a percent. The calculator cannot account for these.

Alternatives and Related Concepts

If you need the gravitational force between two objects (like Earth and the Moon), use the full law of gravitation: F = G × m1 × m2 / r². This calculator gives the acceleration for one object, which is the same as the force divided by the object’s mass.

For a simpler estimate on Earth’s surface, you can use the constant 9.81 m/s². The variation with altitude is small for low heights: at 10 km (cruising altitude of a jet), g drops to about 9.77 m/s², a change of only 0.3%.

For other planets, you can look up their mass and radius and enter them here. For example, Mars has a mass of 6.39 × 10²³ kg and a radius of 3389 km, giving a surface gravity of about 3.71 m/s², which is 0.38 g.

Frequently Asked Questions

Why is the calculator showing 9.82 instead of 9.81?

Because the calculator uses the spherical Earth model with a mass of 5.972 × 10²â�´ kg and a radius of 6371 km. The standard value 9.80665 m/s² is a defined constant that accounts for Earth’s rotation and shape. The difference is about 0.1%, which is negligible for most purposes.

Does gravitational acceleration change with altitude?

Yes, it decreases with the square of the distance from the planet’s center. Doubling your distance from the center (e.g., going from the surface to 6371 km up) reduces gravity to one quarter. At typical airplane altitudes (10 km), the change is only about 0.3%.

What is the gravitational acceleration on the Moon?

The Moon’s mass is 7.342 × 10²² kg and its radius is 1737 km. Plugging those in gives about 1.62 m/s², or 0.165 g. That is why astronauts on the Moon weigh about one-sixth of their Earth weight.

Why is gravity weaker at the equator than at the poles?

Two reasons: Earth bulges at the equator, so the surface is farther from the center, and the equator’s rotation creates a centrifugal force that partially counteracts gravity. The combined effect makes equatorial gravity about 0.5% lower than at the poles.

Does the calculator account for air resistance?

No. The calculator gives the gravitational acceleration only. In a vacuum, a feather and a hammer fall at the same rate. With air resistance, the actual acceleration of a falling object is lower, and depends on its shape, size, and speed.

Can I use this for black holes or neutron stars?

Yes, the formula works for any spherical mass, but the result may be extreme. For a neutron star with 1.4 solar masses and a radius of 10 km, the surface gravity would be about 1.86 × 10¹² m/s². The calculator handles large numbers, but keep in mind that real neutron stars are not perfectly spherical.

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in M | Planet mass | kg | 5.972e24 | 0
in R | Planet radius | km | 6371 | 0
in h | Height above surface | km | 0 | 0
out g | Gravitational acceleration | m/s² | (6.674e-11 * M) / ((R * 1000 + h * 1000)^2) | 3
out g_force | g-force | g | ((6.674e-11 * M) / ((R * 1000 + h * 1000)^2)) / 9.80665 | 3