<p>To enlarge on Rob's comment.</p>

<p>When $m_2 = 1 \mathrm{g},\ F = 9813338.722204$ and therefore a one gram mass would accelerate at $a =F/m = 9813338.722204/0.001 = 9813338722.204 \mathrm{ms^{-2}}$</p>

<p>When $m_2 = 1 1 \mathrm{kg},\ F = 9813338.722204$ and therefore a one kilogram mass would accelerate at $a =F/m = 9813338.722204/1 = 9813338.722204 \mathrm{ms^{-2}}$</p>

<p>In other words, different masses would accelerate at very different rates. Smaller masses would fall much faster than larger masses. As Rob says, that might have been noticed. Instead we observe that larger masses accelerate at the same rate as smaller ones, because larger masses have a larger force of gravity. </p>

<p>In fact, both masses accelerate at the same rate: $9.81 ms^{-2}$</p>

<p>You seem to be confusing "Force due to gravity (or weight)" with "acceleration due to gravity".</p>

<p>The basic problem with your idea is that it doesn't match even simple observations. Newton's model, however, is a very good fit to observations of gravity. Therefore your model is not just wrong, it very wrong and it is of no use.</p>

CC BY-SA 3.0
2017-08-20T18:37:44.873Z
2017-08-21T06:26:15.410Z
2017-08-21T06:26:15.410Z
James K
<p>Help needed, please. I need the input of the experts here. I have always been troubled by the formula ..F = G * m1 * m2 / r^2.. , which appears to be counter-intuitive. It should be ...F = G ( m1 + m2 ) / r^2.. I present here two examples to support my hypothesis.<br>
A )  Comparing 4 different objects falling to Earth at a height of 100 m.
r^2 = (6371 + .01)^2 km. = 40590915 ... G = 6.67*10^-11 ...m1 = 5.972*10^24 kg<br>
when m2 = 1 gm ,                     F = 9813338.722204<br>
when m2 = 1 kg ,                     F = 9813338.722204<br>
when m2 = 1 trillion kg( 12 zeroes), F = 9813338.722206<br>
when m2 = 1 quintillion  kg ( 18 ) , F = 9813340.365429<br>
     Here, heavier objects do fall faster, but impossible to observe or detect, even 
with our most delicate instruments. Only when m2 is overwhelmingly massive, do we begin to notice the difference.<br>
B ) orbits of Earth and Moon around the Sun</p>

<p>m1 = mass of Sun = 1.989*10^30<br>
                                                                                                                                    m2A = mass of earth = 5.972*10^24 and</p>

<p>m2B = moon = 7.3477*10^22....
average distance of Sun to Earth = 149.6 million km 
Since Moon orbits earth, then average distance to Sun also the same.
r^2 =( 149.6 million + radius of Sun 595500 ) ^2 = 150295500 km.<br>
when m2 = moon , F = 5873.11734</p>

<p>when m2 = Earth , F = 5873. 13476 </p>

<p>Here, both the Earth and Moon are falling towards the Sun at the practically same rate, and as such, the Sun's gravitational influence should not affect the orbit of Moon around the Earth. So we don't need to invoke the Hill Theory to explain why the moon is not pulled away from its orbit.
Please let me know where I've erred , and enlighten me , hopefully with a reply coherent to me. Thanks</p>

CC BY-SA 3.0
2017-08-20T07:00:26.727Z
In example A, the force between the Earth and the three masses is almost the same. So the *accelerations* they experience would differ by 21 orders of magnitude. I think that might have been noticed...
ProfRob
CC BY-SA 3.0
2017-08-20T07:27:59.280Z
In your version of Newton's formula, if m2 is zero, there would still be a force between the two objects, even though one of them is massless. Would you care to explain?
Mick
CC BY-SA 3.0
2017-08-20T09:09:30.027Z
@Mick whilst I don't know what OP is doing exactly, photons and other massless particles are still "attracted" to large bodies of mass
Alex Robinson
CC BY-SA 3.0
2017-08-20T13:29:37.270Z
thanks Mick , If m2 = 0 , it would mean that the gravitational influence is present regardless of whether any object is present. Further, it would mean that Newton was right when he said gravity is instantaneous
Hoang Bac Son
CC BY-SA 3.0
2017-08-20T14:33:25.587Z
I'm voting to close this question as off-topic because this is not about Astronomy
adrianmcmenamin
<gravity>
Is there something wrong about the gravity formula?
Hoang Bac Son
# Is there something wrong about the gravity formula?

(Asked by: username_0 on 2017-08-20)

Help needed, please. I need the input of the experts here. I have always been troubled by the formula ..F = G \* m1 \* m2 / r^2.. , which appears to be counter-intuitive. It should be ...F = G ( m1 + m2 ) / r^2.. I present here two examples to support my hypothesis.  
A ) Comparing 4 different objects falling to Earth at a height of 100 m.
r^2 = (6371 + .01)^2 km. = 40590915 ... G = 6.67\*10^-11 ...m1 = 5.972\*10^24 kg  
when m2 = 1 gm , F = 9813338.722204  
when m2 = 1 kg , F = 9813338.722204  
when m2 = 1 trillion kg( 12 zeroes), F = 9813338.722206  
when m2 = 1 quintillion kg ( 18 ) , F = 9813340.365429  
Here, heavier objects do fall faster, but impossible to observe or detect, even
with our most delicate instruments. Only when m2 is overwhelmingly massive, do we begin to notice the difference.  
B ) orbits of Earth and Moon around the Sun

m1 = mass of Sun = 1.989\*10^30  
m2A = mass of earth = 5.972\*10^24 and

m2B = moon = 7.3477\*10^22....
average distance of Sun to Earth = 149.6 million km
Since Moon orbits earth, then average distance to Sun also the same.
r^2 =( 149.6 million + radius of Sun 595500 ) ^2 = 150295500 km.  
when m2 = moon , F = 5873.11734

when m2 = Earth , F = 5873. 13476

Here, both the Earth and Moon are falling towards the Sun at the practically same rate, and as such, the Sun's gravitational influence should not affect the orbit of Moon around the Earth. So we don't need to invoke the Hill Theory to explain why the moon is not pulled away from its orbit.
Please let me know where I've erred , and enlighten me , hopefully with a reply coherent to me. Thanks

**Tags:** gravity

> **Comments:**
> **username_1**: In example A, the force between the Earth and the three masses is almost the same. So the *accelerations* they experience would differ by 21 orders of magnitude. I think that might have been noticed...  
> **username_2**: In your version of Newton's formula, if m2 is zero, there would still be a force between the two objects, even though one of them is massless. Would you care to explain?  
> **username_3**: @username_2 whilst I don't know what OP is doing exactly, photons and other massless particles are still "attracted" to large bodies of mass  
> **username_0**: thanks username_2 , If m2 = 0 , it would mean that the gravitational influence is present regardless of whether any object is present. Further, it would mean that Newton was right when he said gravity is instantaneous  
> **username_4**: I'm voting to close this question as off-topic because this is not about Astronomy  

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## 1 answers

### Answer by username_5 on 2017-08-20. Score: 2

To enlarge on Rob's comment.

When $m\_2 = 1 \mathrm{g},\ F = 9813338.722204$ and therefore a one gram mass would accelerate at $a =F/m = 9813338.722204/0.001 = 9813338722.204 \mathrm{ms^{-2}}$

When $m\_2 = 1 1 \mathrm{kg},\ F = 9813338.722204$ and therefore a one kilogram mass would accelerate at $a =F/m = 9813338.722204/1 = 9813338.722204 \mathrm{ms^{-2}}$

In other words, different masses would accelerate at very different rates. Smaller masses would fall much faster than larger masses. As Rob says, that might have been noticed. Instead we observe that larger masses accelerate at the same rate as smaller ones, because larger masses have a larger force of gravity.

In fact, both masses accelerate at the same rate: $9.81 ms^{-2}$

You seem to be confusing "Force due to gravity (or weight)" with "acceleration due to gravity".

The basic problem with your idea is that it doesn't match even simple observations. Newton's model, however, is a very good fit to observations of gravity. Therefore your model is not just wrong, it very wrong and it is of no use.


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