Is gravity's acceleration constant

[mewmew]

Ok, I have a few questions about gravity. First, is gravity's acceleration constant through the 9.8 m/s^2? I am sure that it is, but that raises a few questions for me. Instantaneously, if you throw something in the air straight up, isn't at time 0 the gravity not have taken any effect on the objects velocity as the velocity is at its max so to say, and the instant it reaches max height the object has a velocity of 0, even though its accelerating at 9.8 m/s^2 the entire time, This seems pretty weird? And lastly, why is gravity constant exactly, even though the object is moving away from the earth. Gravity just seems like a really weird force to me and I am trying to better understand exactly how it works on objects. Any help would be greatly appreciated.

 

[Integral]

To show that the acceleration due to gravity is constant do the following:
Let m be the mass of some object on the surface of the earth.
Let M= 6e24 kg be the mass of the earth.
Let r =6.4e6 m be the radius of the earth
Let G = 6.67e-11 nt-m²/kg²be the Gravitational constant.

we have from Newton that the force on a mass is:
F=ma= \frac { G M m} {r^2}

Notice that m appears on both sides of the equation so:

a = g = \frac {GM} {r^2}

Plug in the above numbers and you should see you familiar acceleration number.

Notice that it depends on r, the radius of the earth. Actually you need to interpret this as your distance from the center of the earth, as you increase r, g deceases.



So the acceleration due to gravity is only constant at a FIXED distance from the center of the earth

 

[anikmartin]

Graivty

The force of gravity does vary with distance from the center of the object, in this case the center of the Earth. The force of gravity between any two objects is inversely proportional to the square of the distance between the two objects. Just like light spreading as sperical waves, the intensity of this light also inverserly proportional to the square of the distance of the light source. (Equation for sphere is 4 pi r^2) Note the equation, which you might not have learned yet,

Gravitational Force = G(m1)(m2)/r^2. (G is the gravitational constant obtained through measurements, a very small number)

The reason why we consider the acceleration due to gravity as constant on small objects moving relative to the Earth, such as in your problem, is becuase the mass of the Earth is so large compared to the small object and small distance (from the center of the Earth) in question. So for these problems, the approximate or average value of 9.8m/s^2, will do. But once you start to deal with such problems as satellites above the Earth's atmosphere (large distance) or the force between the moon and the Earth (both large masses, and large distance) then this equation must be used.

For your qustion as to why there is accelertion when V=0, you can answer this question if you just simply look at the equations. As the ball is going up with initial velocity v, the acceleration due to gravity is constantly subtracting from its initial velocity, hence (v = v(i) - gt). So when the ball is going up away from the ground, g is negative, and at time t it will have gt less velocity from its initial velocity until v(i) = gt, leaving net velocity = 0. So now the ball is at its maximum height, v=0, but it still has acceleration, remember the acceleration due to gravity arises from a natural force between masses, so did the Earth dissappear? no, then g MUST still be acting. So at v=0, g is still acting on the ball, it is still subtracting from the initial velocity. Note also how it will now have a negative velocity (now going in the direction of g) (though, the sign is arbitrary). It always helps to visualize the equations in motion as t varies.

 

[mewmew]

Thank you both very much, that answers my questions, for now atleast

 

[robphy]

The reason why we consider the acceleration due to gravity as constant on small objects moving relative to the Earth, such as in your problem, is becuase the mass of the Earth is so large compared to the small object and small distance (from the center of the Earth) in question.

The approximate constancy of g comes from the small distance from the surface (not the center) of the Earth ... generally, a small radial component of displacement compared to the radius of the earth.

From Integral's calculation above, there was no assumption concerning the magnitude of the Earth's mass compared to that of our small object.

 

[anikmartin]

Okay, I must have been thinking in terms of the acceleration of the Earth towards the object.
But I see from the equation now. If the moon were a baseball it would still accelerate towards the Earth with the same acceleration as the actual moon itself because r is fixed. The Earth would accelerate toward this baseball with a much smaller acceleration than towards the actual moon.