Acceleration due to Gravity - Newton's 2nd?

  • Thread starter wangdang
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  • #1
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Hi folks,

I know this topic has been discussed countless times before but I could not find an answer to my specific query.
I am struggling to understand why acceleration due to gravity is constant for all masses on Earth.
Newton's 2nd law says that the acceleration of a body is prop. to the influencing force (gravity in this case) and inversely prop. to it's mass:
a = F/m

Therefore why isn't it that a body with twice the mass has half the acceleration?
Could someone also provide an example (proving accel. due to gravity is constant) with real values, as this usually helps me with understanding.

Thanks in advance.
 

Answers and Replies

  • #2
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by doubling the mass you are also doubling the force acting on the mass causing "a" to remain constant

F/m = 2F/2m

remember that F will change as the mass changes, the only variable that should remain constant in a given location will be "a" in this case ~9.8 m/s^2

does that sound right?
 
  • #3
D H
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Newton's second law is F=ma while his universal law of gravitation says that the force due to gravitation is F=GMm/r2. Putting these together says that the acceleration of some object of mass m falling toward the Earth is GM/R2.

However, the Earth is also accelerating toward the mass by Gm/R2. Combining the acceleration of the object toward the Earth and the acceleration of the Earth toward the object yields the relative acceleration

[tex]a_{\text{rel}} =
\frac{G(M_{\oplus}+m)}{R_{\oplus}^{\,2}} =
\frac{GM_{\oplus}}{R_{\oplus}^{\,2}}\left(1+\frac m {M_{\oplus}}\right)
[/tex]

So it is not quite true that the acceleration is constant. However, because the fraction m/M is extremely small compared to one, we have [tex]1+m/M_{\oplus} \approx 1[/tex]. The acceleration of the Earth toward the object is not observable.

How small is that effect? Consider the Apollo 11, which at almost 3 million kilograms at launch is one of the heaviest objects ever lifted from the surface of the Earth. 3 million kilograms is about 5×10-19 Earth masses. So even for something as massive as the Apollo 11 at launch, its gravitational acceleration toward the Earth's surface is, to within 18 decimal places of accuracy, equal to the gravitational acceleration of those little chunks of ice that fell off the Saturn V during the launch of the spacecraft.
 
  • #4
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Thanks very much for the replies,

D H, I am not sure what you mean by 'relative acceleration'. Is that the overall acceleration of the 'system'? Could you also please explain your derivations? I understand that:

F = G (Mm/R^2)
is the force in which both masses experience.

Then combining Newton's 2nd you get:
ma = G (Mm/R^2)
a = G (M/R^2) <---- this is the acceleration of the test mass

Repeating this process for Earth's mass:
a = G (m/R^2)

And this is where I am confused. You know the acceleration of both masses but how do you "combine them".

I think I do understand your conclusion though.
Please correct me if i'm wrong:
Relative acceleration is proportional to m / M. Therefore a larger mass will yield a greater acceleration, but considering the denominator in the fraction (the Earth's mass), the increase is negligible.
 
  • #5
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The acceleration that you obtain using Newtons law is based on an inertial frame of reference. In the case of M and m accelerating toward each other, that inertial frame of reference would be the center of mass of M and m. The relative acceleration that you obtain using the equation in DH's post is based on a non-inertial frame of reference, which would be either M or m.

In other words, if you were standing on either M or m and you used some kind of instrument to measure the acceleration of the other mass toward you, then you would be measuring the relative acceleration. If you positioned yourself at the center of mass between M and m then you would measure the accelerations of M and m as predicted by Newtons law.

wangdang said:
And this is where I am confused. You know the acceleration of both masses but how do you "combine them".
assuming a = G (M/R^2) and a' = -G (m/R^2) then

[itex]a_{rel} = a - a'[/itex]

D H, what does the [itex]\oplus[/itex] mean? / never mind, I looked it up. :) Although I'm not sure how you are applying it here. If it's the astronomical symbol for Earth then why do you use it with R?
 
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