Newton's second law and how to calculate the force on the mo

AI Thread Summary
The discussion centers on applying Newton's second law, F=ma, to calculate force on the Moon, where gravity is 1.622 m/s². A participant proposes a formula F=m(g/6) but seeks a more generalized approach without redefining gravity. Suggestions include using Newton's law of gravitation, which relates gravitational force to mass and distance, rather than relying solely on Earth's gravity as a template. The conversation highlights the simplicity of gravitational calculations, emphasizing that only the total mass and radius of celestial bodies are necessary for determining gravitational acceleration. Overall, the thread explores methods to adapt gravitational formulas for different celestial environments.
Farhad-.-
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I, earlier were studying Newton's second law F=ma, or rather the free fall of objects, F=mg.
A thought occurred to me, how do I apply this formula to the moon with the Gravity of 1.622m/^2 and came up with F=m(g/6), however, g/6 = 1.63m/s^2

Normally I'd just use F=ma, but assuming people one day would live there, they'd use g = 1.622m/s^2, which I think many would find confusing. So I tried to make a generalized formula. Is there any way to improve upon the formula?
 
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Farhad-.- said:
I, earlier were studying Newton's second law F=ma, or rather the free fall of objects, F=mg.
A thought occurred to me, how do I apply this formula to the moon with the Gravity of 1.622m/^2 and came up with F=m(g/6), however, g/6 = 1.63m/s^2

Normally I'd just use F=ma, but assuming people one day would live there, they'd use g = 1.622m/s^2, which I think many would find confusing. So I tried to make a generalized formula. Is there any way to improve upon the formula?

What do you mean "apply it to the moon"? Do you mean objects falling on the moon, or the moon-Earth system?
 
Student100 said:
What do you mean "apply it to the moon"? Do you mean objects falling on the moon, or the moon-Earth system?
Objects falling towards the moon*
 
Farhad-.- said:
Objects falling towards the moon*

Simply redefine g. That's all that needs to be done.
 
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Student100 said:
Simply redefine g. That's all that needs to be done.
Yeah, but what I was asking is there any way to use Earth's gravity as a template without actually redefining g, which now I think about, I did not specify, I got the formula f=m(g/6). But is there a better way of doing it? My formula skills aren't that great :/
 
Farhad-.- said:
Yeah, but what I was asking is there any way to use Earth's gravity as a template without actually redefining g, which now I think about, I did not specify

Why not use Newton's law of gravitation in that case?

$$F=\frac{Gm_1m_2}{r^2}$$

Although, g in F=mg is just a dummy variable, it doesn't have an intrinsic meaning until we give it one, so I'm not really understanding where you're going.
 
Student100 said:
Why not use Newton's law of gravitation in that case?

$$F=\frac{GMm}{R^2}$$
We are going through that in friday, I have a basic concept of how to use that but not indept unfortunately.
 
Farhad-.- said:
We are going through that in friday, I have a basic concept of how to use that but not indept unfortunately.

It doesn't use Earth as a template either, (not sure why you'd want to), it uses the nature of the system and the gravitational constant.
 
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Free fall acceleration on Earth is just ##g=G\frac{M_{Earth}}{R^2_{Earth}}##, here ##M_{Earth}## is Earth mass, and ##R_{Earth}## - its radius. For Moon you get ##g=G\frac{M_{Moon}}{R^2_{Moon}}##. Now if you want to express free fall acceleration for Moon in terms of Earth g you just need to take relation ##g_{Moon}/g_{Earth}##. So you will get ##g_{Moon}=\frac{M_{Moon}}{M_{Earth}}\frac{R^2_{Earth}}{R^2_{Moon}}g_{Earth}##
 
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Domullus said:
Free fall acceleration on Earth is just ##g=G\frac{M_{Earth}}{R^2_{Earth}}##, here ##M_{Earth}## is Earth mass, and ##R_{Earth}## - its radius. For Moon you get just ##g=G\frac{M_{Moon}}{R^2_{Moon}}##. Now if you want to express free fall acceleration for Moon in terms of Earth g you just need to take relation ##g_{Moon}/g_{Earth}##. So you will get ##g_{Moon}=\frac{M_{Moon}}{M_{Earth}}\frac{R^2_{Earth}}{R^2_{Moon}}g_{Earth}##
It's interesting that the formula for the gravitational acceleration at or above the surface is true, even if the density of the planet varies with depth (i.e a massive iron core or shell) It's just the total mass and the radius that count. It makes life a lot simpler :smile:.
 
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  • #11
sophiecentaur said:
It's interesting that the formula for the gravitational acceleration at or above the surface is true, even if the density of the planet varies with depth (i.e a massive iron core or shell) It's just the total mass and the radius that count. It makes life a lot simpler :smile:.
Yeah I know, I had the same thought when I was thinking about it ^•^
 
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