Why is gravity, on earth, approximatley 9.81 m/s[SUP]2[/SUP].What

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SUMMARY

The acceleration due to gravity on Earth is approximately 9.81 m/s², derived from Newton's law of universal gravitation, represented by the equation F_G = G(m_1 m_2)/r². This acceleration is independent of the mass of the falling object, as it cancels out in the equations, meaning all objects experience the same gravitational acceleration regardless of their mass. The Earth's mass and the distance from its center are the primary factors influencing this value, with variations being negligible at different altitudes.

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why is gravity, on earth, approximatley 9.81 m/s2.

What causes it to be so?

thanks
 
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Well, its an observational property of all matter in the universe that it attracts other matter by an equation which Newton found (http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation)
<br /> F_G = G \frac{m_1 m_2}{r^2}<br />
For two masses (m1 and m2), separated by a distance (r); given 'Newton's gravitational constant' (G).

Now, according to one of Newton's other laws
<br /> F = ma<br />
If you combine these equations you find out that the acceleration due to gravity only depends on the mass of the Earth and your distance away from its center... if you plug in the values you get ~9.81 m/s^2
 


jsmith613 said:
why is gravity, on earth, approximatley 9.81 m/s2.

What causes it to be so?

thanks

Mass of the Earth and the Newton law that F=m a=-G m M/R^2. In classical physics the acceleration does depend only on the distance to the center of the earth. (This means that the acceleration is on a mountain smaller.)
That m does cancel out in this equation is very important, it is unique in physics and let Einstein think about the General Theory of Relativity.


Jens
 


zhermes said:
Well, its an observational property of all matter in the universe that it attracts other matter by an equation which Newton found (http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation)
<br /> F_G = G \frac{m_1 m_2}{r^2}<br />
For two masses (m1 and m2), separated by a distance (r); given 'Newton's gravitational constant' (G).

Now, according to one of Newton's other laws
<br /> F = ma<br />
If you combine these equations you find out that the acceleration due to gravity only depends on the mass of the Earth and your distance away from its center... if you plug in the values you get ~9.81 m/s^2

So essetisally, because the Earth's mass is so large, the mass of the other objects is next to insignificant. Because the value of r is also large, changing the value by an amount will make little differernce.

Therefore will the value of gravity be in the range 9 - 10

(M1 = earth, M2 = object)

correct?
 


jsmith613 said:
So essetisally, because the Earth's mass is so large, the mass of the other objects is next to insignificant.
Not quite; it doesn't matter how large one body's mass is---because it cancels out in the equations, it has no influence on the acceleration.

jsmith613 said:
Because the value of r is also large, changing the value by an amount will make little differernce.
That's exactly right. The Earth's radius is something like 6000km, while the highest mountain is only about 10km... so the change in gravity will be about 0.3% (or something like that, so very small).

jsmith613 said:
Therefore will the value of gravity be in the range 9 - 10
Yeah, if you plug in the usual values, that's what you get.
 


Not quite; it doesn't matter how large one body's mass is---because it cancels out in the equations, it has no influence on the acceleration.

I don't quite understand what you mean. Please explain


Question 2: Using the equation G = M1 * M2 / r^2 does that mean I cause the Earth to accelerate upwards at a rate of 9.81 m/s^2. If this is the case, why does the Earth not move upwards dramatically due to the upward accealration of every landmass on earth.

Please explain. Thanks
 


No. That equation gives you the force. Acceleration is a=f/m so the larger the mass, the smaller the acceleration for the same force.
 


russ_watters said:
No. That equation gives you the force. Acceleration is a=f/m so the larger the mass, the smaller the acceleration for the same force.


Does that not mean, therefore that Gravity should change based on force and mass for each object, if we assume a = f/m
 


jsmith613 said:
Does that not mean, therefore that Gravity should change based on force and mass for each object, if we assume a = f/m

Simply put, the only variables in the calculation are the masses of the 2 objects and the distance between them. (Of course, this assumes an isolated system or one in which other objects are too distant to influence it.)

Other forces may act to overcome some or all of the influence of gravity, say a velocity of one object greater than the escape velocity of the larger object, but otherwise, that's really and truly all there is to it.
 
  • #10


russ_watters said:
No. That equation gives you the force. Acceleration is a=f/m so the larger the mass, the smaller the acceleration for the same force.
jsmith613 said:
Does that not mean, therefore that Gravity should change based on force and mass for each object, if we assume a = f/m
Note that Russ said 'for the same force'. The force of gravity does change based on the mass. In fact the force is proportional to the mass, which makes the acceleration of a free falling object due to gravity the same for all masses, since the mass drops out of the equation.
 
  • #11


The acceleration gravitational field depends only on the mass of the "source".

Given Earth with mass M and body on its surface with mass m, the force between them is:
F = G\frac{Mm}{r^2}

Now to calculate the acceleration of Earth you divide the force by Earth's mass
a = \frac{F}{M}
which is
a = G\frac{m}{r^2}

The body feels acceleration
a = \frac{F}{m}
a = G\frac{M}{r^2}

Now you see that acceleration of bodies on surface of Earth is in fact the same because it does not depends on body's mass.

Gravity is really nice force!
 
  • #12
jsmith613 said:
Does that not mean, therefore that Gravity should change based on force and mass for each object, if we assume a = f/m
gravitational force, yes: that's what you measure with a bathroom scale.
 
  • #13


Now to calculate the acceleration of Earth you divide the force by Earth's mass
a = \frac{F}{M}
which is
a = G\frac{m}{r^2}

Why is this so,
 
  • #14


Does all this mean that no matter the mass of the smaller object, it will always have the same acceleration?
 
  • #15


jsmith613 said:
Why is this so,

I understand it know

THanks everyone
 
  • #16


Drakkith said:
Does all this mean that no matter the mass of the smaller object, it will always have the same acceleration?

Yes it does.
 
  • #17


zhermes said:
Yes it does.

Is this because as the mass increases, the attraction due to gravity increases as well, but so does the amount of force required to move it? I think that's right, just wanting to make sure.
 
  • #18


Drakkith said:
Is this because as the mass increases, the attraction due to gravity increases as well, but so does the amount of force required to move it? I think that's right, just wanting to make sure.
Yes! That's exactly it.

And as someone said above, this is a very interesting feature of the laws of physics, in Newton's force equation there is an inertial mass
<br /> F_{net} = m_{inertial} a<br />
and in his gravitational equation, there is a gravitational mass
<br /> F_g = G\frac{M m_{grav}}{r^2}<br />

And it turns out, they seem to be the same.
m_{inertial} = m_{grav}

That's what einstein called the "Equivalence principle," which led to his formulation of 'general relativity.' These days we take it for granted that those two masses are equal, but there is no fundamental physical law that says it has to be so, its an observational fact.
 
  • #19


Awesome.
 

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