# Acceleration and gravity

#### divindoubles

Does the accepted constant acceleration of gravity (approx. 32fpsps)refer to the earth's gravity, or to the force of gravity in general? Also, what would the linear acceleration be in fpsps be at 10x earth's gravity if one were travelling away from earth in a dragless environment (not taking into account the gravitational effects of any celestial bodies)? This is a research question for a short story. Thanks. DD

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#### Integral

Staff Emeritus
Gold Member
According to Newton
F=ma

Gravitational force is

F=GMm/r^2

combine the force of gravity with Newtons law to get the acceleration due to gravity

ma = GMm/r^2

a=GM/r^2

Where G it the gravitational constant
M is the Mass of the earth
r is the radius of the earth
m is the mass of some object at the surface.

Now all of the constants can be found on the web, you can look up and compute the above number, it will be g. Note the dependence on the Mass and radius of the earth, you can play with these values to find "g" at any distance from earth.

#### HallsofIvy

Homework Helper
In other words, 32 feet per second per second (and 9.8 meters per second per second) is the acceleration due to gravity at the earth's surface. It is different on other bodies and at different distances from the earths surface (according to the formula Integral gave you). It even varies slightly on the earth's surface due to differences in altitude (and so distance from the earth center) and varying density of the earth itself.

#### divindoubles

OK. Now that we know that the force of gravity causes objects to accelerate toward one another at an acceleration of +/-32fpsps, my question is; if we were to accelerate away from earth at 32fpsps, ignoring the minimal gravitational effects of the celestial bodies, would the inertial force felt by the occupants of our theoretical 'rocket' be equivalent to 1g or would it be more or less?

#### LURCH

Originally posted by divindoubles
OK. Now that we know that the force of gravity causes objects to accelerate toward one another at an acceleration of +/-32fpsps, my question is; if we were to accelerate away from earth at 32fpsps, ignoring the minimal gravitational effects of the celestial bodies, would the inertial force felt by the occupants of our theoretical 'rocket' be equivalent to 1g or would it be more or less?
At the surface, during lift-off and imediately after, the 32fps/s of acceleration would be added to the 1G already felt by the occupants when the craft is sitting still. So the passengers would feel 2G's while close to Earth and accelerating away. But once they were far enough away so that Earth's pull becomes negligable, or at a point between Earth and some other attractor where the two pulls are equal, then they will feel only 1G. Assuming, of course, that they continue to accelerate at 32fps/s. But this could only happen if they "throttle back", decreasing the amount of force being used to accelerate.

If force remains constant, 2G's of acceleration will continue to be felt, but the rate of acceleration (once beyond Earth's practical influence) will be 64fps/s.

#### HallsofIvy

Homework Helper
Originally posted by DivindoublesOK. Now that we know that the force of gravity causes objects to accelerate toward one another at an acceleration of +/-32fpsps

No, we don't know that. That is the exact opposite of what you were told. The force of gravity causes an object, on the surface of the earth, to accelerate toward the earth at 32 fpsps.

The earth accelerates toward that object at a much slower acceleration (in fact, unnoticable) while two objects will accelerate toward each other at a rate depending on the mass of the other object and inversely proportional to the square of the distance between them.

THAT is what you were told.

#### (Q)

Hmmm... I always thought it was only the Earth that was accelerating upwards to whatever objects were on the surface; that the objects were not accelerating.

#### chroot

Staff Emeritus
Gold Member
Originally posted by (Q)
Hmmm... I always thought it was only the Earth that was accelerating upwards to whatever objects were on the surface; that the objects were not accelerating.
Q, old buddy.. I hope you're kidding.

- Warren

#### Albrecht

According to Newton
F=ma
Gravitational force is
F=GMm/r^2
This is the traditional way to understand gravity. It assumes that every object has an inertial mass and a gravitational mass. Both masses are - nobody knows why - exactly proportional to each other.

There is another way. We now very exactly that the speed of light is reduced in a gravitational potential by

c = c0 (1-(GM)/(r^2*c0))

where c0 is the velocity of light in an area without a gravitational potential.

This dependency of c from r causes a normal classical refraction when e.g. a photon passes the sun. This classical refraction explains the normal gravitational acceleration of the photon towards the sun when passing, and yields similarly the additional acceleration which, according to Einstein, is caused by the curvature of space-time.

We know from particle physics (e.g. the Dirac function of the electron) that those particles have an internal oscillation with c. If this process of refraction is applied to this internal motion this explains quantitatively the acceleration of a free falling object.

The point which is special for this kind of treatment: The gravitational acceleration which we observe is the original physical effect. The acceleration has nothing to do with the mass of the object. From this point of view there is nothing like a gravitational mass, and the equivalence principle (of Newton and Einstein) is obsolete.

For the details you may look into
http://www.ag-physics.org/gravity

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