### 7 Answers

- JimLv 75 months agoFavourite answer
No, not really, the gravitation Force is always changing as you drop.

Force = G (m₁ * m₂)/r²

The radius changes as you drop, causing the force to change. And since F=ma, the acceleration due to gravity changes. We choose to 'ignore' this for most science since most things happen at or very near the surface.

Did you know, the max force is at the earth's surface. If you go higher or lower, the force decreases!

- Anonymous5 months ago
If close to Earth surface and disregarding air friction , then it can be considered "constant" and equal to 9.80665 according to SI (International System of Units) approximation : a nice approximation is 9.806 (neither 9.8 nor 9.81)

- Jeffrey KLv 65 months ago
If you ignore air resistance and the distance it falls is small compared to the size of the Earth, then the acceleration is constant.

9.8 m/sec^2

If it is falling from space, a large distance relative to Earth's radius, then acceleration is not constant. Gravity gets stronger as it gets closer to Earth. You must calculate an orbit.

- ZirpLv 75 months ago
in a vacuum, yes, pretty much

in air, no, because resistance is proportional to the speed SQUARED

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- Markus ImhofLv 75 months ago
Towards what?

In any case- if you're talking about acceleration due to gravity, then no. Gravity depends on the distance between the two bodies involved and varies with the inverse of the square of the distance.

For everyday use, when you're moving up to maybe 10 km (30,000 feet) above Earth's surface, that doesn't matter much - Earth's radius is (on average) 6371 km, adding 10 km to that would be a change of 0.16%, changing gravity by 0.3% (ok, that's still 10 times the variability of Earth's gravity at the surface - https://en.wikipedia.org/wiki/Gravity_anomaly ).

It does become important, though, when you want to compute trajectories of satellites or spacecraft.

If you're talking about Earth, though, you'll also have to take air resistance into account, which somewhat complicates things.