What level of acceleration does gravity give?

Question

I was thinking about gravity and isn't it a meters per second per second per second acceleratory force? I mean, as you get close to an object's center of mass the force of gravity gets stronger, meaning that the acceleration accelerates. But then, if you have an apple on the earth, and it falls, the vectors of gravitational force acting upon it are now wider! The "sides" of the earth (from the apple's perspective) are pulling on it more sideways than before, so it actually cancels out more or less. Versus when you are far away, such as the moon, and the sides of the earth (from your perspective) don't really pull you in much of a different direction than the center of mass of the earth. So, maybe when you're far away it is a double acceleration, but when your close its not? Sure, why not. But wait a minute... This is all from a newtonian perspective. Einstein realized what gravity really is. A bend in spacetime. In which case, this would no longer apply! From an einsteinian perspective, gravity is more like a slide. In this case, its a curved slide that gets steeper as you go "down". From this view, the sides of the earth no longer matter because if you're on a slide, and your going north and down, and to your left the slide goes in a different direction than north, you aren't going to be affected by that other part of the slide. You will continue north and downwards no matter what the other parts of the slide look like. So was my original assumtion correct? Or is Newton the way to go on this? Thank you.

Answer 1

Pretty deep question George. Perhaps you would do better to see Einstein's perspective as a 'fabric' in spacetime, the surface of the Earth as a huge spherical baloon with planar, or square, area (but no measurable thickness). If a light source at the center of the Earth was turned on the energy would disperse omni-directionally. The energy at the entire 'inside' surface of that baloon being equal to the total original energy released from the source: with luminosity decreasing by the inverse square law. Gravity acts in an opposite manner. The total energy at the surface equal to that at the center of mass: increasing per second per second, per surface area, as it moves omni-directionally inward. Per second per second per second ... defies this relative perception of reality.

Answer 2

g = 9.8 m/s^2 is the approximate acceleration due to gravity at the surface of the Earth.

 

It's different on different planets.

 

More generally, the force due to gravity between two bodies is, in Newtonian mechanics, F = G m_1 m_2 / d^2. There's a constant, G, the product of the two masses, and the distance between them.

 

Since F = m a, we can equate m_1 g = G m_1 m_earth / r_earth^2, so

 

g = G m_earth / r_earth^2.

 

Check this out:

http://www.wyzant.com/resources/files/260517/newton_s_universal_law_of_gravitation

 

And there’s an entertaining and informative video on:
http://www.learner.org/resources/series42.html#program_descriptions

Click on the little box labelled VoD to the right of:
2. The Law of Falling Bodies

So far we've discussed the acceleration due to gravity on or above the surface of the Earth; what about inside it? There the acceleration due to gravity decreases to zero at the center of the Earth! See:

http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/earthole.html.

Answer 3

Hi George,

From the information given in your question, Newton is the way to go on this.

 

Near the surface of the Earth, the acceleration due to gravity is ≈ 9.8 m/s² (that's 9.8 meters per second per second). If you drop an apple from an airplane, skyscraper, or roof of your house, the acceleration due to gravity will always be ≈ 9.8 m/s².

 

I respectfully disagree with Vivian:

  a) Galileo demonstrated (in his famous Leaning Tower of Pisa experiment in 1589) that heavy and light objects take the same time to fall to the ground. But for the friction effect of air, a feather and a cannon ball would take the same time to fall from the tower.

 

  b) The earth acceleration rate due to gravity is NOT the same universally. Earth's gravity decreases with altitude according to the "inverse square law" - the greater your altitude, the lesser is the earth's gravitational effect.

 

The inverse square law formula for gravity on earth is: g ≈ 9.8 (4000/(4000+h))². where "h" is the altitude (in miles) above the surface of the earth (radius about 4,000 mi). Here are some examples:

• In an airplane at 30,000 feet (5.7 miles), g ≈ 9.77 m/s² (almost identical to surface of earth).
• At about 235 miles (altitude of ISS satellite), g ≈ 8.74 m/s².
• At about 1,000 miles (inner space), g ≈ 6.3 m/s².
• At about 125,000 miles (half way to the moon), g ≈ 0.01 m/s².

 

High school physics classes will never ask you about relativity, accounting for air friction, or high altitudes.

Don't over-engineer the problem and worry about all those vectors vectors.

bob

 

 

 

Answer 4

Hi George;

Acceleration due to gravity is 9.8 meters-per-second-every-second.

While the earth acceleration rate is the same universally, the rate of final velocity varies with weight.  If a heavy object and light object are dropped from the same height, the heavy object has the potential to reach a higher velocity.

Otherwise, what is your question?

Answer 5

Change in distance per second per second per second is jerk.

 

Acceleration is change in distance per second per second.

 

Acceleration due to gravity is roughly 9.8 meters per second per second, at least on the surface of the earth.